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sdas [7]
3 years ago
15

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

m of the stone about the center of the circle is:
Physics
1 answer:
Illusion [34]3 years ago
3 0
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
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Which officeholders are elected by the electoral college
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Explanation:i did it before

4 0
3 years ago
A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0
3 years ago
Any help would be highly appreciated, preferably with steps too!
qwelly [4]

Answer:

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Explanation:

5 0
3 years ago
A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car
jolli1 [7]

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

8 0
2 years ago
A penny is dropped from the Statue of Liberty
Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

vf=vi+at,

vf =unknown velocity  vi= initial velocity vi=0m/s  a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

vf=0m/s+(9.8m/s^2)*0.30seconds

Step 2: Solve

vf=2.94m/s

b)

Use the equation below to solve for the displacement at 0.30 seconds

x=vit+\frac{1}{2} at^{2}

Step 1: Substitute the same variables with the knowns

x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

5 0
3 years ago
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