I need help please pleaseeeee

A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizont

kolbaska11 [484]

Answer:

The magnitude of the acceleration of the box is 2.01 m/s².

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following horizontal forces:

Fr = friction force.

Fx = Horizontal component of the applied force, F.

And we have the following vertical forces:

Fy = vertical component of the applied force.

N = normal force exerted on the box.

W = weight of the box.

According to Newton´s second law:

∑F = m · a

Then, in the horizontal direction:

Fx - Fr = m · a

Where "m" is the mass of the box and "a" its acceleration.

Fx can be obtained by trigonometry (see figure):

Fx = F · cos 30°

Fx = 90.0 N · cos 30°

Fr is calculated as follows:

Fr = μ · N

Where μ is the coefficient of friction and N the normal force.

So, we have to find the magnitude of the normal force.

Using Newton´s second law in the vertical direction:

∑F = N + Fy - W = m · a

Notice that the box has no vertical acceleration, then:

N + Fy - W = 0

Solving for N:

N = W - Fy

The weight is calculated as follows:

W = m · g

Where g is the acceleration due to gravity:

W = 20.0 kg · 9.8 m/s² = 196 N

And the vertical component of the applied force can be obtained by trigonometry:

Fy = F · sin 30°

Fy = 90.0 N · sin 30°

The normal force will be:

N = W - Fy = 196 N - 90.0 N · sin 30°

N = 151 N

Now, we can calculate the friction force:

Fr = μ · N

Fr = 0.250 · 151 N

Fr = 37.8 N

And now, we can obtain the acceleration of the box:

Fx - Fr = m · a

(Fx - Fr) / m = a

(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a

a = 2.01 m/s²

The magnitude of the acceleration of the box is 2.01 m/s².

8 0

3 years ago

E) Thermal energy is released during

vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

$Q_{net,in} - W_{net, out} = \Delta U$ (1)

Where:

$Q_{net, in}$ - Net input heat, measured in joules.

$W_{net, out}$ - Net output work, measured in joules.

$\Delta U$ - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then $\Delta U < 0$ , which is caused by three scenarios:

(i) $Q_{net,in} < 0$ , $W_{net, out} < 0$ , $|Q_{net,in}|>|W_{net,out}|$

(ii) $Q_{net, in} > 0$ , $W_{net,out} > 0$ , $|Q_{net,in}|$

(iii) $Q_{net,in}< 0$ , $W_{net, out}>0$

In the case $Q_{net,in} > 0$ , $W_{net, out}$ , the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

$Q_{net, in} = \Delta U$ , $Q_{net, in} < 0$ (2)

If $Q_{net, in} < 0$ , then $\Delta U < 0$ . Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0

3 years ago

What are graphs used for?

kari74 [83]

A line that joins dots plotted on a graph paper is called a line graph. It is used to show the variation of a quantity with respect to another. A line graph represents two pieces of information that are usually related.

Taking the perspective of a student learning astronomy or physics, graphs are useful because they can summarize a LOT of information into one picture. When scientists do research graphs are often the only way to represent data that has been collected.

plz mark me as brainliest if this helped :)

3 0

3 years ago

I need help for a review sheet

siniylev [52]

Tin is Sn, atomic number 47 is Silver, Mass of sodium is 22.9 u

3 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$