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jasenka [17]
3 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]3 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]3 years ago
4 0

Answer:

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

Explanation:

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Rank the six combinations of electric charges on the basis of the electric force acting on q1. define forces pointing to the rig
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Question 26 (1 Point)
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The book continue to move due to its inertia

Explanation:

We can answer this question by using Newton's first law, which states that:

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4 0
2 years ago
Read 2 more answers
A (10.0+A) g ice cube at -15.0oC is placed in (125 B) g of water at 48.0oC. Find the final temperature of the system when equili
liq [111]

Answer: Final temperature is 34.15°C.

Explanation: When two objects have different temperature, they will exchange heat energy until there is no more net energy transfer between them. At that state, the objects are in <u>thermal</u> <u>equilibrium</u>.

So, when in equilibrium, the total heat flow must be zero, i.e.:

Q_{1}+Q_{2}=0

In our case, there will be a change in state of ice into water, so total heat flow will be:

m_{1}c_{1}(T_{f}-T_{i})+m_{2}c_{2}(T_{f}-T_{i})+mL=0

where

m₁ is mass of ice

m₂ is mass of water

c₁ is specific heat of ice

c₂ is specific heat of water

T_{f} is final temperature

T_{i} is initial temperature

L is latent heat fusion

Temperature is in Kelvin so the transformation from Celsius to Kelvin:

For ice:

T = -15 + 273 = 258K

For water:

T = 48 + 273 = 321K

Solving:

21(2.09)(T_{f}-258)+158(4.186)(T_{f}-321)+21(333)=0

43.89T_{f}-11323.62+661.4T_{f}-212305.55+6993=0

705.3T_{f}=216636.17

T_{f}= 307.15K

In Celsius:

T_{f}= 34.15°C

Final temperature of the system when in equilibrium is 34.15°C

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