1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alisiya [41]
3 years ago
14

In a level parking lot, a man with a mass of 100 kg gives a shove to a boy with a mass of 50 kg who is on roller skates. The boy

and the man are both at rest initially. During the shove, the man remains at rest but the boy on roller skates accelerates backward at 2 m/s2 . What force did the boy exert on the man during the shove?
Physics
1 answer:
natita [175]3 years ago
4 0

Answer:

100N

Explanation:

Newton's third law states that whenever an object exerts a force on a second object, it exerts a force of equal magnitude and direction but in the opposite direction on the first. It is often stated as follows: Each action always opposes an equal but opposite reaction.

The subject 1 of 100kg is making a force F, to move an object from 50Kg to 2m / s ^ 2. This Force the object of 50Kg will reflect it in the opposite direction by Newton's third law.

Once the parameter of the force that both are experiencing is clarified, Newton's second law is applied to their respective calculation.

F = ma = 50kg * 2m / s ^ 2 = 100N

That is the force the boy exert on the man during the shove.

You might be interested in
1.a bag is dropped from a hovering helicopter. the bag has fallen for 2 s. what is the ball's velocity at the instant its hittin
omeli [17]

1. The bag's velocity immediately before hitting the ground.

Recall this kinematics equation:

Vf = Vi + aΔt

Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)

a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)

Δt = 2s

Plug in the values and solve for Vf:

Vf = 0 + 9.81×2

Vf = 19.62m/s

2. The height of the helicopter.

Recall this other kinematics equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (bag is dropped starting from rest)

a = 9.81m/s² (acceleration due to gravity of the earth)

Δt = 2s

Plug in the values and solve for d:

d = 0×2 + 0.5×9.81×2²

d = 19.62m

3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s

Reuse the equation from question 2:

d = ViΔt + 0.5aΔt²

Given values:

d = 19.6m (height of the helicopter obtained from question 2)

Vi = 2m/s

a = 9.81m/s² (acceleration due to earth's gravity)

Plug in the values and solve for Δt:

19.6 = 2Δt + 0.5×9.81Δt²

4.91Δt² + 2Δt - 19.6 = 0

Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):

Δt = 1.8s, Δt = −2.2s

The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.

Δt = 1.8s

Now we can use this new value of Δt to get the velocity before hitting the ground:

Vf = Vi + aΔt

Given values:

Vi = 2m/s

a = 9.81m/s²

Δt = 1.8s (result from previous question)

Plug in the values and solve for Vf:

Vf = 2 + 9.81×1.8

Vf = 19.66m/s

4 0
4 years ago
what is the force of friction between an 80kg box and the ground on Earth if the coefficient is 0.2?​
Reil [10]

Answer:

160N

Explanation: When 80kg mass is one group . It's reaction force acting on a ground.

Weight of the object = 80*10

= 800 N

Here we are given cofficient of static friction its 0.2. It should be smaller than 1

Friction force = Reaction * Friction Cofficient

Reaction = 800N ( Considering Vertical Equilibrium )

F = 800* 0.2

F = 160N

3 0
3 years ago
Read 2 more answers
Identify the equation used to calculate the perpendicular force (F⊥) acting on a block on an inclined plane.
pychu [463]
Using geometrical arguments, we can see that the angle of the inclined plane \theta is equal to the angle between Fg and the perpendicular force.

But the perpendicular force is the projection of Fg along the perpendicular axis, and Fg=mg, so the correct answer is
<span>C) F=mg cosΘ </span>
6 0
3 years ago
Read 2 more answers
What does WNBA stand for
Alexeev081 [22]

Answer:The Women's National Basketball Association,

Explanation:Branliest:)

6 0
3 years ago
Read 2 more answers
A 62.0-kg skier is moving at 6.30 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.90
Klio2033 [76]
Here are the missing questions:
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?
Part A
The initial kinetic energy of the skier is:
E_{k0}=m\frac{v_0^2}{2}
Part of this energy is then used to do work against the force of friction. Force of friction on the horizontal surface can be calculated using following formula:
F_f=mg\mu
The work is simply the force times the length:
W_f=F_f\cdot L=mg\mu L
So when the skier passes over the rough patch its energy is:
E=E_{k0}-W_f
When the skier is going down the skill gravitational potential energy is transformed into the kinetic energy:
E_p=E_{k1}\\ mgh=E_{k1}
So the final energy of the skier is:
E_f=E_{k0}-W_f+E_{k1}\\ E_f=m\frac{v_0^2}{2}-mg\mu L+mgh=1856.86$J
This energy is the kinetic energy of the skier:
E_f=m\frac{v_f^2}{2}\\ v_f=\sqrt{\frac{2E_f}{m}}=7.74\frac{m}{s}
Part B
We know that skier lost some of its kinetic energy when crossing over the rough patch. This energy is equal to the work done by the skier against the force of friction.
E_{int}=W_f\\&#10;E_{int}=mg\mu L=894.1$J

4 0
4 years ago
Other questions:
  • "a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1
    5·2 answers
  • A certain cloud contains 220 water droplets per cubic centimeter. If 1cubic meter = 1,000,000cubic centimeters, how many drops a
    11·1 answer
  • You make a simple instrument out of two tubes which looks like a flute and extends like a trombone. One tube is placed within an
    7·1 answer
  • Name the steps of how the universe was formed
    14·1 answer
  • The principles used to solve this problem are similar to those in Multiple-Concept Example 17. A 205-kg log is pulled up a ramp
    11·1 answer
  • What is the momentum of A 200 kg shark swimming at 15 m/s
    9·1 answer
  • The density of mercury is 13600 kg/m3 at 0°C. what is its density at temperature 100°C​
    9·1 answer
  • At its peak, a tornado is 74.0 m in diameter and carries 335-km/h winds. What is its angular velocity in revolutions per second?
    5·1 answer
  • True or False<br><br> An object will slow down and stop as long as no forces act upon it
    14·2 answers
  • Full moon is located______
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!