a Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a time of 15 s. what is the magnitude of the wheels angular acceleration?
b. what is the magnitude of the tangential acceleration after the maximum operational speed is reached?​

Answer:

The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air

Explanation:

No explanation

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Explanation:

Answer:

The border is 8km above sea level.

Explanation:

We know that:

Density = 1.25 kg/m^3

Pressure = 10^5 N/m^2

g = 10m/s^2

Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.

This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)

Then the total mass inside that volume is:

M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)

The weight of this mass is:

W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to  10^5 N/m^2

then:

P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

Whit this equation we can find the value of H.

10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

10^5 N =  (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H

8,000 m^3 = 1m^2*H

(8,000 m^3)/(1m^2) =H

8,000 m = H

And we want this answer in km, knowing that 1,000m = 1km

8,000m = 8km = H

The border is 8km above sea level.

Height of boundaries is 8.2 km

Normal density = 1.25 kg/m³

1 atm = 101325 N/m²

Height of boundaries

Pressure = Height × Density × Gravitational acceleration

101325 = Height × 1.25 × 9.8

101325 = Height × 12.25

Height of boundaries = 101325 / 12.25

Height of boundaries = 8271.42 m

Height of boundaries = 8.2 km

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Answer:

The second bulb will have thicker filament

Explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]

Resistance of the second bulb:

[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]

Resistivity of the tungsten filament is given by the following equation;

[tex]\rho = \frac{RA}{L}[/tex]

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

[tex]A = \pi r^2[/tex]

where;

r is the thickness of each filament

[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]

Therefore, the second bulb will have thicker filament

Answer:

10.1 m/s

Explanation:

By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.

And by the law of conservation of momentum,

initial momentum of squid + expelled water = final momentum of squid + expelled water.

Now, the initial momentum of the system is zero.

So, 0 = final momentum of squid + expelled water

0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled

So, MV + mv = 0

MV = -mv

v = -MV/m

= -6.50 kg × 2.40 m/s ÷ 1.55 kg

= -15.6 kgm/s ÷ 1.55 kg

= -10.1 m/s

So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s

Answer:

0.54 sec

Explanation:

Answer:

Explanation:

Speed = 14.2 m/s

Distance = 514 m

Time = Distance / Speed

Time = 514 / 14.2

Time = 36.19 seconds

Answer:

4.25 rad/s

Explanation:

Given that.

Mass, m = 51.9 kg

Radius, r1 = 2.4 m

Moment of inertia, I = 215.24 kgm^2

Angular velocity, ω = 2.1 rad/s

Radius, r2 = 0.864 m

To start with, we are going to use the Conservation of angular momentum to solve the question, which is

l(initial) = l(final)

[I₁ + I₂](initial)*ω(initial) = [I₁ + I₂](final)*ω(final)

Making ω(final) the subject of formula, we have

ω(final) = [I₁ + I₂](initial)*ω(initial) / [I₁ + I₂](final)

ω(final) = [215.24 + (51.9)(2.4)²](2.1) / [215.24 + (51.9)(0.864)²]

ω(final) = [215.24 + 298.944]2.1 / [215.24 + 38.74]

ω(final) = 514.184 * 2.1 / 253.98

ω(final) = 1079.786 / 253.98

ω(final) = 4.25 rad/s

= 5.273 rad/s

Answer:

B. 3.1 × 10^5 V

Explanation:

Answer:

B

Explanation:

e2021

Answer:

true

Explanation:

with increased temperature particles move faster as they gain kinetic energy

Answer:

because they are found freely in nature uncombined so they are highly reactive with other elements

Answer:

Approximately [tex]261\; \rm K[/tex], if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.

Explanation:

Let [tex]P_1[/tex] and [tex]P_2[/tex] denote the pressure of this gas before and after the changes.

Let [tex]V_1[/tex] and [tex]V_2[/tex] denote the volume of this gas before and after the changes.

Let [tex]T_1[/tex] and [tex]T_2[/tex] denote the temperature (in degrees Kelvins) of this gas before and after the changes.

Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the quantity (number of moles of gas particles) in this gas before and after the changes.

Assume that this gas is an ideal gas. By the ideal gas law, the ratios [tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex] and [tex]\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex] should both be equal to the ideal gas constant, [tex]R[/tex].

In other words:

[tex]R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex].

[tex]R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Combine the two equations (equate the right-hand side) to obtain:

[tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Rearrange this equation for an expression for [tex]T_2[/tex], the temperature of this gas after the changes:

[tex]\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1[/tex].

Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: [tex]n_2 = n_1[/tex], [tex](n_2 / n_1) = 1[/tex].

