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jonny [76]
3 years ago
13

Which of the following would cause electromagnetic induction?

Physics
1 answer:
Ganezh [65]3 years ago
4 0

Answer:

The loop of coils -  Electromagnetic induction is caused a changing magnetic field moving thru a closed loop(s) of coils

You might be interested in
How is air temperature related to high and low pressure air
vitfil [10]

Answer: So, I looked at it to see what was the correct one, and the correct answer is Cool air near surface forms high-pressure areas, warm air forms low pressure areas. I hope this helps :D :)

Explanation:

3 0
3 years ago
A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0
3 years ago
A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.
Mekhanik [1.2K]
<h3>Answer:</h3>

7.53 m

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

<u>Time taken by the ball to reach the ground:</u>

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

<u>Solving for the time taken:</u>

s = ut + 1/2(at²)                 [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5                  [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds     [taking the square root of both sides]

<u>Horizontal distance covered by the ball:</u>

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

<u>Distance covered:</u>

s = ut + 1/2 (at²)                            [seconds equation of motion]

s = ut                                            [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

5 0
3 years ago
You are observing traffic in a single lane of a highway at a specific location. You measure the average headway and average spac
HACTEHA [7]

Answers:

1) flow of traffic =   1198.8 veh/h

2) average speed = 34.09 mi/h

3)density of traffic = 34.34 veh/mi

Explanation:

1) to find flow of traffic

we use the relation

q=1/h

where q is the traffic flow and h is average time headway.

h= 3s       (given)

insert the value

q=1/3=0.333veh/s=1198.8veh/h

2) to find average traffic speed

use the relation

u=S/h

where u is average speed S is average spacing and h average time

S= 150 ft   (given)

so inserting the values

u= 150/3*3600/5280=34.09 mi/h

3) density of traffic

K=q/u

where K is density of traffic q is flow of traffic and u is average speed

inserting values from above solved parts

K=1198.8/34.09=34.34 veh/mi

6 0
3 years ago
An object is dropped from an airplane and takes 72 seconds to hit the ground. What is the magnitude of its VF Y ?
Dafna1 [17]

The final vertical velocity of the object is 705.6 m

Explanation:

Here we have to analyze the vertical motion of the object, which is a free fall motion, with constant acceleration of

g=9.8 m/s^2

towards the ground due to gravity.

Therefore, we can use the following suvat equation:

v=u+at

where

v is the final vertical velocity

u is the initial vertical velocity

a=g is the acceleration

t is the time

Since the object is dropped,

u = 0

Therefore, the magnitude of the vertical velocity of the object after t = 72 seconds is:

v=0+(9.8)(72)=705.6 m/s

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

3 0
3 years ago
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