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kaheart [24]
2 years ago
8

You leave your home at 1pm. At 3pm, you are 100 km east of your house. What was your average velocity in km/hr

Physics
1 answer:
Rasek [7]2 years ago
4 0

Answer:

50km/h

Explanation:

Average velocity = change in distance (or distance travelled) divided by/ the change in time (or time taken.)

The change in distance has been given as 100km.

The change in time is 3pm-1pm = 2 hours.

Therefore the average velocity was 100/2 = 50km/h (to the east).

Hope this helped!

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Answer:

Erosion is the process of eroding or being eroded by wind, water, or other natural agents.

Explanation:

"the problem of soil erosion"

the gradual destruction or diminution of something.

"the erosion of support for the party"

are examples of how it could be used.

5 0
3 years ago
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A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.
Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2

the angular speed is given by:

\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s

to calculate how many radians have the wheel turned we need the apply the following formula:

\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad

the distance is given by:

d=\theta*r

d=381rad*(35.5*10^{-2}m)\\d=135m

4 0
3 years ago
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th
Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

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v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

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The brightness of a star is determined
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100% C . By size and distance
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