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jasenka [17]
4 years ago
11

A boat of mass 225 kg drifts along a river at a speed of 21 m/s to the west. what impulse is required to decrease the speed of t

he boat to 15 m/s to the west?
Physics
2 answers:
zzz [600]4 years ago
8 0
The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
J=\Delta p=m \Delta v
where
m=225 kg is the mass of the boat
\Delta v=v_f-v_i=15 m/s-21 m/s=-6 m/s is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
J=m\Delta v=(225 kg)(-6 m/s)=-1350 N s
and the negative sign means the direction of the impulse is against the direction of motion of the boat.
kondaur [170]4 years ago
4 0

Answer:

as the other person explained, the answer is 1350 kg*m/s east

Explanation:

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Answer:

v = 1.98 mph

Explanation:

Given that,

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Putting the values, we get :

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7 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

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\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
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Given;

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4 0
2 years ago
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Answer:

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