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svp [43]
3 years ago
12

What is the force required to propel an object with mass of 12.0 kg at a rate of 27.0 m/s^2

Physics
2 answers:
Llana [10]3 years ago
8 0
Heyyy mate

here is your answer

given Mass of object= 12.0kg

rate of propel= 27.0m/s^2


according to second law of Newton

force F = m.a

so F= 12×27=324N

hope it help you
enyata [817]3 years ago
4 0

Newtons second law --> F = ma

So we have mass which = 12.0

And acceleration = 27.0

So the force = (12)(27)

F = 324 N

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Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
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Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

s_A = -6 + 9t

s_B = 5 - 8t

When A an B meets, they are at the same position and at the same time. So

s_A = s_B

-6 +9t = 5 - 8t

17t = 5 + 6 = 11

t = 11/17 = 0.647 s

s_A = -6 + 9*0.647 = -0.176 m

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5 0
3 years ago
41. 2072 Set E Q.No. 11 A source of sound produces a note of
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3 years ago
In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a
xz_007 [3.2K]

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

m_1u_1+m_2u_2=m_1v_1+m_1v_2

Where :

m_1, m_2 = masses of object collided

u_1,u_2 = initial velocity before collision

v_1,v_2 = final velocity after collision

We have :

Two equal-mass carts roll towards each other.

m_1=m_2=M

Initial velocity of m_1=u_1=2 m/s

Initial velocity of m_2=u_2=-1 m/s (opposite direction)

Final velocity of m_1=v_1=v (same direction )

Final velocity of m_2=v_2=v  (same direction)

M\times 2 m/s+M(-1 m/s)=Mv+Mv

1 m/s=2v

v = 0.5 m/s

rg135

The speed of the carts after their collision is 0.5 m/s.

7 0
4 years ago
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