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3 years ago

A sled weighing 300 N is moved at constant speed over a horizontal floor by a force of 50 N applied parallel to the floor

1 answer:

3 0

If the sled is moving at a constant speed, that mean there needs to be an equal but opposite force countering the 50 N force. This opposite force is friction, and is equal to the weight times a factor:

F = n*W

Let n be the coefficient of friction.

Now you simply fill in the formula and you get:

50 N = n*300N or n = 50N / 300N = 1/6

So, the final answer is 1/6 or 0.1667

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Phyics !!! Cannon ball question

natima [27]

Answer:

Zero; no force is required to keep it going

Explanation:

Since the cannon ball is fired into frictionless space, there will be nothing to stop it, so it will keep going and going.

3 0

3 years ago

A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.

pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

$h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t = 6.55 \ s$

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

$X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m$

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0

3 years ago

What will be the average velocity of a body falling in free fall on Earth for 3 s?

SpyIntel [72]

Answer:

29.4m/s

Explanation:

Given parameters:

Time = 3s

Unknown:

Average velocity = ?

Solution:

To solve this problem, we use the expression below:

v = u + gt

v is the average velocity

u is the initial velocity = 0m/s

g is the acceleration due to gravity = 9.8m/s²

t is the time

So;

v = 0 + (9.8 x 3) = 29.4m/s

6 0

3 years ago

Exercises

Crank

$\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}$

$\\ \rm\Rrightarrow u=\dfrac{190}{-14}$

$\\ \rm\Rrightarrow u=13.6cm$

Real

5 0

3 years ago

A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficien

Liula [17]

Answer:

Yes it will move and a= 4.19m/s^2

Explanation:

In order for the box to move it needs to overcome the maximum static friction force

Max Static Friction = μFn(normal force)

plug in givens

Max Static friction = 31.9226

Since 36.6>31.9226, the box will move

Mass= Wieght/g which is 45.8/9.8= 4.67kg

Fnet = Fapp-Fk

= 36.6-16.9918

=19.6082

=ma

Solve for a=4.19m/s^2

7 0

3 years ago