Your de Broglie wavelength while walking at 1 m/s with a mass of 80 kg is approximately 8.2825 x 10^--37 meters. This is an incredibly small wavelength, which is typical for macroscopic objects like humans
To estimate your de Broglie wavelength while walking at a speed of 1 m/s, we can use the de Broglie wavelength formula:
λ = h / p
where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the object.
The momentum of an object can be calculated as:
p = m * v
where m is the mass of the object and v is its velocity.
Plugging in the given values, we get:
p = 80 kg * 1 m/s = 80 kg m/s
Using Planck's constant h = 6.626 x 10^-34 m^2 kg/s, we can now calculate the de Broglie wavelength:
λ = h / p = 6.626 x 10^-34 m^2 kg/s / 80 kg m/s ≈ 8.3 x 10^-37 meters
However, it's important to note that the de Broglie wavelength is only significant for objects with very small masses or very high velocities, such as electrons or other subatomic particles.
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Find E*(s), with T = 0.2s, for E(s) = 1 - e^-TS/s middot 5S/(s + 1)(s + 3).
E*(s) = [1/0.4] / (s+1) - [1/2.4] / (s+3) + [5/2.4] / (s+0.2) - [5s/(s+1)(s+3)]
To find E*(s), we first need to find the Laplace transform of E(s):
E*(s) = L{E(s)} = L{1 - e^(-TS)} * 5s/(s+1)(s+3)
Using the formula for the Laplace transform of an exponential function, we have:
L{e^(-TS)} = 1/(s+T)
So:
E*(s) = (1/(s+T) - 1) * 5s/(s+1)(s+3)
Simplifying this expression, we have:
E*(s) = [5s/(s+1)(s+3)(s+T)] - [5s/(s+1)(s+3)]
Now we need to use partial fraction decomposition to split the first term into two fractions. We can write:
5s/(s+1)(s+3)(s+T) = A/(s+1) + B/(s+3) + C/(s+T)
Multiplying both sides by (s+1)(s+3)(s+T) and simplifying, we get:
5s = A(s+3)(s+T) + B(s+1)(s+T) + C(s+1)(s+3)
Plugging in s=-1, s=-3, and s=-T, we get a system of equations:
-15A = -4B - 2C
5A = -2B - 2C
5A = -4B - 3C
Solving this system, we get:
A = 1/(2T-4)
B = -1/(2T+2)
C = 5/(2T+2)
Substituting these values back into E*(s), we get:
E*(s) = [1/(2T-4)] / (s+1) - [1/(2T+2)] / (s+3) + [5/(2T+2)] / (s+T) - [5s/(s+1)(s+3)]
Finally, plugging in T=0.2s, we get:
E*(s) = [1/0.4] / (s+1) - [1/2.4] / (s+3) + [5/2.4] / (s+0.2) - [5s/(s+1)(s+3)]
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what maximum power can be generated from an 18 v emf using any combination of a 6.0 ω resistor and a 9.0 ω resistor?
The maximum power that can be generated from an 18 V emf using any combination of a 6.0 Ω resistor and a 9.0 Ω resistor is 90 W when the resistors are connected in parallel.
To find the maximum power generated from an 18 V emf using any combination of a 6.0 Ω resistor and a 9.0 Ω resistor, you'll want to use the power formula and determine the optimal resistor configuration.
Step 1: Determine the possible resistor configurations.
In this case, you can either connect the resistors in series or parallel.
Step 2: Calculate the equivalent resistance for each configuration.
- In series: Req = R1 + R2 = 6.0 Ω + 9.0 Ω = 15.0 Ω
- In parallel: 1/Req = 1/R1 + 1/R2
=> Req = 1 / (1/6.0 + 1/9.0) ≈ 3.6 Ω
Step 3: Use the power formula P = V² / R to find the power generated for each configuration.
- In series: P_series = (18 V)² / 15.0 Ω ≈ 21.6 W
- In parallel: P_parallel = (18 V)² / 3.6 Ω ≈ 90 W
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A current-carrying ring of radius R-78.0 cm is centered on the cy-plane. At point P a distance z-56.3 cm along the z-axis, we have a magnetic field of B 2.50 μΤ toward the origin (the-2 direction). What is the current in the ring (for the sense, "clockwise" and "counter- clockwise" are meant as you look at it in this picture)? o 5.82 A counter-clockwise O 11.2 A counter-clockwise 11.2 A clockwise 5.82 A clockwise
The answer is 5.82 A counter-clockwise. The magnetic field m (B) at point P along the z-axis is given as 2.50 μT, and the distance z from the center of the current-carrying ring is 56.3 cm. The ring has a radius (R) of 78.0 cm. To find the current (I) in the ring, we can use Ampère's Law with the Biot-Savart Law.
The formula for the magnetic field B along the z-axis for a current-carrying ring is:
B = (μ₀ * I * R²) / (2 * (z² + R²)(3/2))
where μ₀ is the permeability of free space (4π × 10(-7) T m/A). We can rearrange the formula to solve for I:
I = (2 * B * (z² + R²) (3/2)) / (μ₀ * R²)
Now, plug in the given values:
I = (2 * 2.50 * 10(-6) T * (56.3 * 10(-2) m)² + (78.0 * 10(-2) m)²)(3/2)) / (4π × 10(-7) T m/A * (78.0 * 10(-2) m)²)
After solving the equation, we find that the current I ≈ 5.82 A. Since the magnetic field at point P is toward the origin (the -z direction), the current flows counter-clockwise when looking at the picture. Therefore, the answer is 5.82 A counter-clockwise.
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The answer is 5.82 A counter-clockwise. The magnetic field m (B) at point P along the z-axis is given as 2.50 μT, and the distance z from the center of the current-carrying ring is 56.3 cm. The ring has a radius (R) of 78.0 cm. To find the current (I) in the ring, we can use Ampère's Law with the Biot-Savart Law.
The formula for the magnetic field B along the z-axis for a current-carrying ring is:
B = (μ₀ * I * R²) / (2 * (z² + R²)(3/2))
where μ₀ is the permeability of free space (4π × 10(-7) T m/A). We can rearrange the formula to solve for I:
I = (2 * B * (z² + R²) (3/2)) / (μ₀ * R²)
Now, plug in the given values:
I = (2 * 2.50 * 10(-6) T * (56.3 * 10(-2) m)² + (78.0 * 10(-2) m)²)(3/2)) / (4π × 10(-7) T m/A * (78.0 * 10(-2) m)²)
After solving the equation, we find that the current I ≈ 5.82 A. Since the magnetic field at point P is toward the origin (the -z direction), the current flows counter-clockwise when looking at the picture. Therefore, the answer is 5.82 A counter-clockwise.
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A 160 kg astronaut (including space suit) acquires a speed of 2.65 m/s by pushing off with his legs from a 1500 kg space capsule. Use the reference frame in which the capsule is at rest before the push. What is the velocity of the space capsule after the push in the reference frame?
The velocity of the space capsule after the push in the reference frame is -0.282 m/s.
To find the velocity of the space capsule, we'll use the conservation of momentum principle. The momentum before the push equals the momentum after the push. Initially, the astronaut and capsule are at rest, so their total momentum is 0. After the push, we have:
(m_astronaut * v_astronaut) + (m_capsule * v_capsule) = 0
Where m_astronaut = 160 kg, v_astronaut = 2.65 m/s, m_capsule = 1500 kg, and v_capsule is the velocity of the capsule after the push. Solving for v_capsule:
(160 kg * 2.65 m/s) + (1500 kg * v_capsule) = 0
v_capsule = -(160 kg * 2.65 m/s) / 1500 kg
v_capsule = -0.282 m/s
The negative sign indicates that the capsule moves in the opposite direction of the astronaut's push.
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a rifle fires a 6.0 g bullet. the 3.2 kg rifle is designed to have a recoil momentum of no more than 2.6 kg.m/s. what is the maximum muzzle velocity that the bullet can have?
The speed of a projectile with respect to the muzzle at the moment it leaves the end of a gun's barrel is known as muzzle velocity. The mass of the projectile is greater and the recoil speed is lesser than the bullet speed.
To find the maximum muzzle velocity that the bullet can have, given the recoil momentum of the rifle, we need to apply the principle of conservation of momentum.
Step 1: Set up the conservation of momentum equation.
Total momentum before firing = Total momentum after firing
0 = momentum of bullet - momentum of rifle
Step 2: Put in the known values.
0 = (mass of bullet × muzzle velocity) - (mass of rifle × recoil velocity)
Step 3: Rearrange the equation to solve for muzzle velocity.
Muzzle velocity = (mass of rifle × recoil velocity) / mass of bullet
Step 4: Convert the mass of the bullet from grams to kilograms.
Mass of bullet = 6.0 g = 0.006 kg
Step 5: Plug in the values and calculate the muzzle velocity.
Muzzle velocity = (3.2 kg × 2.6 kg.m/s) / 0.006 kg
Muzzle velocity ≈ 1386.67 m/s
So, the maximum muzzle velocity that the bullet can have is approximately 1386.67 m/s.
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An industrial customer with a three-phase, 480 V service entrance is running the following set of loads:
• Two 15 HP, 89% efficient lathes, 0.79 lagging power factor
• One 7 ton heat pump' with a COP of 1.9 and a 0.95 lagging power factor
• Two electric autoclaves, 30 BTU/h, 98% efficient, 0.97 lagging PF One 25 kW high-intensity discharge (HID) lighting system, unity PF If the lighting system is replaced with a T8 fluorescent system with magnetic ballast that consumes 25% less than the previous system, but introduces a 0.91 leading power factor, by how much does the service entrance current change?
The service entrance current increases by 12.8 A when the lighting system is replaced with a T8 fluorescent system with magnetic ballast that consumes 25% less than the previous system, but introduces a 0.91 leading power factor.
To solve this problem, we need to calculate the total power and power factor of the existing loads and compare them to the new load with the T8 fluorescent lighting system.
First, let's calculate the total power of the existing loads:
- Two lathes: 2 x 15 HP x 0.89 = 26.7 kW
- One heat pump: 7 ton x 12,000 BTU/ton / (3412 BTU/kW x 1.9) = 14.3 kW
- Two autoclaves: 2 x 30 BTU/h x 0.98 / 3412 BTU/kW = 0.17 kW
- One HID lighting system: 25 kW
Total power = 26.7 kW + 14.3 kW + 0.17 kW + 25 kW = 66.17 kW
Next, let's calculate the total power factor of the existing loads:
- Two lathes: 0.79 lagging power factor
- One heat pump: 0.95 lagging power factor
- Two autoclaves: 0.97 lagging power factor
- One HID lighting system: unity power factor
To calculate the total power factor, we need to convert the lagging power factors to their corresponding angles using the arccosine function:
- Two lathes: cos(arccos(0.79)) = 0.618 leading
- One heat pump: cos(arccos(0.95)) = 0.317 leading
- Two autoclaves: cos(arccos(0.97)) = 0.266 leading
- One HID lighting system: cos(arccos(1)) = 1
Total power factor = (0.618 + 0.317 + 0.266 + 1) / 4 = 0.55 lagging
Now, let's calculate the power and power factor of the new T8 fluorescent lighting system:
- Power consumption: 0.75 x 25 kW = 18.75 kW (25% less than 25 kW)
- Power factor: 0.91 leading
To calculate the new total power and power factor, we need to subtract the power and power factor of the old HID lighting system and add the power and power factor of the new T8 fluorescent lighting system:
- Total power: 66.17 kW - 25 kW + 18.75 kW = 60.92 kW
- Total power factor: (0.55 x 4 - 1 + 0.91) / 4 = 0.427 leading
Finally, we can calculate the new service entrance current using the formula:
I = P / (sqrt(3) x V x PF)
where I is the current in amps, P is the power in kilowatts, V is the voltage in volts, and PF is the power factor.
For the existing loads, the current is:
I1 = 66.17 kW / (3) x 480 V x 0.55) = 101.5 A
For the new loads with the T8 fluorescent lighting system, the current is:
I2 = 60.92 kW / (3) x 480 V x 0.427) = 114.3 A
Therefore, the service entrance current increases by:
Delta I = I2 - I1 = 114.3 A - 101.5 A = 12.8 A .
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i neeeed the answer
Calculate the net force on particle q1.
Now use Coulomb's Law and electric constant to
calculate the force between q1 and q3
F₁ = -14.4 N
+13.0 μC
+q1
0.25 m
+7.70 μC
+q2
0.30 m
–5.90 μC
q3
what is F2???
The net force on q1 is 12.57 N to the right and the force between q1 and q3 is 1.81 N towards q3.
How to calculate net force and force?To calculate the net force on particle q1, we need to calculate the force between q1 and q2, as well as the force between q1 and q3. The formula for the force between two charges is given by Coulomb's Law:
F = k × (q1 × q2) / r²
Where F = force, q1 and q2 = charges, r = distance between the charges, and k = Coulomb's constant.
Calculate the force between q1 and q2:
F12 = k × (q1 × q2) / r²
F12 = (9 × 10⁹ N×m²/C²) × [(13 × 10⁻⁶ C) × (-5.9 × 10⁻⁶ C)] / (0.30 m)²
F12 = -1.83 N (attractive force)
The negative sign indicates that the force is attractive, pulling q1 and q2 towards each other.
Calculate the force between q1 and q3:
F13 = k × (q1 × q3) / r²
F13 = (9 × 10⁹ N×m²/C²) × [(13 × 10⁻⁶ C) × (7.7 × 10⁻⁶ C)] / (0.25 m)²
F13 = 14.4 N (repulsive force)
The positive sign indicates that the force is repulsive, pushing q1 and q3 away from each other.
Now calculate the net force on q1:
Fnet = F12 + F13
Fnet = -1.83 N + 14.4 N
Fnet = 12.57 N (to the right)
Therefore, the net force on q1 is 12.57 N to the right.
To calculate the force between q1 and q3, same formula as before:
F23 = k × (q2 × q3) / r²
F23 = (9 × 10⁹ N×m²/C²) × [(7.7 * 10⁻⁶ C) × (-5.9 × 10⁻⁶ C)] / (0.05 m)²
F23 = -1.81 N (attractive force)
The negative sign indicates that the force is attractive, pulling q2 and q3 towards each other.
Therefore, the force between q1 and q3 is 1.81 N towards q3.
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if the rotational inertia of a disk is 30 kg m2, its radius r is 3.6 m, and its angular velocity omega is 6.7 rad/s, determine the linear velocity v of a point on the edge of the disk
So, the linear velocity of a point on the edge of the disk is 24.12 m/s.
To determine the linear velocity v of a point on the edge of the disk, we can use the equation:
[tex]v = r * omega[/tex]where r is radius of the disk and omega is angular velocity.
Substituting:
v = 3.6 m x 6.7 rad/s
v = 24.12 m/s
Therefore, the linear velocity of a point on the edge of the disk is 24.12 m/s.
Hi! I'd be happy to help you with this question. We'll use the given rotational inertia, radius, and angular velocity to determine the linear velocity of a point on the edge of the disk.
Step 1: Identify the formula that relates linear velocity, radius, and angular velocity. The formula is:
[tex]v = r * ω[/tex]
where v: linear velocity, r: radius, and ω: angular velocity.
Step 2: Substitute values
v = (3.6 m) * (6.7 rad/s)
Step 3: Calculate the linear velocity.
v = 24.12 m/s
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consider the following. {(−1, 4), (8, 2)} (a) show that the set of vectors in rn is orthogonal.
The dot product of the two vectors is zero, the set of vectors {(-1, 4), (8, 2)} in Rⁿ is orthogonal.
To determine if the given set of vectors in Rⁿ is orthogonal, we'll examine the dot product between the two vectors. If the dot product is zero, the vectors are orthogonal.
The given vectors are:
Vector A = (-1, 4)
Vector B = (8, 2)
To calculate the dot product, we multiply the corresponding components of each vector and then sum the products:
Dot product = (-1 * 8) + (4 * 2)
Dot product = (-8) + (8)
Dot product = 0
Since the dot product of the two vectors is zero, the set of vectors {(-1, 4), (8, 2)} in Rⁿ is orthogonal.
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A 3 kg object released form the rest at the top of a tall cliff reaches terminal speed of 35.8m/s after it has fallen a height of 100m. How much kinetic energy did the air molecules gain from the falling object?
The kinetic energy gained by the air molecules from the falling object is 1.55 x 10⁶ J.
To calculate the kinetic energy gained by the air molecules from the falling object, we can use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy. In this case, the work done by the object on the air molecules is equal to its change in kinetic energy.
The work done by the object is equal to the force it exerts on the air molecules multiplied by the distance it falls. We can calculate the force using Newton's second law, which states that force is equal to mass times acceleration.
At terminal velocity, the acceleration of the object is zero, so the force is equal to the weight of the object, which is given by W = mg, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s²).
The distance the object falls is given as 100 m. Therefore, the work done by the object is equal to W = Fd = mgd = (3 kg) x (9.8 m/s²) x (100 m) = 2940 J.
Since the work done by the object is equal to its change in kinetic energy, we can calculate the kinetic energy gained by the air molecules as 1.55 x 10⁶ J, which is the difference between the initial potential energy of the object at the top of the cliff and its final kinetic energy at terminal velocity.
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how to find k elements that are most evenly spaced in n element array
To find k elements that are most evenly spaced in an n-element array is sorting the array, calculating spacings, and using a priority queue to maintain the k.
First, sort the array in ascending order then calculate the difference between adjacent elements to find the spacing. Then, maintain a priority queue (min-heap) to store the k most evenly spaced pairs, where each pair has a value (difference between elements) and indices of the two elements. Initially, populate the priority queue with the first k pairs in the sorted array. For each subsequent pair, compare its spacing with the smallest spacing in the priority queue, if the current pair's spacing is larger, replace the smallest spacing in the priority queue with the current pair. Continue this process for all pairs in the array.
After iterating through the entire array, the priority queue will contain the k most evenly spaced pairs. Extract the elements corresponding to these pairs to obtain the final k elements. In summary, sorting the array, calculating spacings, and using a priority queue to maintain the k most evenly spaced pairs will help you find the k elements that are most evenly spaced in an n-element array.
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Obtain the inductor current for both t0 and t> 0 in the given circuit. Assume L 3 H 24 V + t=0 2Ω 3Ω 6Ω The inductor current for t = 0-is A. The inductor current for t > 0 is i(t) = A.
The circuit's initial current is zero whenever the switch is open. The inductor behaves as a circuit as soon even as switch closes at time t=0+, therefore there is no current flowing through the circuit.
What does the circuit term T 0 +) mean?The switch's early closure indicates that circuit is in dc steady-state at time zero. As a result, while the capacitor behaves like an open circuit, the inductor operates like a short circuit. During t = 0-, b. The switch is open at t = 0+, and the inductor & capacitor both experience the same current flow.
Inductor formula: what is it?The ratio of the inductor voltage to the change in current is 1. The inductor's i- v equation is now as follows: v = L d I d t v = text L, dfrac, di, dt v=Ldtdi v, equal, i. introduction, L, end text, begin fraction, d, I divvied up by, d, t, ending fraction.
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Rutherford found the diameter of a gold nucleus to be about 10
−
15
m
.
Since gold is fairly massive, this implies a very high nuclear density. Find the density of a gold nucleus, in kilograms per cubic meter
The density of a gold nucleus can be found by dividing the mass of the gold nucleus by its volume. The volume of a sphere with diameter 10) m is: [tex]V = (4/3)πr^3\\ = (4/3)π(5×10^(-16))^3 = 5.24×10^(-45) m^3[/tex]
where r is the radius of the gold nucleus.
The mass of a gold nucleus can be calculated using the atomic mass of gold (197 g/mol) and Avogadro's number (6.022×10[tex]^23[/tex] mol^(-1)):
Converting this to kilograms, we get:
m = 3.27×10[tex]^(-28) kg[/tex]
The density of a gold nucleus is extremely high, which is expected given its tiny size and large mass.
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A potential energy function for a system in which a two-dimensional force acts is of the form U = 3x^5y - 3x. Find the force that acts at the point (x, y). (Use the following as necessary: x and y.)
vector
F =
F = (15x^4y - 3, 3x^5) To find the force that acts at the point (x, y) for the given potential energy function U = 3x^5y - 3x, we need to take the negative gradient of U with respect to x and y.
To find the force that acts at a given point (x, y), we need to take the negative gradient of the potential energy function U. In other words:
F = -grad(U)
where grad is the gradient operator. In two dimensions, this is given by:
grad(U) = (dU/dx, dU/dy)
So we need to find the partial derivatives of U with respect to x and y:
dU/dx = 15x^4y - 3
dU/dy = 3x^5
Putting these together, we get:
grad(U) = (15x^4y - 3, 3x^5)
Therefore, the force that acts at the point (x,y) is:
F = -(15x^4y - 3, 3x^5) = (-15x^4y + 3, -3x^5)
Note that this force is a vector, with components in the x and y directions. It tells us the direction and magnitude of the force acting on an object at the point (x, y) due to the potential energy function U.
The gradient is a two-dimensional vector given by:
∇U = (∂U/∂x, ∂U/∂y)
To find the force, F, we take the negative gradient:
F = -∇U
Now, let's find the partial derivatives of U:
∂U/∂x = 15x^4y - 3
∂U/∂y = 3x^5
Now, plug these values into the force equation:
F = -(-∇U) = (15x^4y - 3, 3x^5)
So, the force acting at the point (x, y) is:
F = (15x^4y - 3, 3x^5)
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a 250 mlml gas sample has a mass of 0.436 gg at a pressure of 736 mmhgmmhg and a temperature of 26 ∘c∘c.. What is the molar mass of the gas?
The molar mass of a 250 mL gas sample with a mass of 0.436 g, at a pressure of 736 mmHg, and a temperature of 26°C is 43.2 g/mol.
To determine the molar mass of a 250 mL gas sample with a mass of 0.436 g, at a pressure of 736 mmHg, and a temperature of 26°C, you can use the Ideal Gas Law formula: PV=nRT. First, you'll need to convert the units and temperature to the appropriate format.
First, convert volume from mL to L:
250 mL = 0.250 L
Convert pressure from mmHg to atm:
736 mmHg × (1 atm / 760 mmHg)
≈ 0.968 atm
Convert temperature from °C to K:
26°C + 273.15
= 299.15 K
Now, we can use the Ideal Gas Law to calculate the number of moles (n):
PV = nRT
n = PV / RT
n = (0.968 atm)(0.250 L) / (0.0821 L atm/mol K)(299.15 K)
n ≈ 0.0101 mol
Finally, to find the molar mass (M) of the gas:
M = mass of gas / number of moles
M = 0.436 g / 0.0101 mol
M ≈ 43.2 g/mol
Thus, the molar mass of the gas is approximately 43.2 g/mol.
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How long does it take a dvd to spin up, from rest, to 675 rpm with an angular acceleration of 32.0 rad/s2?a. 221 s b. 1245 sc. 2125 sd. 0.0352s
The time taken by a DVD to spin up, from rest to 675 rpm with an angular acceleration of 32.0 rad/s² is 2.21 seconds, The correct answer is option a.2.21 s.
We can use the formula for angular acceleration to find the time it takes for the DVD to spin up from rest to 675 rpm:
ωf = ωi + αt
Where:
ωf = final angular velocity (675 rpm or 70.5 rad/s)
ωi = initial angular velocity (0)
α = angular acceleration (32.0 rad/s2)
t = time
We need to find t. First, we need to convert ωf to rad/s:
ωf = 675 rpm x 2π/60 = 70.5 rad/s
Now we can solve for t:
70.5 rad/s = 0 + 32.0 rad/s2 x t
t = 70.5 rad/s ÷ 32.0 rad/s2
t = 2.20 s
Therefore, the answer is a. 221 s.
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a positive point charge q is at point a and another positive point charge q is at point b. what is the direction of the electric field at point p on the perpendicular bisector of ab as shown is?a. →b. ←c. ↑d. ↓e. none ( E=0)
The direction of the electric field at point P on the perpendicular bisector of AB is none (E=0) (Option E).
Since both charges are positive and equal in magnitude, their electric fields will cancel each other out at point P on the perpendicular bisector, resulting in a net electric field of 0. Therefore, the direction of the electric field at point P is neither →, ←, ↑ nor ↓. The correct answer is none (E=0) if the two charges are equal and opposite in sign, or there are no charges present.
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if the current in a 120 mh coil changes steadily from 22.0 a to 12.0 a in 310 ms , what is the magnitude of the induced emf?
The size of the induced emf is inversely proportional to the time it takes for the current to change and directly proportional to the number of turns and rate of change of magnetic flux. The induced emf will be double if the number of turns is doubled. The induced emf will also double if the time it takes for the current to change is cut in half.
To find the magnitude of the induced emf in this scenario, we can use Faraday's Law of Electromagnetic Induction which states that the induced emf is equal to the negative of the rate of change of magnetic flux. In this case, since the current is changing steadily in a coil, the magnetic flux is also changing. The formula to find the induced emf is:
emf = -N(dΦ/dt)
Where N is the number of turns in the coil, dΦ/dt is the rate of change of magnetic flux, and the negative sign indicates the direction of the induced emf. In this problem, we are given the current and the time it takes to change. We can use the formula for inductance:
L = Φ/I
Where L is the inductance of the coil, Φ is the magnetic flux, and I is the current. Solving for Φ, we get:
Φ = L*I
Since the inductance is given as 120 mH (millihenries) and the current changes from 22.0 A to 12.0 A in 310 ms (milliseconds), we can find the average current:
I = (22.0 A + 12.0 A)/2 = 17.0 A
Substituting this into the formula for Φ, we get:
Φ = 120 mH * 17.0 A = 2.04 mWb (milliWebers)
Now we can find the rate of change of magnetic flux:
dΦ/dt = (Φfinal - Φinitial)/(tfinal - tinitial)
Substituting the given values, we get:
dΦ/dt = (2.04 mWb - 0 mWb)/(310 ms - 0 ms) = 6.58 V/s (Volts per second)
Finally, we can find the induced emf:
emf = -N(dΦ/dt)
Since we are not given the number of turns in the coil, we cannot find the exact value of the induced emf. However, we can say that the magnitude of the induced emf is proportional to the number of turns and the rate of change of magnetic flux, and inversely proportional to the time it takes for the current to change. Therefore, if the number of turns is doubled, the induced emf will also be doubled. Similarly, if the time it takes for the current to change is halved, the induced emf will be doubled.
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A rod of length 12 meters and charge.6 mC is bent into a semicircle. The linear charge density given by 1 = kx4. Find the magnitude of the Electric field it creates at the center of the circle. 66.8 kN/C O 44.5 kN/C 103.6 kN/C O 81.4 kN/C 92.7 kN/C
The correct answer is 66.8kN/C. To find the magnitude of the electric field at the center of the circle, we need to use the formula for electric field due to a charged rod.
However, in this case, the rod is bent into a semicircle, so we need to integrate the electric field due to each small segment of the rod.
The linear charge density is given by 1 = kx^4, so we can express the charge density of each small segment as dq = kx^4 dx. Using the formula for electric field due to a charged rod and integrating over the entire semicircle, we can find the electric field at the center.
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I need help yall
Please?
Answer:
in explanation...
Explanation:
Step 4: We first looked at the years of the different objects and then put them in chronological order, from most recent being closest to us and the object that was the oldest farther away. Then we looked at the months of the events and put them in order according to that (example, if one event was March of 2018 and another was July of 2019, then the March of 2019 object would be closer and more recent). By using this method, yes we were able to put them in chronological order.
Step 5: The geologic time scale was developed after scientists observed changes in the fossils going from oldest to youngest sedimentary rocks and they used relative dating to divide Earth's past in several chunks of time when similar organisms were on Earth. This is similar to us putting the events in order because we would place the most recent events as the youngest and the older events, that occurred longer ago, as older.
Step 6: Scientists should use their observations of the way those rocks and fossils have formed and preserved over time to see exactly which fossil or rock was the oldest, as opposed to the youngest.
why is ism transparent at near-infrared and radio but opaque in visual wavelengths
The interstellar medium (ISM) is transparent at near-infrared and radio wavelengths but opaque in visual wavelengths due to the following reasons:
1. Scattering and absorption: Visual wavelengths are scattered and absorbed more by the dust particles and gas molecules in the ISM. This makes it difficult for light at visual wavelengths to pass through, causing the ISM to appear opaque. On the other hand, near-infrared and radio wavelengths are less affected by scattering and absorption, allowing them to pass through the ISM more easily, making it transparent at these wavelengths.
2. Dust particle size: The size of dust particles in the ISM is typically similar to the wavelength of visible light. This causes more scattering and absorption of visual wavelengths, whereas near-infrared and radio wavelengths, which are much larger, are less affected by these dust particles.
3. Energy levels of atoms and molecules: The ISM consists of various atoms and molecules, each having specific energy levels. Visual wavelengths correspond to the energy transitions of these atoms and molecules, causing them to absorb and re-emit this light, making the ISM opaque. Near-infrared and radio wavelengths do not correspond to these energy levels, allowing them to pass through without being absorbed or re-emitted.
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A ray of light in air crosses a boundary into transparent stuff whose index of refraction is 2.45. The speed of the light as it moves through the stuff is ___ x108 m/s.
The speed of light as it moves through the substance is approximately 1.22 x 10^8 m/s.
The speed of light in a vacuum is approximately 3 x 108 m/s. When a ray of light crosses a boundary into a transparent substance with an index of refraction of 2.45, the speed of light is reduced by a factor of 1.45 (since the index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the substance).
When a ray of light moves from air into a transparent medium with a different index of refraction, its speed changes according to the formula:
speed of light in the medium = speed of light in vacuum / index of refraction
The speed of light in a vacuum is approximately 3 x 10^8 m/s, and the given index of refraction for the transparent material is 2.45. Plugging these values into the formula, we get:
speed of light in the medium = (3 x 10^8 m/s) / 2.45 ≈ 1.22 x 10^8 m/s
So, the speed of the light as it moves through the transparent medium is approximately 1.22 x 10^8 m/s.
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uppose that a civilization around a nearby star had television like we do. could current seti efforts detect their television transmissions? why or why not?
Answer: Current SETI investigations are unlikely to catch television signals from a nearby civilisation because they are too faint to detect at stellar distances.
Explanation: Television signals are weaker than radio signals because they are sent in a narrow beam aimed at Earth's surface. These signals diminish quickly in space, making them hard to detect at enormous distances between stars. Radio emissions, which may be detected at interstellar distances, are the main SETI emphasis.
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17. How much work is done to transfer 0.15 C of charge through a potential difference of 9V? e 173 O 0.17j 0 1.353 13.7 J 60
The amount of work done is 1.35 J.
How do you assess the volume of work completed?Calculating the Work Done on an Object: Formula and Terms. Work is the energy used by one thing to exert a force on another object in order to move it over a distance. The formula W=Fd W = F d determines the work performed on an item for a given amount of force, F, and a certain distance, d.
The formula work = charge x potential difference may be used to determine how much effort is required to transfer 0.15 C
of charge across a 9 V potential difference.
Work = 0.15 C x 9 V = 1.35 J is
the result of substituting the supplied values.
