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vampirchik [111]
3 years ago
14

During a safety test, a car hits a wall and stops in 0.55 s. The net force on the car is 6500 N during the collision. What is th

e magnitude of the change in momentum of the car? Round answer to two significant digits. kg m
Physics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

ΔP = 3600 kg m/s.

Explanation:

Change in momentum (ΔP) of the car is the amount of impulse on it. Impulse (I) can be define as the product of force and the time at which it acts.

i.e I = ΔP

But, I = Ft

Ft = ΔP

ΔP = 6500 x 0.55

     = 3575 Ns

     = 3600 kg m/s^{2} x s

ΔP = 3600 kg m/s

The magnitude of the change in momentum of the car is 3600 kgm/s.

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If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance). If instead, you throw it downward, its downward acceleration after release is 9.8 m/s2.

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A net force of 5.0 newtons moves a 2.0-kilogram object a distance of 3.0 meters in 3.0 seconds. how much work is done on the obj
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What is the boiling point (in °C) of a solution of 7.94 g of I2 in 69.2 g of toluene, assuming the I2 is nonvolatile? (For tolue
erik [133]

Answer: The boiling point of  solution is 112.16^0C

Explanation:

Elevation in boiling point:

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of toluene = 110.63^oC

k_b = boiling point constant of toluene =3.40^oC/m

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

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w_1 = mass of solvent (toluene) = 69.2 g

M_2 = molar mass of solute (I_2)= 254g/mol

Now put all the given values in the above formula, we get:

(T_b-110.63)^oC=1\times (3.40^oC/m)\times \frac{(7.94g)\times 1000}{254\times (69.2g)}

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Therefore, the boiling point (in °C) of a solution is 112.16

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