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3 years ago

How are magnetic field lines like electric field lines?

Physics

2 answers:

RUDIKE [14]3 years ago

8 0

Answer:

The correct answer is C. They never cross.

Explanation:

Hope this helps

olasank [31]3 years ago

3 0

Answer:

Option D, they always point away from a north pole

Explanation:

Magnetic fields and electric fields are similar in following ways –

a) Electric field is produced by two opposite charges while magnetic field is produced by two opposite poles i.e north and South Pole. However, they are also produced by charges that are moving.

b) Like charges repel and unlike charges attract each other. Similarly, like poles repel and unlike poles attract each other

c) Electric field is pointed towards the positive charge while magnetic field is away from the north pole and move toward the south pole.

Hence, option D is correct

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A horse pulls on an object with a force of 300 newtons and does 12,000 joules of work. How far was the object moved?

liubo4ka [24]

We know, work done = Force * displacement
Here, w = 12,000 J
f = 300 N

Substitute their values into the expression:
12,000 = 300 * d
d = 12,000 / 300
d = 120/3
d = 40 m

In short, Your Answer would be 40 meters

Hope this helps!

6 0

3 years ago

2) What are the directions of the velocity and acceleration of an object in

zaharov [31]

Motion because the motion is the range

7 0

4 years ago

Read 2 more answers

A rock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. Assume

Andreyy89

Answer:

(a) 5.7 s

(b) 39 m/s

Explanation:

(a) u = 18 m/s

At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.

use first equation of motion.

v = u + g t

0 = 18 - 10 x t

t = 1.8 s

Let the maximum height attained by the ball when it thrown from 50 m height is h'.

Use third equation of motion

v^2 = u^2 + 2 g h'

0 = 18^2 - 2 x 10 x h'

h' = 16.2 m

Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m

Let t' be the time taken by the ball to hit the ground as it falls from maximum height.

use third equation of motion

H = ut + 1/2 x g t'^2

76.2 = 0 + 1/2 x 10 x t'^2

t' = 3.9 s

Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s

(b) Let v be the velocity with which the ball strikes the ground.

v^2 = u^2 + 2 g H

v^2 = 0 + 2 x 10 x 76.2

v = 39 m/s

4 0

4 years ago

Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when i

LenaWriter [7]

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

$v^{2} = u^{2} + 2as$

a = $v^{2} - 2gh$

or, h = $\frac{v^{2}}{2g}$

= $\frac{(26)^{2}}{2 \times 9.8}$

= $\frac{676}{19.6}$

= 34.5 m/s

Hence, it rises 34.5 m/s above the level where it leaves your hand.

(b) Time to reach maximum height is as follows.

v = u + at

or, v - gt = 0

t = $\frac{v}{g}$

= $\frac{26}{10}$

= 2.6 sec

Therefore, it will take 2.6 sec to reach its highest point.

(c) Time taken by the ball to ascent is equal to the time it has taken to descent.

Therefore, time taken by the ball to return to the level where it left your hand after it reaches its highest point? is also 2.6 sec.

(d) Speed of the ball will be 26 m/s in the downward direction. Hence, the velocity will be -26 m/s.

3 0

3 years ago

The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m

lions [1.4K]

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

8 0

3 years ago