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4 years ago

6

A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally

rotating at 55 rpm. What was the initial rotation rate in radians per second?

1 answer:

nadezda [96]4 years ago

3 0

Answer: 5.76 rads/s

Explanation:

The initial rotation is 55 rpm

1 rev = 2π radians

55 revs = 55 × 2π/1 = 345.58 radians/min

345.58 rads/min = 345.58rads/60s = 5.76 rads/s

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A hockey puck, mass of 0.115 kg, moving at 35 m/s strikes a ray thrown on the ice by a fan. The ray has a mass of 0.265 kg. The

emmainna [20.7K]

We can use conservation of momentum. MaVa + MbVb = (Ma+Mb)Vf.

Plugging in values:
(.115)(35) + (.265)0 = (.380)(Vf)

Now solve for Vf.
Vf= .115*35 / .380 = 10.6 m/s

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3 years ago

How many atoms of each element are in the chemical formula NH4NO3?

Igoryamba

<span>2 Nitrogen, 4 Hydrogen, 3 Oxygen

9 atoms per molecule.

NH4, ammonium (not to be confused with ammonia NH3) is a 1+ ion and NO3 is 1-.
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3 years ago

Read 2 more answers

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

6 0

3 years ago

Earth gravity acts upon objects with a steady of

stepladder [879]

Answer:

D.

9.8 meters per second squared

Explanation:

This is the right answer

4 0

4 years ago

What is an object's change in position, or displacement, and the change in time over which that displacement occurred used to ca

yulyashka [42]

The answer is a average velocity

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3 years ago