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3 years ago

2 kids are running down the hall at 2 m/s, one is 40kg and the other is 50kg (p=ma)

Physics

1 answer:

Georgia [21]3 years ago

7 0

The net force acting on the two (2) kids is equal to 180 Newton.

<u>Given the following data:</u>

Mass of kid 1 = 40 kilogram.

Mass of kid 2 = 50 kilogram.

Acceleration = 2 $m/s^2$

To determine the net force acting on the two (2) kids:

First of all, we would calculate the total mass of the two (2) kids:

$Total\;mass = 50+40$

Total mass = 90 kg.

<h3>How to calculate net force.</h3>

In this exercise, you're required to calculate the magnitude of the net force that is acting on these two (2) kids running down the hall. Thus, we would apply Newton's Second Law of Motion.

Mathematically, Newton's Second Law of Motion is given by this formula;

$Net\;force = mass \times acceleration$

Substituting the given parameters into the formula, we have;

$Net\;force = 90 \times 2$

Net force = 180 Newton.

Read more on acceleration here: brainly.com/question/1121817

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Which objects have both potential and kinetic energy? Check all that apply.

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A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.

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4 years ago

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A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will

Aleks [24]

Answer:

We know that for a pendulum of length L, the period (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

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pi = 3.14

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g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

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3 years ago

A creature moves at a speed of 7.86 furlongs/fortnight (not a very common unit of speed). If 1.0 furlong = 220 yards 1 fortnight

AVprozaik [17]

Answer:

2 0.00130719 m/s

Explanation:

Now we have

7.86 furlongs/fortnight

7.86 x 220 yards / fortnight

1729.2yards / fortnight

1647.8 yards / 14 days

123.5yards / day

123.5 x 0.9144 meter / day

112.meters / day

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5 0

3 years ago

A car is moving with a constant velocity of 25 m/s. Which of the following is true?

S_A_V [24]

b) the net force on the car is zero.

Explanation:

Let's analyze each option one by one:

a) the force from the engine is greater than all the forces of friction. --> FALSE. In fact, the car is moving at constant velocity: this means that its acceleration is zero,

a = 0

and so Newton's second law becomes

$\sum F = ma = 0$

where $\sum F$ is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.

b) the net force on the car is zero. --> TRUE, for what we said at point A)

c) the inertia is changing. --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.

d) the forces of friction are proportional to the acceleration of the car. --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is

$F_f = \mu mg$

where $\mu$ is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.

e) All of the above. --> FALSE

Learn more about net force and Newton's second law:

brainly.com/question/3820012

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3 years ago

A 2 200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.20 m before coming into contact

Kisachek [45]

Answer:

669042 N upward

Explanation:

$W_g$ = Work done by gravity on pile driver

$W_b$ = Work done by beam

u = Initial velocity

v = Final velocity

m = Mass of pile driver = 2200 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height the pile driver falls = 4.2 m

d = Bend in beam = 14 cm

Total work done = Change in Kinetic energy

$W_g+W_b=\frac{1}{2}m(v^2-u^2)\\\Rightarrow mg(h+d)+Fd=0\\\Rightarrow F=-\frac{mg(h+d)}{d}\\\Rightarrow F=-\frac{2200\times 9.81(4.2+0.14)}{0.14}\\\Rightarrow F=-669042\ N$

The force the beam exerts on the pile driver while the pile driver is brought to rest is 669042 N upward

8 0

3 years ago