Moving a neutral wire in a(n) ____ field will induce a(n) ____

Describe a solar eclipse. Be sure to include the positions of the sun, moon, and earth.

Alex777 [14]

It’s the type of eclipse that occurred when the moon passes between the sun and earth, and when the moon fully or partially blocks the sun.

7 0

3 years ago

I will be so thankful if u answer correctly!!

olga_2 [115]

<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.

In this case, a frictional force $F_{r}$ is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F - $F_{r}$

m = 50kg

a = 0 [At constant velocity, acceleration is zero]

Substitute these values into equation (i) as follows;

F - $F_{r}$ = m x a

F - $F_{r}$ = 50 x 0

F - $F_{r}$ = 0

F = $F_{r}$ -------------------(ii)

Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²

In this case, the same frictional force $F_{r}$ is opposing the movement of the box.

∑F = 1.5F - $F_{r}$

m = 50kg

a = 0.1m/s²

Substitute these values into equation (i) as follows;

1.5F - $F_{r}$ = m x a

1.5F - $F_{r}$ = 50 x 0.1

1.5F - $F_{r}$ = 5 ---------------------(iii)

Substitute $F_{r}$ = F from equation (ii) into equation (iii) as follows;

1.5F - F = 5

0.5F = 5

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

4 0

2 years ago

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

3 years ago

A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?

poizon [28]

Answer:

$\boxed {\boxed {\sf 7.84 \ Joules}}$

Explanation:

The formula for potential energy is:

$PE=m*g*h$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

$m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m$

Substitute the values into the formula.

$PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)$

Multiply the first two numbers.

0.4 kg*9.8 m/s²= 3.92 kg*m/s²
If we convert the units now, the problem will be much easier later on.
1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

$PE=(3.92 \ N )(2 \ m)$