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jeka57 [31]
3 years ago
14

At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coeffi

cient of friction was estimated to be 0.38. a) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.b) Why does the car's mass not matter?
Physics
1 answer:
LiRa [457]3 years ago
6 0

Answer:

a. V = 19.1m/s

b. The mass of the car does not matter

Explanation:

A.

KE = 1/2mv² = fd --------(1)

Fd = umgd ---------(2)

Therefore,

1/2mv² = umgd ---------(3)

M will cancel itself out from both sides of equation 3.

Then we will have:

1/2v² = ugd

Then we cross multiply to make v² the subject of the formula

V² = 2ugd

V = √2ugd -------(4)

U = 0.38

g = 9.81

d = 98

When we input these values into equation 4, we will have:

V = √2x0.38x9.81x98

V = √730.6488

V = 27.03m/s

B.

The mass of the car does not actually matter as the mass was cancelled out on the both sides of equation 3

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7. How long does it take a ball rolling down a hill to change its speed from 3 m/sec to 34.5 m/sec
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The time elapsed is 9 seconds

Explanation:

The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:

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u = 3 m/s is the initial velocity

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a=3.5m/s^2 is the acceleration

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Compare the electric and gravitational force between the electron and the proton in a hydrogen atom. The electric charge of the
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The gravitational attraction between electron and proton is 10−40 whereas electrostatic force of attraction between a proton and an electron is 10-8.

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The electric charge of the electron and proton are the same i.e. -1.60x10-19C whereas their gravitational force is different due to difference in mass.

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In a container of negligible mass, 0.400 kg of ice at an initial temperature of -29.0 ∘C is mixed with a mass m of water that ha
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Answer:

1 kg

Explanation:

The container has negligible mass and no heat is loss to the surrounding.

Mass of ice = 0.4kg, initial temperature of ice = -29oC, final temperature of the mixture = 26oC, mass of water (m2) = ?kg, initial temperature of water = 80oC, c ( specific heat capacity of water ) = 4200J/kg.K, Lf = heat of fusion of water = 3.36 × 10^5 J/kg

Using the formula:

Quantity of heat gain by ice = Quantity of heat loss by water

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Quantity of heat loss by water = m2cΔT

Quantity of heat loss by water = m2 ×4200× (80 - 26) = m(226800)

since heat gain = heat loss

22.68 × 10^4 = 226800 m2

divide both side by 226800

226800 / 226800 = m2

m2 = 1 kg

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