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3 years ago

Identify the forces acting on the cars and explain why the cars do not slide down the hill

1 answer:

5 0

<span>Examples of outside forces acting on a car is gravity, wind, and other cars. Cars do not slide down hills because their weight, combined with the friction of their tires against the road, hold them in place. </span>

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Weight of a man is less at coal mine.Why?

dmitriy555 [2]

If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.

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2 years ago

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A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of

Neko [114]

Answer: 185.5672566

Explanation: The friction is not relevant

Normal reaction is the force perpendicular to the surface.

this force resists the downwards forces applied which are gravity and a component of the applied force.

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3 years ago

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As the distance between the sun and earth decreases, the speed of the planet

mixas84 [53]

Answer:

Explanation:

Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.

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2 years ago

A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m = 0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m $U^{2}$

K.E = 1/2 x 0.17 x $6^{2}$

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/ $s^{2}$

To calculate the final velocity, let us use third equation of motion.

$V^{2}$ = $U^{2}$ + 2as

$V^{2}$ = $6^{2}$ + 2 x 0.3 x 61

$V^{2}$ = 36 + 36

$V^{2}$ = 72

V = $\sqrt{72}$

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

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1 year ago

A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.

ValentinkaMS [17]

A) $2.03 m/s^2$

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

$mg sin \theta - \mu_k R = ma$ (1)

where

m is the mass

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

$\theta=22.5^{\circ}$

$\mu_k = 0.19$ is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

$R-mg cos \theta =0$ (2)

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2$

B) 5.94 m/s

We can solve this part by using the 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 displacement

Here we have

v = ?

u = 0 (it starts from rest)

$a=2.03 m/s^2$

s = 8.70 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s$

6 0

3 years ago