[tex]\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}[/tex].

Answer:

1. Wavelength = 3.2 m

2. Amplitude = 0.6 m

Explanation:

1. Determination of the wavelength.

The wavelength of a wave is defined as the distance between two successive crest. This implies that for every complete vibration, there is one wavelength.

From the diagram given above, we can see that the wave makes 2½ vibrations.

This means that there are 2½ equal wavelength of the wave. Therefore, the wavelength can be obtained as follow:

Length (L) = 8 m

Wavelength (λ) =?

2½ λ = L

5/2 λ = 8

5λ / 2 = 8

Cross multiply

5λ = 2 × 8

5λ = 16

Divide both side by 5

λ = 16 / 5

λ = 3.2 m

Therefore, wavelength of the wave is 3.2 m.

2. Determination of the amplitude.

The amplitude of a wave is defined as the maximum displacement of the wave from the origin.

From the diagram given above, the distance between the maximum and minimum displacement is given as 1.2 m. Thus, we can obtain the amplitude of wave as follow:

Distance between the maximum and minimum displacement (D) = 1.2

Amplitude (A) =?

A = ½D

A = ½ × 1.2

A = 0.6 m

Thus, the amplitude of the wave is 0.6 m

Answer:

[tex]0.842\ \text{lb ft}[/tex]

[tex]0.1052\ \text{lb ft}[/tex]

Explanation:

d = Diameter of wheel = 6 in

r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]

t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]

w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]

[tex]t_2[/tex] = Time taken to slow down = 35 s

[tex]t_1[/tex] = Time taken to reach operating speed = 5 s

[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]

Weight is given by

[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]

Mass is given by

[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]

Moment of inertia is given by

[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]

Angular acceleration while slowing down is given by

[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]

Frictional moment is given

[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]

Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]

Angular acceleration while speeding up is given by

[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]

Motor torque is given by

[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]

Motor torque is [tex]0.842\ \text{lb ft}[/tex].

Answer:

Answer: Oxygen

Explanation:

Oxygen would not be considered as a pollutant

Answer:

Hey there the answer is C. Oxygen

Explanation:

Oxygen is the chemical element with the symbol O and atomic number 8. It is a member of the chalcogen group in the periodic table, a highly reactive nonmetal, and an oxidizing agent that readily forms oxides with most elements as well as with other compounds. After hydrogen and helium, oxygen is the third-most abundant element in the universe by mass. At standard temperature and pressure, two atoms of the element bind to form dioxygen, a colorless and odorless diatomic gas with the formula O 2. Diatomic oxygen gas constitutes 20.95% of the Earth's atmosphere. Oxygen makes up almost half of the Earth's crust in the form of oxides. Hope this helps! Have a great day!

Answer:

239 rpm

Explanation: So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (or wheel revolutions) fit inside the 75,000 cm? In other words, if I were to peel this wheel's tread from the cart and lay it out flat, it would measure a distance of 100π cm. How many of these lengths fit into the entire distance covered in one minute? To find out how many of (this) fit into so many of (that), I must divide (that) by (this), so:

100πcm/rev

75,000cm/min

​750 min rev≈238.7324146RPM

Answer:

The photosphere is the visible "surface" of the sun. So your answer would be C.

Explanation: its right

Answer:

a = 52s²

Explanation:

How to find acceleration

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

Solve

We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)

We first need to solve the velocity equation for time (t):

v = u + at

v - u = at

(v - u)/a = t

Plugging in the known values we get,

t = (v - u)/a

t = (16 m/s - 120 m/s) -2/s2

t = -104 m/s / -2 m/s2

t = 52 s

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Answer:

Most likely (B)

Explanation:

B in the passage is the most representative out of all your choices and it has evidence from the passage  

Hope dis helps Jit!

Sorry i forgot to type C

B and C both measure mass while the others are calculations and are bias

The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:

A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.

When  a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.

When  a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.

Learn more about  gravitational field here:

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Answer:

The human body carries out its main functions by consuming food and turning it into usable energy. Immediate energy is supplied to the body in the form of adenosine triphosphate (ATP). Since ATP is the primary source of energy for every body function, other stored

Explanation:

this what teacher explain to us

Answer:

question #1 is A

Question #2 is C

Explanation:

Answer:

20.0N

Becuase It's the largest

Answer:

20.0

Explanation:

It's the biggest number

Answer:

A. Increases

Explanation:

As altitude decreases, the amount of gas molecules in the air increases - the air becomes less dense. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense.

Answer:

It decreases so it is B

Explanation:

As altitude rises, air pressure drops.