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Romashka-Z-Leto [24]

2 years ago

11

A 150 kg box requires a force of 200 N to be pushed across the floor at a constant velocity 2.0 m/s. What is the Normal Force ac

ting on the box?

1 answer:

lana66690 [7]2 years ago

6 0

Answer:

normal force = mg

= 159× 9.8=1470N

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When a metal bar is heated the increase in length is greater if

lorasvet [3.4K]

Answer:

A

Explanation:

i think the answer is A

6 0

3 years ago

9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

3 years ago

A ball traveling with an initial momentum of 5.1 kgm/s bounces off a wall and comes back in the opposite direction with a moment

Elina [12.6K]

Answer: Change in momentum=9.4kgm/s

Impulse=9.4kgm/s

Explanation:

Change in momentum=5.1-(-4.3)=5.1+4.3=9.4kgm/s

Impulse=Change in momentum

There impulse=9.4kgm/s

4 0

2 years ago

Can anyone help me with this please

Anastasy [175]

Answer:

1270 J

Explanation:

Recall that the mechanical energy of a system is the addition of the Potential energy and the Kinetic energy at any given time.

As the skier descends, potential energy is converted into kinetic energy, but the total mechanical energy should remain the same.

We see that it is not the case, so that difference is what has gone into thermal energy; 19500 J - 18230 J = 1270 J

7 0

3 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

3 years ago