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Mariana [72]
2 years ago
11

A race car has a mass of 710 kg. It starts from rest and travels 40 m in 3.0 s. The car is uniformly accelerated during the enti

re time. What net force is applied to it
Physics
1 answer:
Archy [21]2 years ago
6 0

The net force applied to it is 6311.9 N.

To calculate the net force to the car, first, we need to find the acceleration using the equation of motion.

Force: This as be defined as the product of mass and acceleration of a body. The S.I unit of force is Newton (N)

⇒ Formula

  • S = ut+at²/2................. Equation 1

⇒ Where:

  • u = Initial velocity
  • S = Distance
  • t = Time
  • a = acceleration

From the question,

 ⇒Given:

  • S = 40 m
  • t = 3.0 s
  • u = 0 m/s (at rest)

⇒ Substitute these values into equation 1

  • 40 = 0(3)+a(3²)/2

⇒ Solve for a

  • 9a = 80
  • a = 80/9
  • a = 8.89 m/s²

⇒ To get the force, we use the formula below

  • F = ma    .................... Equation 2

⇒ Where:

  • F = Force applied
  • m = mass of the car

⇒ Given:

  • m = 710 kg
  • a = 8.89 m/s²

⇒ Substitute these values into equation 2

  • F = 710(8.89)
  • F = 6311.9 N

Hence the net force applied to it is 6311.9 N.

Learn more about force here: brainly.com/question/13370981

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Explanation:

Momentum = mass × speed

p = (30 kg) (3 m/s)

p = 90 kg m/s

7 0
2 years ago
a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

#SPJ4

4 0
1 year ago
How far will a 600 kg boat travel in 10 s if there is a constant 900 N force on it and it starts from rest?
natta225 [31]

Answer:

here

Explanation:

There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. ... a 3.25-kg object rightward with a constant acceleration of 1.20 m/s/s if the force of ... of 33.8 kg, how far (in meters) will it move in 1.31 seconds, starting from rest?

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A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

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katrin [286]

Answer:

0.405 seconds

Explanation:

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First,     d = (1/2) gt^2    or     t=   ( 2 d / g)^1/2

= ( 2 × 39 / 9.8)^1/2 = 2.8212 seconds

Then, to fall from 53 down to 2 meters...

 d = (1/2) gt^2    or     t=   ( 2 d / g)^1/2

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So the amount of time it takes for the block to fall from 14 m upto 2 m above the ground

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2 years ago
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