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kobusy [5.1K]
2 years ago
12

Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?

Physics
1 answer:
Tanya [424]2 years ago
5 0

It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

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How much power does it take to lift a 24 kg gift box 6m above the floor in 4 s?
Mrac [35]

Answer:

<h2>5.6kW</h2>

Explanation:

Step one:

given

mass m= 24kg

distance moved= 6m

time taken= 4seconds

Step two:

Required

power

but work done is the force applied at a distance, and the power is the work done time the time taken

Work done= F*D

F=mg

W= mg*D

W=24*9.81*6

W=1412.6J

Power P= work * time

P=1412.6*4

p=5650.5W

P=5.6kW

3 0
1 year ago
A photon with a wavelength of 2.29 × 10^–7 meter strikes a mercury atom in the ground state.
4vir4ik [10]
The photon can be absorbed and the energy of the photon is exactly equal to the energy-level difference between the ground state and the level d.

8 0
2 years ago
A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat
weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2

Now we can find the period of the second pendulum at the same location, which is given by

T=2\pi \sqrt{\frac{L}{g}}

where we have

L = 0.36 m (length of the  second pendulum)

g=17.5 m/s^2

Substituting,

T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s

#LearnwithBrainly

8 0
2 years ago
A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
KengaRu [80]

a) 219.8 rad/s

b) 20.0 rad/s^2

c) 2.9 m/s^2

d) 7005 m/s^2

e) Towards the axis of rotation

f) 0 m/s^2

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

\omega=\frac{\theta}{t}

where

\theta is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

\omega_{max}=35 rev/s

The angle covered during 1 revolution is

\theta=2\pi rad

Therefore, the maximum angular velocity is:

\omega_{max}=35 \cdot 2\pi = 219.8 rad/s

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

\alpha = \frac{\Delta \omega}{t}

where

\Delta \omega is the change in angular velocity

t is the time elapsed

Here we have:

\omega_0 = 0 is the initial angular velocity

\omega_{max}=219.8 rad/s is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

\alpha = \frac{219.8-0}{11}=20.0 rad/s^2

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

Here we have

\alpha =20.0 rad/s^2

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

a_t=(20.0)(0.145)=2.9 m/s^2

d)

The magnitude of the radial (centripetal) acceleration is given by

a_c = \omega^2 r

where

\omega is the angular velocity

r is the radius of the circle

Here we have:

\omega_{max}=219.8 rad/s is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

a_c=(219.8)^2(0.145)=7005 m/s^2

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

\alpha=0, since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

a_t=(0)(0.145)=0 m/s^2

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

v=\omega r

where

v is the linear speed

\omega is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

\omega=219.8 rad/s

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

v=(219.8)(0.145)=31.9 m/s

7 0
2 years ago
Which amplitude of the following longitudinal waves has the greatest energy?
Rashid [163]

Which amplitude of the following longitudinal waves has the greatest energy?

amplitude = 10 cm; wavelength = 6 cm; period = 4 seconds

8 0
2 years ago
Read 2 more answers
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