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Annette [7]
3 years ago
11

A roller of radius 12.5 cm turns at 14 revolutions per second. What is the linear velocity of the roller in meters per second?

Physics
2 answers:
Burka [1]3 years ago
8 0

Answer:

Linear velocity of the roller, v = 11 m/s

Explanation:

It is given that,

Radius of roller, r = 12.5 cm = 0.125 m

Angular velocity of the roller, \omega=14\ rev/s

Firstly, we will convert revolution per second to radian per second i.e.

Angular velocity, \omega=87.96\ rad/s    

We need to find the linear velocity of the roller. It can be calculated by taking the product of angular velocity and the radius of roller.

v=r\times \omega

v=0.125\ m\times 87.96\ rad/s

v = 10.995 m/s

or

v = 11 m/s

So, the linear velocity of the roller is 11 m/s. Hence, this is the required solution.    

Firdavs [7]3 years ago
7 0

12.5 times 14 and convert to meters its 1.75 meters per second

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Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

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Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

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Answer:

4. kinetic energy

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How does an airplane's kinetic energy and potential energy change as it takes off and lands
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The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr
Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

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Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

q=\int\limits {I} \, dt

Here the limits of integration is from 0 to infinite. So,

q=\int\limits {0.88e^{-t/6\times3600s}}\, dt

q=0.88\times(-6\times3600)(0-1)

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

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Substitute the suitable values in the above equation.

N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}

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2 years ago
When a balloon is deflating, why does air leave the balloon?
Gwar [14]

Answer:

<em>When a balloon deflates air moves out of the balloon </em><em>because the pressure inside the balloon is higher than the pressure outside the balloon.</em>

Explanation:

An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.

The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.

This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.

6 0
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