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3 years ago

How far above Earth's surface do Gamma Waves reach?

Physics

2 answers:

velikii [3]3 years ago

8 0

100 miles from earth

lapo4ka [179]3 years ago

8 0

I am pretty sure around 100 miles from Earth

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A 2.75 kg block is pulled across a flat, frictionless floor with a 5.11 N force directed 53.8° above horizontal. What is the tot

Softa [21]

Answer:

F = 3.01 N

Explanation:

Given that,

Mass of a block, m = 2.75 kg

Force applied to the block, F = 5.11 N

It is directed 53.8° above horizontal.

We need to find the total force acting on the block. The force acting on it is given by :

$F=F\cos\theta\\\\F=5.11\times \cos53.8\\F=3.01\ N$

So, 3.01 N of force is acting on the block.

4 0

3 years ago

What is electricity?

igor_vitrenko [27]

A form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current.

5 0

3 years ago

A drum rotates around its central axis at an angular velocity of 19.8 rad/s. If the drum then slows at a constant rate of 3.02 r

Debora [2.8K]

a) Time taken =6.39 s

b) Angle of rotation while coming to rest = 188.17 rad

<u>Explanation:</u>

Initial angular velocity of the drum $\omega_{0} = 19.8 rad/s^{2}$

Angular acceleration of the drum $\alpha = -3.02 rad/s^{2}$

a.) We know that

$\omega =\omega_{0} + \alpha t$

$0 = 19.8 - 3.02 t$

$t = \frac{19.8}{3.02}$

$t = 6.39s$

b.) we know that

$\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}$

$\theta = 19.8 \times 6.39 + \frac{1}{2} \times 3.02 \times 6.39^{2}$

$\theta =188.17 rad$

3 0

4 years ago

Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the video, the mass of t

dimaraw [331]

Answer:

b) one-third as great.

Explanation:

As we know that same heat is supplied in this experiment

so we will have

$Q = ms\Delta T$

now we know that both are initially at same temperature

then their final temperatures are

$T_1 = 40 degree$

$T_2 = 80 degree$

now we have

$m_1s_1 (40 - 20) = m_2s_2(80 - 20)$

so we have

$m_2s_2 = \frac{20}{60} m_1s_1$

so heat capacity of mystery metal is 1/3 times that of water

6 0

3 years ago

Read 2 more answers

All digits shown on the measuring device, plus one estimated digit are considered____

erastovalidia [21]

I think the answer is significant

4 0

3 years ago