Ask question

Ask question

All categories

3 years ago

8

1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts

1 answer:

SashulF [63]3 years ago

5 0

Answer:

by using it's buoyant or floating effect by Archimedes.

the buoyant force act on the astronauts body and make he/ she feels like in low gravity.

the buoyant force equation is

F = Density of liquid x earth gravitational field x volume of astronauts body and suit.

the Weight of astronauts in the pools will be less than in the land or air.

Weight in water = weight in air/land - buoyant force

so the astronauts will feel like in the outer space with low gravity.

You might be interested in

Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running

tangare [24]

Answer:

Since both start with the same vertical velocity from the same position with the same acceleration they will reach the lake at the same time.

6 0

3 years ago

What is the magnitude of the acceleration of an electron at a point where the electric field has magnitude 6377 n/c and is direc

shusha [124]

Use the magnitude acceleration formula .

4 0

4 years ago

Write a question about how changing temperature affects gas

Lera25 [3.4K]

Answer:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

Explanation:

One possible question can be:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

The answer to this question is contained in Charle's law, which states that for a gas at constant pressure, the volume of the gas is proportional to its absolute temperature:

$V\propto T$

Or also written as

$\frac{V}{T}=const.$

By looking at this equation, we can find immediately the answer to our question: as the (absolute) temperature of the gas increases, the volume increases as well, by the same proportion.

3 0

3 years ago

An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t

lana [24]

Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

$q_{h}$ = R* $T_{h}$ in( $V_{2}$ / $V_{1}$ )

$=0.287 * 1200 ln(3)$

$=0.287*1318.33$

$=378.36kJ/kg$

The specific heat capacity:

$q_{L}$ =q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

$n_{TH}$ =1 - ( $T_{L}$ / $T_{H}$ )

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

4 0

3 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago