Answer:
by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Explanation:
Given the data in the question;
drag coefficient of Cd = 1.4
speed v = 6.0 m/s
One model expands to a square 1.8 mm on a side
Area A = 1.8 × 1.8 = 3.24 mm² = 3.24 × 10⁻⁶ m²
distance travelled s = 240 mm = 0.24 m
we know that; density of air e = 1.225 kg/m³
Now,
Dragging force F[tex]_D[/tex] = ( Cd × e × v² × A ) / 2
thermal energy = F[tex]_D[/tex] × s
so
thermal energy = ( 1.4 × 1.225 × (6)² × (3.24 × 10⁻⁶) × 0.24 ) / 2
thermal energy = ( 4.8009024 × 10⁻⁵ ) / 2
thermal energy = 2.4004512 × 10⁻⁵ J
Therefore, by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Suppose the Earth were squeezed down to the size of a small mountain. The mass of the Earth would not change, just its volume and radius. If you were now standing on the surface, would the gravitational force on you be greater than, less than or the same as before the Earth was squeezed?
Explanation:
Much greater. Gravitational force depends on the mass and separation distance. Shrinking the earth means the mass remains the same while the radius gets smaller. Since gravitational force is inversely proportional to the square of the separation distance, as shown by Newton's universal gravitational law
[tex]\:\:\:F_G = G \dfrac{mM_E}{R^2}[/tex]
shrinking the radius even by a factor of 10 will cause your weight, which also happens to be the gravitational force of the earth on you, to be 100 times more.
If the mass of the Earth remains the same and only its volume and radius are decreased, then the average density of the Earth would increase significantly.
What is gravitational force?Gravity, also known as gravitational force, pulls objects with mass towards each other. We frequently consider the force of gravity from Earth.
If the Earth's mass remains constant while its volume and radius are reduced, the average density of the Earth increases significantly.
This is due to the fact that the mass has been compressed into a much smaller volume.
As a result, standing on the surface, the gravitational force on you would be greater than before the Earth was compressed to the size of a small mountain.
When the Earth is compressed, the distance between you and the center of the Earth decreases while the mass remains constant.
Thus, your gravitational force increases.
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If a small child swallowed a safety pin, why
would an X-ray photograph clearly show the
location of the pin?
Answer:
yes
Explanation:
it is in the body system
Answer:
it would show clearly because it is a metal piece in the body.
What is the most commonly used semiconducting material? Nickel Aluminum b. Cobalt d. Silicon
Answer:
i believe the answer is D
Explanation:
Hope it works
Answer:
D. SiliconExplanation:
Silicon is the most widely used type of semiconductor material. Its major advantage is that it is easy to fabricate and provides good general electrical and mechanical properties.caculate the component of a force of 200 ns
at a direction of 60° to the force
Answer:
[tex]F_x = 100N[/tex]
[tex]F_y = 100\sqrt 3 \ N[/tex]
Explanation:
Given
[tex]F = 200N[/tex]
[tex]\theta = 60^o[/tex]
Required
The component of the force in F direction
To do this, we simply calculate the force in the vertical and horizontal direction.
This is calculated as:
[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal
[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical
So, we have:
[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal
[tex]F_x = 200N * \cos(60^o)[/tex]
[tex]F_x = 200N * 0.5[/tex]
[tex]F_x = 100N[/tex]
[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical
[tex]F_y = 200N * \sin(60^o)[/tex]
[tex]F_y = 200N * \frac{\sqrt 3}{2}[/tex]
[tex]F_y = 100\sqrt 3 \ N[/tex]
A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, carrying a 30 A current. What acceleration (in g's) does this interaction give the airplane?
Answer:
[tex]a=0.2*10^{-5}g[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=180=>0.18kg[/tex]
Charge [tex]Q=18mC=18*10^-^3C[/tex]
Velocity [tex]v=2.2m/s[/tex]
Length of Wire [tex]L=8.6cm=>0.086[/tex]
Current [tex]I=30A[/tex]
Generally the equation for Magnetic Field of Wire B is mathematically given by
[tex]B=\frac{\mu_0*I}{2\pi*l}[/tex]
[tex]B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}[/tex]
[tex]B=6.978*10^{-5}T[/tex]
Generally the equation for Force on the plane F is mathematically given by
[tex]F=qvB[/tex]
Therefore
[tex]ma=qvB[/tex]
[tex]a=\frac{qvB}{m}[/tex]
[tex]a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}[/tex]
[tex]a=2.37*10^{-5}[/tex]
Therefore in Terms of g's
[tex]a=\frac{2.37*10^{-5}}{9.8}[/tex]
[tex]a=0.2*10^{-5}g[/tex]
find the resistance of wire of
0.65m Radius 0.25
and
resistivity 3x10-6 OHM
Complete Question:
Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.
Answer:
Resistance = 9.95 Ohms
Explanation:
Given the following data;
Length = 0.65 m
Radius = 0.25 mm to meters = 0.00025 m
Resistivity = 3 * 10^{-6} ohm-metre.
To find the resistance of the wire;
Mathematically, resistance is given by the formula;
[tex] Resistance = P \frac {L}{A} [/tex]
Where;
P is the resistivity of the material. L is the length of the material.A is the cross-sectional area of the material.First of all, we would find the cross-sectional area of the wire.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.00025)²
Area = 3.142 * 6.25 * 10^{-8}
Area = 1.96 * 10^{-7} m²
Now, to find the resistance of the wire;
[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]
[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]
Resistance = 9.95 Ohms
4. You currently have TWO 9 volt batteries connected to a 82 resistor. How
much current is flowing through the circuit? *
1.125 A
5 A
72 A
2.25 A
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Explanation:
Total voltage = 9 × 2 = 18v
Resistance = 82 Ω
Ohm's law::
V = IR
18v = 82 Ω × I
18v /82 /Ω = I
18/82 Ampere is the current
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 13.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 25 s interval
Answer:
The tub turns 37.520 revolutions during the 25-second interval.
Explanation:
The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:
[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)
Then, we expand the previous expression by kinematic equations for uniform accelerated motion:
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)
Where:
[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.
[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.
And each acceleration is determined by the following formulas:
Acceleration
[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)
Deceleration
[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)
Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.
If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:
[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]
[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]
[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]
[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]
[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]
[tex]\Delta n = 37.520\,rev[/tex]
The tub turns 37.520 revolutions during the 25-second interval.
Scientific theories are deductive in nature.?
Answer:
deductive reasoning usually follows steps .
That is, how we predict what the observations should be if the theory were correctidentify the word being referred to choose your answer from the words below
Answer:
1:Rotation
2:Axis
3:Aphelion
4:orbit
Two forces and are applied to an object whose mass is 11.8 kg. The larger force is . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when points due east and points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of and (b) the magnitude of .
Can you please fill in whatever goes in the blanks ?
Without them, the question makes no sense and has no answer.
Two forces (___) and (___) are applied to an object whose mass is 11.8 kg. The larger force is (___) . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when (___) points due east and (___) points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of (___) and (b) the magnitude of (___) .
3. Four charges having charge q are placed at the corners of a square with sides of length L. What is the magnitude of the force acting on any of the charges
Answer:
Fr = 1.91 * 9*10⁹*q²/L²
Explanation:
Let´s say that the corners of the square are A B C and D
We are going to find out the force on the charge placed on B ( the charge placed in the upper right corner.
As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.
Force due to charge placed in A
module Fₓ = K* q² / L² in the direction of x
Force due to charge placed in C
module Fy = K* q²/L² in the direction of y
Force due to the charge placed in D
That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.
d² = L² + L² = 2*L²
d = √2 * L
The module of the force due to charge place in D
F₄₅ = K*q²/ 2*L²
To get the force we need to add first Fₓ and Fy
Fx + Fy = F₁
module of F₁ = √ Fx² + Fy² the direction will be the same as the diagonal of the square then:
F₁ = √ ( K* q²/L² )² + ( K* q²/L² )²
F₁ = √ 2 * K*q²/L²
And now we add forces F₁ and F₄₅ to get the net force Fr on charge in point B.
The direction of Fr is the direction of the diagonal and is of rejection
the module is
Fr = F₁ * F₄₅
Fr = √ 2 * K*q²/L² + K*q²/ 2*L²
Fr = ( √ 2 + 0,5 ) * K*q² /L²
K = 9*10⁹ Nm²C²
Fr = 1.91 * 9*10⁹*q²/L²
We don´t know units of L and q
An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?
a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min
Answer:
T = 188.5 s, correct is C
Explanation:
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]
let's calculate
v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]
v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
v = x / T
T = x / v
the distance of a circle with radius r = 0.6 m
x = 2π r
we substitute
T = 2π r / v
let's calculate
T = 2π 0.6/0.02
T = 188.5 s
reduce
t = 188.5 s ( 1 min/60 s) = 3.13 min
correct is C
A thin, uniform rod has length L and mass M. A small uniform sphere of mass m is placed a distance x from one end of the rod, along the axis of the rod.
a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. (Hint: Use the power series expansion for ln(1+x).)
b) Use Fx=−dU/dx to find the magnitude of the gravitational force exerted on the sphere by the rod.
Answer:
Explanation:
From the given information:
a)
Let's have an imaginary view of the rod located at a given distance r from he the mass (m) of the sphere.
Then the equation for the potential energy as related to the small area of the dr of the rod can be computed as:
[tex]dU = -\dfrac{GMm}{L}*\dfrac{dr}{r}[/tex]
where,
G = gravitational constant
[tex]U = - \int^{x+L}_{x}Gm\dfrac{M}{L}*\dfrac{dr}{r}[/tex]
[tex]U = - \dfrac{GMm}{L}\int^{x+L}_{x}\dfrac{dr}{r}[/tex]
By taking the integral within the limit
[tex]U = - \dfrac{GMm}{L} \Big[In \ r\Big]^{x+L}_{x}[/tex]
[tex]\mathbf{\implies - \dfrac{GmM}{L} In \Big(\dfrac{{x+L}}{{x}}\Big)}[/tex]
b)
By using [tex]F= -\dfrac{dU}{dx}[/tex], the magnitude of the gravitational force can be determined as follows:
Here, we have:
[tex]F = -\dfrac{d}{dx} \Big [\dfrac{-GmM}{L}In(\dfrac{x+l}{x}) \Big ] \\ \\ = \dfrac{GmM}{L}\times \dfrac{x}{x+L}\times (0-\dfrac{L}{x^2}) \\ \\ By \ solving \\ \\ \mathbf{ =-\dfrac{GmM}{x(x+L)}}[/tex]
From above, the negative sign indicates an attractive force
how to get infinite thanks could you show me please ☺️
Answer: divide by zero, or square root of a negative
Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.
A couple of ways to get that...
You try to divide by zero. In other words, a problem that asks you to perform something like this: 5/0= or 23/(4-4)= Such a problem will give you an error on a calculator because the answer is infinity or doesn't exist. Another way is to try to get the square root of a negative number. That answer doesn't exist as a real number, so [tex]\sqrt{-4\\}[/tex] will also give you an error on a calculator.
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. The plate separation is now increased to a distance of 2d. What would be the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?
Answer:
The answer is "Option D".
Explanation:
Please find the complete question in the attached file.
As plate separation increased to 2d the capacitance get halred but the change remain same
[tex]\therefore V=\frac{Q}{C}[/tex]
The voltage doubles are now electric field remain same because both the distance and voltage get doubled.
[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]
So,
[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]
Darwin believed that emotional expressions began as ________ that came to have evolutionary value because they ________. Select one: a. physiological reactions; increased the efficiency of bodily reactions b. communication devices; increased the efficiency of bodily reactions Incorrect c. physiological reactions; convey emotional states to other members of the species d. random mutations; convey expectations to other members of the species
Answer:
c. physiological reactions; convey emotional states to other members of the species
Explanation:
Darwin believed that emotional expressions began as physiological reactions that came to have evolutionary value because they convey emotional states to other members of the species. These reactions are used by many different types of species to convey various things.
Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
From the given information:
The elastic potential energy can be calculated by using the formula:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
Making K the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the frequency of the small oscillation can be determined by using the formula:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom = 1.66 × 10⁻²⁶ kg
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Which of the following best describes the upper respiratory tract?
O It takes air in from outside the body.
O It is where oxygen and carbon dioxide are exchanged.
O It is located inside the thorax.
O It is not directly involved in respiration.
How can magnetic levitation be improved?
How can a small spark start a huge explosion
Answer:
The explosion is set off by an electrostatic spark. When the mixture ignites, the rapid increase in temperature brings about a huge increase in gas pressure. If the burning vapour were to be confined the resulting rise in pressure could destroy the chamber with a loud explosion.
Explanation:
What is the main way in which heat transfer occurs in liquids and gases?
The large scale structure of the universe has been carefully mapped using redshift surveys of a very large number of galaxies. Answer the following: Suppose that a large telescope with modern equipment can measure the redshift to a galaxy in just 10 minutes. And suppose we want to spend no more than a year mapping the distribution of galaxies. How many redshifts can be surveyed
Answer:
26280
Explanation:
In current time, good telescope can measure redshift to a galaxy in 10 minutes.
Thus, in one year that has on an average 365 days, the total time taken to measure redshifts is = ( 365 *12 *60) minute
= 262800 minutes .
Hence, the number of redshifts observed in a year = (262800/10) = 26280
Electricity is the result of moving electrons, so it's classified as
A. Kinetic Energy
B. Gravitational Energy
C. Potential Energy
D. Elastic Energy
What effect does increased blood flow have on the body when performing exercises? A. delivers more sugar to organs B. delivers more energy to muscles C. delivers more oxygen to the body D. delivers more protein to muscles Please select the best answer from the choices provided. O A . OB ос OD Next Submit Save and Exit Mark this and return
A small object with mass 0.200 kg moves with constant speed in a vertical circle of radius 0.500 m. It takes the object 0.500 s to complete one revolution. (a) What is the translational speed of the object
Answer:
6.28 m/s.
Explanation:
Given that,
The mass of the object, m = 0.2 kg
The radius of the circle, r = 0.5 m
It takes the object 0.500 s to complete one revolution.
We need to find the translational speed of the object. Let it is v. We know that,
[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]
So, the transalational speed of the object is 6.28 m/s.
Which Circut Model is more efficient, why?
Answer:
right one is more efficient
Explanation:
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees rest on the flowers. The charged bees both generate an electric field, and while the charged bees are resting on the flowers, the net electric field at some distance between them is zero.
Required:
Do the bees have the same or opposite signs of charge?
Answer:
the charge of the bees must be of the same sign
Explanation:
The electric field is given by the relation
E = k q / r²
This electric field has outgoing direction if the charge is positive and incoming towards the charge if it is negative.
The force generated by this field on a test charge is
F = q E
Since the charge is a scalar, the direction of the force is the same as the electric field.
In this case the two flowers are at a certain distance and the two charged bees land on them, so the force on a test charge is the vector sum of the force that each bee creates, so that this force is subtracted from the two bees must have the same charge sign.
The force created by the bee on the left goes to the right and the force created by the bee on the right goes to the left, so the forces are subtracted,
Consequently the charge of the bees must be of the same sign
A motor has an output of 1000 watts. When the motor is working a full capacity, how much time will it require to lift a 50 Newton weight 100 meters?
The time required to lift the weight is 5 seconds.
What is time?Time is the measure of past or present events or occurrences. The S.I unit of time is seconds (s).
To calculate the time required to lift the weight, we use the formula below.
Formula:
P = Fd/t.................. Equation 1Where:
P = PowerF = Weightd = distance.t = timemake t the subject of the equation.
t = Fd/P................ Equation 2From the question,
Given:
F = 50 Nd = 100 mP = 1000 WSubstitute these values into equation 2
t = (50×100)/1000t = 5 seconds.Hence, The time required to lift the weight is 5 seconds.
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50 POINTS‼️‼️‼️‼️‼️
A 4.88 x 10-6 C charge moves 265 m/s
perpendicular (at 90°) to a magnetic
field of 0.0579 T. What is the magnetic
force on the charge?
Answer: 0
Explanation: Trust
The magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.
The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:
F = q * v * B * sin(θ)
Where:
F is the magnetic force,
q is the charge of the particle (in this case, 4.88 x 10^(-6) C),
v is the velocity of the particle (in this case, 265 m/s),
B is the magnetic field strength (in this case, 0.0579 T),
θ is the angle between the velocity vector and the magnetic field vector (in this case, 90 degrees).
Plugging in the values:
F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * sin(90°)
Since sin(90°) is equal to 1, the equation simplifies to:
F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * 1
Calculating the value:
F = 6.47 x 10^(-4) N
Therefore, the magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.
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