A force is directly proportional to what ?

A neighbor's child wants to go to a neighborhood carnival to experience the wild rides. The neighbor is worried about safety as

seraphim [82]

Answer:

No, the ride is not safe

Explanation:

From the diagram attached, it is seen that

$\sum f_{y} = 0$

$\tau cos \theta - W = 0$ ................(1)

$\sum f_{x} = 0$

$-ma + \tau sin \theta = 0$ .......(2)

$a = w^{2} r$

From the diagram, $r = Lsin \theta$

$a = w^{2} L sin \theta$

$w = \frac{2\pi }{T}$

$a = \frac{4\pi ^{2}L sin \theta }{T^{2} }$ ............(3)

Put (3) into (2)

$\tau sin \theta = \frac{4\pi ^{2}m L sin \theta }{T^{2} }$

$\tau = \frac{4\pi ^{2}m L }{T^{2} }$

w = weight of chair + weight of child = 50 + 10 = 60 lb

g = 32 ft/s²

m = w/g = 60/32 = 1.875

L = 30 ft

T = 3 secs

$\tau = \frac{4\pi ^{2}*1.875*30 }{0.3^{2} }$

$\tau = 246.74 lbs$

since 246.74 lbs > 200 lbs, it is not safe because the stationary chair will creak

6 0

3 years ago

Read 2 more answers

The spring of a toy car stores 10 J of potential energy. Only 8 J of energy changes to kinetic energy as the car moves. What hap

son4ous [18]

Answer:

It has been converted into thermal energy due to friction

Explanation:

According to the law of conservation of energy, energy cannot be created nor destroyed, but only transformed from one form into another.

Applied to this problem, it means that the total initial energy of the spring-toy system must be conserved.

Therefore:

- At the beginning, the total energy stored in the spring is 10 J

- After the toy is released, the total energy must still be 10 J.

In reality, we are told that the kinetic energy of the car is only 8 J. The other 2 J have not been destroyed, but they have been converted into thermal energy, due to the presence of frictional forces that act against the motion of the toy car.

3 0

3 years ago

Pleaseee HElpp!!!!!!!

lutik1710 [3]

Coulomb's Law

Given:

F = 3.0 x 10^-3 Newton

d = 6.0 x 10^2 meters

Q1 = 3.3x 10^-8 Coulombs

k = 9.0 x 10^9 Newton*m^2/Coulombs^2

Required:

Q2 =?

Formula:

F = k • Q1 • Q2 / d²

Solution:

So, to solve for Q2

Q2 = F • d²/ k • Q1

Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9 Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)

Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)

Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)

Then, take the reciprocal of the denominator and start multiplying

Q2 = 1080 • 1 Coulombs/297

Q2 = 1080 Coulombs / 297

Q2 = 3.63636363636 Coulombs

Q2 = 3.64 Coulumbs

6 0

3 years ago

Read 2 more answers

The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10

loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

The average force exerted on the limestone floor is approximately 1.6013 × 10⁻² N

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, h = 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

$\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}$

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

$\displaystyle Average \ force, \, F_{ave} \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}$

The average force exerted on the limestone floor by the droplets of water is $F_{ave}$ ≈ 1.6013 × 10⁻² N

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

3 0

2 years ago

How much work is accplished when a force of 250 n pushes a box accross the floor for a distance of 50 meters

7nadin3 [17]

You just multiply these two numbers, it's 1250J

4 0

3 years ago