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3 years ago

10

A car of mass 2 320 kg is descending a sloping road, inclined at 10.0° to the horizontal. The driver sees a hazard and brakes to

make an emergency stop. The distance travelled while braking is 22.5 m.
The initial speed of the car was 13.4 m s−1. Calculate the average braking force. Assume air resistance is negligible.

1 answer:

strojnjashka [21]3 years ago

7 0

f = 1.32×10^4 N

Explanation:

We can use the work-energy theorem to find the work done by the braking force f:

W = ∆KE + ∆PE

= (KEf - KEi) + (PEf - PEi)

= [(1/2)mvf^2 - (1/2)mvi^2] + (mhf - mghi)

At the bottom of the slope, vf = 0 and hf = 0 and hi = dsin10° (d = braking distance) so work W becomes

W = -[(1/2)mvi^2 + mgdsin10°]

= -m[(1/2)vi^2 + gdsin10°]

= -(2320kg)[(1/2)(13.4m/s)^2 + (9.8 m/s^2)(22.5m)sin10]

= -2.97×10^5 J

Since W = fd, where f is the braking force, we can now solve for f:

f = W/d = (-2.97×10^5 J)/(22.5 m)

= -1.32×10^4 N

Note: the negative sign means that it is a dissipative force.

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Answer:

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Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

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The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0

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therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

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T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

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