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MissTica
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.
Physics
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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A solid cylindrical object has a mass of 2.0 kg, a diameter
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Answer:

I = 0.0025 kg.m²

Explanation:

Given that

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Now by putting the all values

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3 years ago
The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con
ira [324]

The Period of the resulting shm will be T=39.7

<u>Explanation:</u>

<u>Given data</u>

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

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k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

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<u>By newton's second law.</u>

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

<u>using the k=mω value</u>

k=mω

ω=k/m

ω=1200/3

ω=400

<u>Using F = -kx value</u>

x = F/-k

x=(-10.16)/1200

x=0.00847m

<u>Restoring the  torque value </u>

τ = -dmgsinΘ    where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

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α=-1.781

<u>Rotational to linear form</u>

a = αr  

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

<u>Time Period</u>

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

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Answer:

In the reaction you would have 15.0 mols of Y and X.

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