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3 years ago

How many minutes in 1 hour

Physics

2 answers:

fgiga [73]3 years ago

8 0

Answer:

Explanation:

Every 60 minutes is an hour

Alexeev081 [22]3 years ago

8 0

Answer:

60; minute in 1 HR is the answer

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An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he

labwork [276]

Answer:

New moment of inertia will be $I=1.92kgm^2$

Explanation:

It is given initially angular velocity $\omega =6rev/sec=6\times 2\pi =37.68rad/sec$

Moment of inertia $I=0.4kgm^2$

Angular momentum is equal to $L=I\omega =37.68\times 0.4=15.072kgm^2/sec$

Now angular velocity is decreases to $\omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec$

As we know that angular momentum is conserved

So $15.072=I\times 7.85$

$I=1.92kgm^2$

So new moment of inertia will be $I=1.92kgm^2$

4 0

3 years ago

If you jumped out of a plane, you would begin speeding up as you fall downward. Eventually, due to wind resistance, your velocit

MrRa [10]

Answer:

Mg or your weight.

Explanation:

When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.

5 0

3 years ago

Calculate the most probable speed of an ozone molecule in the stratosphere

Marysya12 [62]

Answer:

$v_{mp}=305.83 m/s$

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

$v_{mp}= \sqrt{\frac{2RT}{M} }$

plugging the values we get

$v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }$

$v_{mp}=305.83 m/s$

7 0

3 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$