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mylen [45]
1 year ago
15

A car will skid to halt at a uniform rate of -9.4 m/s^2. If you measure skid marks that are 34 m long, with what speed was the c

ar going just before the driver slammed the brakes?
Physics
1 answer:
Klio2033 [76]1 year ago
6 0

The speed with which the car going just before the driver slammed the brakes is 25.28 m / s, if the car skids to halt at a uniform rate of -9.4 m/s^2.

v² = u² + 2 a s

v = Final velocity

u = Initial velocity

a = Acceleration

s = Displacement

v = 0

a = - 9.4 m / s²

s = 34 m

0 = u² + ( 2 * - 9.4 * 34 )

u² = 639.2

u = 25.28 m / s

Deceleration is the slowing down of an object. It can be negative acceleration or positive acceleration. Deceleration is negative if the object moves in positive direction and it is positive if the object moves in negative direction.

Therefore, the speed with which the car going just before the driver slammed the brakes is 25.28 m / s

To know more about deceleration

brainly.com/question/17618273

#SPJ1

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Answer:

(a) F = 1500 N.

(b) Ratio force to the antelepe's weight = 3.40

Explanation:

Force : This can be defined as the product of mass and the distance moved by a body. Its S.I unit is Newton. It can be represented mathematically as

F = Ma

Where F= force, M = mass (Kg) and a = Acceleration (m/s²)

Weight: This can be defined as the force on a body due to gravitation field. It is also measured in Newton (N). It can be represented mathematically as

W = Mg

Where W = weight of the body, M = mass of the body (Kg), g = Acceleration due to gravity.

(a)

F = Ma

Where M = 45kg,

a = unknown.

But we can look for acceleration Using one of the equation of motion,

v² = u² + 2gs

Where v= final velocity(m/s), u = initial velocity (m/s) g = 0 m/s, g = 9.8m/s² and s = height = 2.5m.

∴ v² = 2gs

 v = √2gs = √(2×9.8×2.5)

v= √49 = 7m/s

With the force applied, the impala’s velocity must increase from 0 m/s to 7 m/s in 0.21 second

∴ a = (v-u)/t

 a = (7-0)/0.21 = 7/0.21

  a = 33.33 m/s².

F = 45 × 33.33 ≈ 1500

F = 1500 N.

(b)

Where F = Force = 1500 N

and W = Weight = Mg = 45 × 9.8 = 441 N

∴Ratio force to the antelepe's weight = F/W = 1500/441 = 3.40

 Ratio force to the antelepe's weight = 3.40

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How much pressure is applied to the
den301095 [7]

Answer:

306.12N/m^2

Explanation:

Mass = 30 kg

g = 9.8m/s2

F = mg

F = 30 x 9.8

F = 294N

L = 0.98m

Since the stilts is square, then the area is L^2

A = L^2

A = 0.98^2

A = 0.9604m^2

P =?

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P = 306.12N/m^2

The pressure exerted by the man is 306.12N/m^2

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