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salantis [7]
2 years ago
14

NEED HELP!!! ANSWER THESE 4 QUESTIONS FOR 20 POINTS!!!! PLEASE ANSWER!! I WILL GIVE YOU BRAINEST

Physics
1 answer:
MatroZZZ [7]2 years ago
5 0

Answer:

1)A

2)C

3)A

4)C

Pls give brainliest

I ACTUALLY NEED IT!!!

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Differentiate between sound waves and seismic waves?
algol13

The only real difference is that common seismic waves travel through the ground and sound waves travel through the air. If you had a pipe attached to granite and you were listening to it, you might detect both.

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3 years ago
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YO CAN ANYONE DO THE BLANK COLUMN AND THE QUESTION PART RQ PLS!!
VikaD [51]

Answer:

stryo:  1

wood: 1

ice: 1

brick: 2

aluminum: 2.7

Explanation:

d= mass/ total volume

(fyi: for aluminum, they did the subtraction wrong to find the total volume. it is actually 5 or 5.00)

6 0
3 years ago
A stack of 55 business cards is 1.85 cm tall. Use this information to determine the thickness of one business
Sindrei [870]
To find the answer, take 55 and divide it by 1.85 to get the thickness of one card. In this case the answer would be 29.72973 cm. each.
8 0
3 years ago
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A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s
3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer.  The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\  340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

8 0
2 years ago
What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.1 g that has been charged to -
Liula [17]

Answer:

Electric field, E=6.86\times 10^6\ N/C

Explanation:

It is given that,

Mass of sphere, m = 2.1 g = 0.0021 kg

Charge, q=-3\ nC=-3\times 10^{-9}\ C

We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,

ma=qE

a = g

E=\dfrac{mg}{q}

E=\dfrac{0.0021\ kg\times 9.8\ m/s^2}{-3\times 10^{-9}\ C}

E=6860000\ N/C

or

E=6.86\times 10^6\ N/C

Hence, the magnitude of electric field that balances its weight is 6.86\times 10^6\ N/C. Hence, this is the required solution.

4 0
3 years ago
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