As a battery is used to charge a capacitor, does the overall charge inside the battery get smaller, greater, or stay the same?

A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so

shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

$T_y+N=m.g$

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$

Recalling that

$F_r=\mu N$

$Tcos\theta-\mu N=m.a$

Replacing N from [1]

$Tcos\theta-\mu (m.g-Tsin\theta)=m.a$

Operating

$Tcos\theta-\mu m.g+\mu Tsin\theta=m.a$

Solving for T

$T(cos\theta+\mu sin\theta)=m.a+\mu m.g$

$\displaystyle T=\frac{m.a+\mu m.g}{cos\theta+\mu sin\theta}$

$\displaystyle T=m\frac{a+\mu g}{cos\theta+\mu sin\theta}$

We don't know the value of m, thus we'll plug in the rest of the data

$\displaystyle T=m\frac{0.763+0.10\cdot 9.8}{cos36.8^o+0.10 sin36.8^o}$

$T=2.0252m$

Dividing by the weight of the box m.g

$T/(m.g)=2.0252/9.8=0.2066$

Thus, the magnitude of the pulling force is 20.66% of the gravitational force acting on the box

7 0

3 years ago

Hii:) anyone able to explain and help me with this question?

Assoli18 [71]

Answer:

No

Explanation:

The equation of state for ideal gases tells that:

$pV=nRT$

where

p is the gas pressure

V is the gas volume

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature

In this problem, we have a fixed mass of gas. This means that the number of moles of the gas, $n$ , does not change; also, the volume V remains the same, and R is a constant, this means that

$p\propto T$

So, as the pressure increases, the temperature increases.

However, here we want to understand what happens to the average distance between the molecules.

We have said previously that the number of moles n does not change: and therefore, the total number of molecules in has does not change either.

If we consider one dimension only, we can say that the average distance between the molecules is

$d=\frac{L}{N}$

where L is the length of the container and N the number of molecules. Since the volume of the container here does not change, L does not change, and since N is constant, this means that the average distance between the molecules remains the same.

4 0

3 years ago

The greater the change in the speed of light in a different media the greater the angle of what?

valkas [14]

Answer:

Idk

Explanation:

8 0

3 years ago

Explain how wind is the primary source for wave energy

siniylev [52]

Waves get their energy from the wind.