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2 years ago

Wind and moving water provide ______________________ energy.

Physics

1 answer:

Volgvan2 years ago

5 0

None of the above, it should be kinetic.

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***PLEASE HELP WITH ANSWER AND EXPLANATION: Imagine the current in a current-carrying wire is flowing into the screen. What is t

aalyn [17]

Magnetic field direction is given by right hand thumb rule.

If we put our thumb in the direction of current then curl of fingers will show the magnetic field direction around the wire.

Now here since current is going into the screen so we will put our thumb into the screen and then the curl of fingers is clockwise around it.

The magnetic field is clockwise.

So this would be the direction of magnetic field

5 0

2 years ago

Read 2 more answers

If a car is moving on a straight line with a velocity of 40 m/s and it changes its velocity to 60 m/s in 4 seconds, calculate it

Bogdan [553]

Answer:

5m/s²

Explanation:

Initial Velocity= 40m/s

Final Velocity=60m/s

Time = 4s

a=(v-u)/t

a=60-40/4

a=20/4

a=5m/s²

8 0

2 years ago

3. Using the F, m, a triangle, calculate the boy's mass. Use the "force

aliya0001 [1]

Answer: 30kg

Explanation:

F = 300N

m = ?

a = 10 m/s/s

m = f/a = 300N/10 m/s/s

m = 30kg

5 0

2 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

2 years ago

Read 2 more answers

If a specimen was being viewed using a 20x objective lens and 10x ocular lens, what would be the total magnification

Paraphin [41]

Answer:

As Per Given Information

20x objective lens was used by specimen

10x ocular lens was also used by him.

we have to find the total magnification.

For calculating the total magnification we 'll simply do multiplication

Total Magnification = 20x × 10x

Total Magnification = 200x

So , the total magnification will be 200x .

6 0

1 year ago