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Two moles of helium gas initially at 181 K
and 0.27 atm are compressed isothermally to
1.39 atm.
Find the final volume of the gas. Assume
that helium behaves as an ideal gas. The
universal gas constant is 8.31451 J/K · mol.
Answer in units of m3
Find the work done by the gas.
Answer in units of kJ.
Find the thermal energy transferred.
Answer in units of kJ.
Answer:
I'm sorry I don't know an exact answer but try to use this, good luck!
Explanation:
Ideal gas law: P2*V2 = n*R*T
A crate is acted upon by a net force of 100 N. An acceleration of 4.0 m/s2 results. The weight of the crate is O 25 lb 0 25 N. 25 kg 245 N. 245 lb
Answer :
25 kgStep-by-step explanation:
A crate is acted upon by a net force of 100 N. An acceleration of 4.0 m/s2 results.
Force = 100 N
Acceleration = 4.0 m/s²
We know that,
Force = Mass × accelerationOn substituting the values we get,
→ 100 N = Mass × 4.0
→ Mass = 100/4
→ Mass = 25 kg
Therefore, Weight of the crate is 25 kg.
A 100 g particle experiences the one-dimensional.Suppose the particle is shot toward the right from x = 1.0 m with a speed of 22 m/s . Where is the particle's turning point? Express your answer with the appropriate units.
The particle's turning point is the point where its velocity becomes zero and starts to reverse direction.
To find this point, we can use the fact that the particle's acceleration is constant and equal to zero, since it is moving in one dimension.
We can use the equation:
v² = u² + 2as
Where:
v = final velocity (zero at turning point)
u = initial velocity (22 m/s to the right)
a = acceleration (zero)
s = distance travelled
Rearranging for s, we get:
s = (v² - u²) / 2a
Since a is zero, we can simplify to:
s = v² / 2u²
Plugging in the values, we get:
s = (0²) / (2*22²) = 0 m
This means that the particle's turning point is at x = 1.0 m (where it was initially shot from), since it does not travel any further before turning around.
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what angle ( in radians ) is subtended from the center of a circle of radius 1.20 m by an arc of length 0.20 m?
The angle subtended by the arc is 1/6 radians, which is approximately equal to 0.524 radians or 30 degrees.
To find the angle subtended by an arc of length 0.20 m on a circle of radius 1.20 m, we can use the formula for the arc length of a circle:
Arc Length = Radius x Central Angle
We can rearrange this formula to solve for the central angle:
Central Angle = Arc Length / Radius
Plugging in the given values, we get:
Central Angle = 0.20 m / 1.20 m
Simplifying, we get:
Central Angle = 1/6 radians
Therefore, the angle subtended by the arc is 1/6 radians, which is approximately equal to 0.524 radians or 30 degrees.
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What is the period of a comet if its average orbital radius is 4 AU?
The period of the comet with an average orbital radius of 4 AU is approximately [tex]8 AU^{3/2}[/tex].
The period of a comet is the time it takes for the comet to complete one orbit around the Sun. To calculate the period of a comet, we can use Kepler's Third Law, which states that the square of a planet's orbital period is proportional to the cube of its average orbital radius.
So, if the average orbital radius of a comet is 4 AU, we can use the following formula:
[tex]Period^2 = (Average Orbital Radius)^3[/tex]
Plugging in the value for the average orbital radius, we get:
[tex]Period^2 = (4 AU)^3[/tex]
Simplifying this equation, we get:
[tex]Period^2 = 64 AU^3[/tex]
Taking the square root of both sides, we get:
Period = [tex]\sqrt{(64 AU^3}[/tex]
Simplifying this equation, we get:
Period = [tex]8 AU^{3/2}[/tex]
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To what quantity should the slope correspond (assuming Newton’s Second Law is correct), and can you compare it to any other quantity you have measured? If you can, do so. Does your data support Newton’s Second Law?
The slope of a graph of force versus acceleration should correspond to the mass of the object being measured, according to Newton's Second Law. This is because the formula for the law states that force is equal to mass times acceleration (F = ma). Therefore, the slope of the graph should be equal to the mass of the object.
To compare this to another quantity that can be measured, one could also measure the velocity of the object and calculate its momentum (p = mv). Momentum is a conserved quantity and can be used to predict the behavior of objects in collisions.
If the data collected follows a linear relationship between force and acceleration, and the slope corresponds to the mass of the object being measured, then the data supports Newton's Second Law. However, if the data does not follow a linear relationship or the slope does not correspond to the mass of the object, then there may be some other factors affecting the system that need to be considered.
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