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wolverine [178]
3 years ago
11

I have 2 questions

Physics
1 answer:
fenix001 [56]3 years ago
8 0
1. your ear drums feel like they pop when you go deep enough under water. kinda feels like bricks. and it hurts if you stay under to long.

2. this could be a reaction from fluid pressure yes. can could be avoided by getting wax ear plugs however even with ear plugs going deep enough under water can still make your ears "pop" due to pressure but at high surfaces it would avoid that.

Please rate me brainliest?

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A person sitting in a parked car hears an approaching ambulance siren at a frequency f1. as it passes him and moves away, he hea
natali 33 [55]

Answer:E

Explanation:

This can be explained by Doppler effect which gives the relation between apparent Frequency and actual frequency when the source of sound is moving

f'=f\frac{v+v_o}{v-v_s}

where f'=apparent\ frequency

f=actual\ frequency

v=velocity\ of\ sound

v_o=velocity\ of\ observer

v_s=velocity\ of\ source

here v_o =0 as observer is standing

when Ambulance is approaching then velocity of sound and velocity of ambulance have relative velocity thus denominator is less than and apparent frequency is more while when ambulance is going away then velocity of sound waves and velocity of observer is in opposite direction thus denominator is less than 1 and apparent frequency is less.

Thus f_2

6 0
4 years ago
Sort the characteristics of solids liquids and gases into the correct colums
boyakko [2]

Solids have molecules that move slowly and are close together and are very attracted to eachother.

Liquids have molecules that move freely and are slightly attracted to eachother.

Gases have molecules that move, but aren't attracted to eachother.

8 0
3 years ago
How can tidal force from the moon affect our earth?​
VARVARA [1.3K]

Answer:

sorry but I can understand the question

5 0
3 years ago
Read 2 more answers
Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring c
klemol [59]

Answer:

v_0=17.3m/s

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2

Since we only care about the velocity v_0, we can keep only the second and third parts of the equation and solve:

mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s

So, the speed of the ball just after the launch is 17.3m/s.

4 0
3 years ago
What is the mean of the data set 3cm, 7cm, 7cm, 7cm, 9cm, 11cm, 12cm
zzz [600]
The correct answer for the mean would be 8

I hope that this helps !
6 0
4 years ago
Read 2 more answers
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