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algol [13]
2 years ago
14

A drone is flying horizontally when it runs out of battery and begins to free fall from 20m. No drag. If it lands 60m away (in t

he x-direction) from where it began to fall, what was it's horizontal velocity while it fell
Physics
1 answer:
rewona [7]2 years ago
3 0

Answer:

Explanation:

Remark

At the time it takes to drop 20 m is the same time it takes to travel 60 m horizontally.

Givens

h = 20 m

hd = 60 m

g = 9.81

vi = 0

Formula

d = vi*t + 1/2 a * t^2                  We are solving for t

Solution

When the battery fails, the vertical initial velocity is 0. So we have to find the time it would take to drop 20 meters

d = 0*t + 1/2 * 9.81 a* t^2

20 = 4.91 * t^2                          Divide by 4.91

20/4.91 = 4.91 t^2 / 4.91    

4.073 = t^2                              Take the square root of both sides.

t = 2.02 seconds

Horizontal

d = 60 m

t = 2.02 seconds

v = ?

Note: there is no horizontal deceleration or acceleration

v = d/t

v = 60/2.02

Answer: v = 29.73 m/s

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Answer:

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Squids are the fastest marine invertebrates, using a powerful set of muscles to take in and then eject water in a form of jet pr
tatuchka [14]

Answer:

0.25 m/s

Explanation:

This problem can be solved by using the law of conservation of momentum - the total momentum of the squid-water system must be conserved.

Initially, the squid and the water are at rest, so the total momentum is zero:

p_i = 0

After the squid ejects the water, the total momentum is

p_f = m_s v_s + m_w v_w

where

m_s = 1.60 kg is the mass of the squid

v_s is the velocity of the squid

m_2 = 0.115 kg is the mass of the water

v_w = 3.50 m/s is the velocity of the water

Due to the conservation of momentum,

p_i = p_f

so

0=m_s v_s + m_w v_w

so we can find the final velocity of the squid:

v_s = -\frac{m_w v_w}{m_s}=-\frac{(0.115 kg)(3.50 m/s)}{1.60 kg}=-0.25 m/s

and the negative sign means the direction is opposite to that of the water.

8 0
3 years ago
A snowstorm was predicted in Chicago. Identify the possible upper air temperature, surface temperature, and air pressure of Chic
OverLord2011 [107]
The correct answer for this question is this one:

<span>A snowstorm was predicted in Chicago. The possible upper air temperature, surface temperature, and air pressure of Chicago on that day. Normal atmospheric pressure is 29.9 inches of mercury.  </span><em>I'm pretty sure the answer is 40 for upper air, 29 for surface temp, and 30 for air pressure. </em>Hope this helps answer your question and have a nice day ahead.

7 0
3 years ago
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A proton of mass 1.67×10-27 kg is being accelerated along a straight line at 7.01×1015 m/s2 in an accelerator. If the proton has
Masja [62]

Answer:

v_f=7*10^7\frac{m}{s}

Explanation:

The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

v_f^2=v_0^2+2ax\\v_f=\sqrt{v_0^2+2ax}\\v_f=\sqrt{(6.63*10^7\frac{m}{s})^2+2(7.01*10^{15}\frac{m}{s^2})(3.63*10^{-2}m)}\\v_f=7*10^7\frac{m}{s}

5 0
4 years ago
Earth-orbiting astronauts feel weightless in space because _____. Choose all that apply. 1 point They are in free-fall motion. T
STALIN [3.7K]

Answer:

They are in free-fall motion.

Explanation:

The Earth orbiting astronauts are falling at an acceleration that is the same or greater than the acceleration due to gravity i.e., 9.81 m/s². If you are continuously falling at this rate then you will feel weightless.

This same effect is felt while going down in an elevator. When you down in an elevator you feel that you are lighter and feel that something is pushing you up. Earth-orbiting astronauts feel the same effect but the accelration is greater hence they feel weightless.

5 0
3 years ago
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