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3 years ago

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If a car takes a banked curve at less than a given speed, friction is needed to keep it from sliding toward the inside of the cu

rve (a real problem on icy mountain roads). (a) Calculate the minimum speed, in meters per second, required to take a 84 m radius curve banked at 16° so that you don't slide inwards, assuming there is no friction. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 16 km/h?

2 answers:

kvasek [131]3 years ago

7 0

Answer:

Please see attachment

Explanation:

Please see attachment

zubka84 [21]3 years ago

4 0

Answer:

minimum speed is 15.35 m/s

frictional coefficient is 0.26

Explanation:

given data

radius = 84 m

angle = 16°

speed = 16 km/h = 4.43 m/s

to find out

minimum speed and minimum coefficient

solution

we will apply here formula for velocity that is

velocity² = radius × g × tanθ

v² = 84 × 9.8 × tan16

v² = 236.04

v = 15.35 m/s

and

we find first friction force here

friction force 1 = m v² /r

friction force 1 = m (15.35)² / 84 = 2.80 m

and

friction force 2 = m v² /r

friction force 2 = m (4.43)² / 84 = 0.245 m

so total friction force = f1 - f2

total friction force = 2.80 - 0.245 = 2.55 m

so frictional coefficient = friction force /g

frictional coefficient = 2.55 / 9.8

so frictional coefficient is 0.26

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A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

makkiz [27]

Answer:D

Explanation:

Given

mass of object $m=5 kg$

Distance traveled $h=10 m$

velocity acquired $v=12 m/s$

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Initial Energy $=mgh=5\times 9.8\times 10=490 J$

Final Energy $=\frac{1}{2}mv^2+W_{f}$

$=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}$

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3 years ago

A man's higher initial acceleration means that a man can outrun a horse over a very short race. A simple - but plausible - model

Lera25 [3.4K]

Answer:

t_man = 10.16 s, t_horse = 10.73 s, the winner is the man

Explanation:

To solve this problem we are going to find the time of each one separately.

Man we look for distance and time during acceleration

x₁ = v₀ t₁ + ½ a₁ t₁²

as it comes out of rest its initial velocity is zero

x₁ = ½ a₁ t₁²

x₁ = ½ 6.0 1.8²

x₁ = 9.72 m

at this point its speed is

v₁ = v₀ + a t

v₁ = 0 + 6 1.8

v₁ = 10.8 m / s

From here on it continues at constant speed, the distance that the man needs to travel from the point where the man leaves at 100m is

x₂ = 100 - x₁

x₂ = 100- 9.72

x₂ = 90.28 m

the time for this part is

v₁ = x₂ / t₂

t₂ = x₂ / v₂

t₂ = 90.28 / 10.8

t₂ = 8.36 s

the total time for the man is

t_man = t₁ + t₂

t _man = 1.8 + 8.36

t_man = 10.16 s

We repeat the calculation for the horse

distance traveled during the acceleration period

x₃ = v₀ t + ½ a₂ t₃²

as part of rest its initial velocity is zero

x₃ = ½ a₂ t₃²

x₃ = ½ 5.0 4.8²

x₃ = 57.6 m

the velocity at this point is

v₃ = v₀ + a₂ t₃

v₃ = 0 + 5 4.8

v₃ = 24 m / s

the rest of the route is at constant speed, the remaining distance

x₄ = 200 - x₃

x₄ = 200 - 57.6

x₄ = 142.4 m

the time to go through it is

t₄ = x₄ / v₃

t₄ = 142.4 / 24

t₄ = 5.93 s

the total time for the horse is

t_horse = t₃ + t₄

t_horse = 4.8 + 5.93

t_horse = 10.73 s

when we compare the times we see that the man arrives a little before the horse, the winner is the man

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DENIUS [597]

Answer

given,

mass of the drop, m = 0.0014 g

speed of the drop, u = 8.1 m/s

a) Change in momentum is equal to impulse

final velocity of the drop, v = 0 m/s

J = m ( v - u )

J = 0.0014 x 10⁻³ x ( 0 - 8.1 )

J = -1.134 x 10⁻⁵ kg.m/s

impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s

b) time, t = 0.37 m s

impact of force = ?

we know

J = F x t

1.134 x 10⁻⁵ = F x 0.37 x 10⁻³

F = 0.031 N

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