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3 years ago

Please help although I think I have the awnser but correct me if I’m wrong

Physics

1 answer:

iren2701 [21]3 years ago

7 0

I believe you got it correct already

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Match the glacier feature with its description.

scZoUnD [109]

Answer:

kettles: holes left by glaciers.

cirques: three-sided valleys

erratics: large, out-of-place rocks bouldersleft by glaciers.

drumlins: egg-shaped hills

Explanation: APEX

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3 years ago

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PLEASE HELP WILL GIVE BRAINLIEST FILE IS ATTACHED

Ainat [17]

Answer:

chemical, electrical

mechanical, electrical

heat, light

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3 years ago

A wave with a height of 6 would have greater amplitude than a wave with a height of 5 true or false? HELP

BabaBlast [244]

Answer:

it could be either or because it doesnt just depend on the height but it also depends on the pressure but then again the question didnt ask anything about the pressure so the answer should be true

Explanation:

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3 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.

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3 years ago

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A car engine produces a force of 2000N while moving a distance of 200m in a time of 10s. What is the power developed by the engi

andreyandreev [35.5K]

Power = work/time
Work = force * distance

(2000 N * 200 m) / 10 s = 40,000 Joules

3 0

3 years ago