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Kipish [7]
2 years ago
11

Physics question help

Physics
1 answer:
S_A_V [24]2 years ago
8 0

Answer:

not know sorry sorry

Explanation:

sorry sorry

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Which of the following statements is true of electromagnetic waves but NOT mechanical waves.
Neko [114]

Answer:

a

Explanation:

what the heck is a medium

4 0
2 years ago
A VW Beetle goes from 0 to 60.0 mi/h with an acceleration of +2.35 m/s^2. (a) How much time does it take for the beetle to reach
Firlakuza [10]
Im going to tell you what to do but not the result. So pay close attention: the first thing you need to do is convert miles/h to m/s. Then for the part a) <span>divide the final velocity by the initial velocity. That will give you the amount of it will take to accelerate to the final velocity.Now for the part b you </span>use the formula v=vo+at. I hope this can help you
8 0
3 years ago
There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat
Mamont248 [21]

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator =heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec

(b) Total heat absorbed =140 j/sec =140\times 3600=504000j/hour

Let the temperature be \Delta T

Heat absorbed per hour =504000 [tex]=400\times 10^3\times \Delta T

So  \Delta T=\frac{504000}{400000}=1.26K

8 0
3 years ago
What happens when an atom gives up loosely held valence electrons to another atom​
vazorg [7]

Answer:

The number of valence electrons increases to 8 or, the atom gives up loosely held valence electrons.

Explanation:

Please mark brainliest and have a great day!

3 0
3 years ago
A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from
o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}

There is NO initial vertical velocity, so:

\large\boxed{\Delta d= \frac{1}{2}at^2}}

Rearrange to solve for time:

2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}

6 0
3 years ago
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