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rosijanka [135]
3 years ago
12

What is the momentum of a bowling ball with mass 5 kg and velocity 10 m/s

Physics
2 answers:
kondor19780726 [428]3 years ago
8 0
50 kg% that is your answer to the question you asked
Rashid [163]3 years ago
7 0

Answer:

50kg.m/s

Explanation:

to find the momentum you need this formula

p=mv p=momentum m=mass v=velocity

multiply the mass by the velocity

p=5kg x 10m/s

10 x 5 is 50 and the momentum unit is kg.m/s

p=50kg.m/s

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If you notice any mistake in my english, please let me know, because i am not native.
8 0
3 years ago
If the car traveled 754 meters in a straight line between 0 and 36 seconds, what was the
ArbitrLikvidat [17]

Answer:

20.94 m/s

Explanation:

Recall that average velocity is defined as:

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Then, for our case:

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4 0
3 years ago
A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an a
lions [1.4K]

Answer:

Approximately 4.61\times 10^{3}\; {\rm N} upwards (assuming that g = 9.81\; {\rm m\cdot s^{-2}}.)

Explanation:

External forces on this astronaut:

  • Weight (gravitational attraction) from the earth (downwards,) and
  • Normal force from the floor (upwards.)

Let (\text{normal force}) denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}.

Let m denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be (\text{weight}) = m\, g.

Let a denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be (\text{net force}) = m\, a.

Rearrange \begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned} to obtain an expression for the magnitude of the normal force on this astronaut:

\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}.

3 0
2 years ago
Which of the following is not a function of PACs?
Katena32 [7]

Answer:

i think D

hope this helps

let me know if i'm wrong i will change the answer

Explanation:

5 0
3 years ago
Read 2 more answers
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nignag [31]

The position of the particle is given by:

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Differentiate x(t) with respect to t to find the velocity x'(t):

x'(t) = 3t² - 24t + 21

Differentiate x'(t) with respect to t to find the acceleration x''(t):

x''(t) = 6t - 24

5 0
3 years ago
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