Ask question

Ask question

All categories

3 years ago

12

A person applies a force of 200N to move an object initially at rest weighing 400N. What could be the coefficient of static fric

tion between the object and the surface?

1 answer:

SOVA2 [1]3 years ago

5 0

Answer:

$\mu=0.5$

Explanation:

Given that,

Applied force, F = 200 N

Weight of the object, or the normal force, N = 400 N

We need to find the coefficient of static friction between the object and the surface.

The force of friction has the same magnitude as that of applied force. It can be given by :

$F=\mu mg$

mg is weight of the object

$\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{200}{400}\\\\\mu=0.5$

So, the coefficient of static friction between the object and the surface is 0.5.

You might be interested in

The gravitational force between two objects is represented by the variable F. If the distance is decreased by 5 what is the new

Anvisha [2.4K]

The universal law of gravitation states that:

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of distance between them.

It means that if the gravitational force is F, then if the distance is decreased by 5 times, then the new gravitation force is:

F/5² = F/25

6 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago

Compare all the organisms. What trend do you see?

Digiron [165]

When you compare all organisms around us, even the smallest ones, you see that they have Life; You could even see this as a trend of some sort.

5 0

3 years ago

Read 2 more answers

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

1 year ago

Candace is interested in learning more about colors. Which question about colors could be answered through scientific investigat

NeTakaya

Hi!

I believe that there were the following options in this question:

<span>A) Which color bends the most to make a rainbow?
B) What color of scented marker has the best smell?
C) Which flower has the prettiest shade of purple on its petals?
D) What is the best-looking color on a high-performance sports car</span>

Now, options b), c) and d) are about a personal preference and opinion - they are not scientific uqestions.
The correct answer is a) - it asks about the properties of light.

4 0

3 years ago

Read 2 more answers