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3 years ago

The part of the Electromagnetic (Light) Spectrum used to see through the dust

Physics

2 answers:

irakobra [83]3 years ago

5 0

The answer is A. Not C. C is what you use to remotely control something

dimaraw [331]3 years ago

4 0

Answer:

A or C

Explanation:

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A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.

Len [333]

Explanation:

Given

Velocity v = 23.0m/s

Distance S = 3.45m

Required

Time it will take the skier to reach the ground;

Using the equation of motion;

S = ut + 1/2gt²

3.45 = 23t + 1/2(9.8)t²

3.45 = 23t + 4.9t²

4.9t²+23t-3.45 = 0

Factorize;

t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)

t = -23 ±√529+67.62/9.8

t = -23±√596.62/9.8

t = -23±24.43/9.8

t = 1.43/9.8

t = 0.146 secs

Hence take the skier 0.146 secs to reach the ground.

b) Horizontal distance covered is the range;

Range = U√2H/g

Range = 23√2(3.45)/9.8

Range = 23√6.9/9.8

Range = 23√0.7041

Range = 23(0.8391)

Range = 19.29m

Hence the horizontal distance travelled in air is 19.29m

7 0

2 years ago

A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.

Luba_88 [7]

Answer: Option (b) is the correct answer.

Explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

F = $\frac{kqQ}{r^{2}} - kq\frac{(2Q)}{(2r)^{2}}$

= $\frac{4KqQ - 2KqQ}{4r^{2}}$

= $\frac{KqQ}{2r^{2}}$

F = $\frac{KqQ}{2r^{2}}$ > 0

Thus, we can conclude that the test charge move to the right immediately after being released.

7 0

3 years ago

A book falls off a shelf that is 10.0 m tall. What is the velocity at which the book hits the ground?

Elena L [17]

Answer:

14 m/s

Explanation:

The motion of the book is a free fall motion, so it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. Therefore we can find the final velocity by using the equation:

$v^2 = u^2 + 2gd$