Question 10 of 10

If 600 N of force is used to move a car 4 meters, how much work is done?

slamgirl [31]

Answer:

<h3>The answer is 2400 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

force = 600 N

distance = 4 m

We have

work done = 600 × 4

We have the final answer as

Hope this helps you

4 0

3 years ago

Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

<em>"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.</em>"

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "</em>

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "</em>

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago

A ski jumper travels down a slope and leaves

Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

$v_x = 28 m/s$

and the distance covered along the horizontal direction in a time t is

$d_x = v_x t$

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity $u_y = 0$ and constant acceleration $g=9.8 m/s^2$ (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

$d_y = \frac{1}{2}gt^2$

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

$\theta = 36^{\circ}$ below the horizontal

This happens when:

$tan \theta = \frac{d_y}{d_x}$

Substituting and solving for t, we find:

$tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s$

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

$d_x = v_x t = (28)(4.15)=116.2 m$