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3 years ago

why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level

1 answer:

5 0

Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.

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An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$

Now we can calculate the acceleration of the plane, by using Newton's second law:

$a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2$

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

$v^2- u^2 = 2aS$

where

$v=?$ is the final velocity

$u=66.0 m/s$ is the initial velocity

$a=1.69 m/s^2$ is the acceleration

$S=5.00 \cdot 10^2 m$ is the distance travelled

Solving for v, we find

$v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s$

8 0

3 years ago

Which example best illustrates that light behaves like particles?

AlekseyPX

I would say B. Because actual mass would ricochet off the sidewalk.

8 0

3 years ago

Read 2 more answers

An 80 N force causes a spring to compress 0.15 m. What is the spring constant? What is the potential energy of the spring?

Anni [7]

Force = -kx
80N=0.15m * -k
K=-80/0.15=533.333. Spring constant
Energy=1/2kx^2
1/2*(-80/0.15)*80^2=Energy

3 0

3 years ago

What is the main promblem with survey research?

zlopas [31]

Answer:

A major problem in all survey research is that respondents are almost always self-selected. Not everyone who receives a survey is likely to answer it, no matter how many times they are reminded or what incentives are offered.

Explanation:

<em><u>DISADVANTAGES</u></em>

Respondents may not feel encouraged to provide accurate, honest answers.

Respondents may not feel comfortable providing answers that present themselves in a unfavorable manner.

Respondents may not be fully aware of their reasons for any given answer because of lack of memory on the subject, or even boredom.

HAVE A GOOD DAY!

7 0

3 years ago

With the piston head locked in place, will the volume of the gas increase, decrease, or stay the same whenthe piston is placed a

Nuetrik [128]

Answer:

For real gas the volume of a given mass of gas will increase with increase in temperature.

Explanation:

With the piston head locked in place and place above the fire,the volume of the gas will increase,because the volume of a given mass of gas increases with increase temperature.

6 0

3 years ago