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xeze [42]
3 years ago
10

1. A 59 kg person is in a vehicle travelling at 41 m/s. The vehicle runs into a telephone pole. At impact, it

Physics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

a. 12,600 N

b. 1290 kg

Explanation:

a. Impulse = change in momentum

F Δt = m Δv

F (0.192 s) = (59 kg) (0 m/s − 41 m/s)

F = -12,600 N

b. F = mg

12,600 N = m (9.8 m/s²)

m = 1290 kg (or 2,830 lbs)

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A 142 kg boulder is located near the edge of a cliff 25 m above the ground. What is the gravitational potential energy of the bo
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8 0
3 years ago
Read 2 more answers
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

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6 0
3 years ago
Which statement BEST describes the condition of steady-state equilibrium? - System slowly adjusts to long-term changes in input
krok68 [10]

Answer:

System inputs and outputs fluctuate around a stable average so the system does not move far from its average condition.

Explanation:

Steady-state equilibrium can also be called dynamic equilibrium. The main difference between the two is the type of system we are considering. When a system is closed, and it attains equilibrium, there is no transfer of energy. In  the case of an open system, even if the system achieves equilibrium there will be some transfer of energy but it will not deviate far from it's equilibrium point, that is, it will be in a steady-state.

6 0
3 years ago
G How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
sattari [20]

Answer:

686.7N

Explanation:

Given data

Mass= 70kg

We know that the buoyant force experienced by the person is equal to the weight of the person

Hence the weight is

Weight = mass* Acceleration

Weight= 70*9.81

Weight= 686.7N

Therefore the weight is 686.7N

5 0
3 years ago
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