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klio [65]
3 years ago
11

What is the momentum of a 500-kg car moving at 10 m/s?

Physics
2 answers:
nika2105 [10]3 years ago
5 0

Answer: 5000 kg m/s

Explanation:

Formula for momentum,

Momentum = mass * velocity.

Momentum = 500 * 10

Momentum = 5000 kg m/s

Delicious77 [7]3 years ago
3 0

Answer:5000kgm/see

Explanation:

Momentum=mass×velocity

Momentum=500×10

Momentum=5000kgm/s

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2 years ago
a roller coaster begins at the top of a hill if it acelerates at the rate of 2 m/s2 and has a mass of 2000 kg what net force is
AlladinOne [14]

Answer:

\boxed {\boxed {\sf 4000 \ Newtons }}

Explanation:

Force can be found by multiplying the mass by the acceleration.

F=m*a

The mass of the roller coaster is 2000 kilograms and the acceleration is 2 meters per second squared.

m= 2000 \ kg \\a= 2 \ m/s^2

Substitute the values into the formula.

F= 2000 \ kg * 2 \ m/s^2

Multiply.

F= 4000  \ kg*m/s^2

  • 1 kg*m/s² is equal to 1 N
  • Therefore our answer of 4000 kg*m/s² is equal to 4000 Newtons

F= 4000 \ N

The net force acting on the roller coaster is <u>4000 Newtons.</u>

7 0
2 years ago
A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

3 0
3 years ago
Stephanie serves a volleyball from a height of 0.80 m and gives it an initial velocity of +7.2 m/s straight up. how high will th
Papessa [141]
<span>3.78 m Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes. 7.2 m/s / 9.81 m/s^2 = 0.77945 s The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving d = 1/2 A T^2 d = 1/2 9.81 m/s^2 (0.77945 s)^2 d = 4.905 m/s^2 0.607542 s^2 d = 2.979995 m So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height. d = 2.979995 m + 0.8 m = 3.779995 m Rounding to 2 decimal places gives us 3.78 m</span>
7 0
3 years ago
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