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2 years ago

1.

Physics

1 answer:

lys-0071 [83]2 years ago

4 0

Answer:

The rider experiences weight or does not experience weight depends upon the direction of the elevator in which it moves and its acceleration.

Explanation:

Consider the mass of a rider as m. So its actual weight is mg and it acts vertically downward. The apparent weight of the rider is depended on the acceleration of the elevator and its direction of movement.

When elevator moves with a constant speed i.e. its acceleration is zero, the apparent weight of the rider is equal to the actual weight. Thus the rider's sensation is normal.

If the elevator is moving upward with an acceleration a, then the apparent weight of the rider will be more and the rider will experience an increase in weight or the sensation is heavy.

Or when the elevator moves downward with an acceleration a, the apparent weight of the rider is less. And the rider sensation is lighter.

You might be interested in

Which is not correct when describing the formation of rainbows?

umka21 [38]

D- All wave lengths refract at the same angle

Explanation- The amount of refraction increases as the wavelength of light decreases. Shorter wavelengths of light are slowed more and consequently experience more bending than the longer wavelengths.

7 0

2 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

Read 2 more answers

When you are on a roller coaster, you are constantly transforming from Potential to Kinetic energy and back. Explain how these e

andreev551 [17]

Answer:

The two types of energy possessed by the roller coaster are:

- Potential energy: it is the energy possessed by the roller coaster due to its position. It is calculated as

$PE=mgh$

where

m is the mass of the roller coaster

g is the acceleration due to gravity

h is the height of the roller coaster relative to the ground

- KInetic energy: it is the energy possessed by the roller coaster due to its motion. It is calculated as

$KE=\frac{1}{2}mv^2$

where

v is the speed of the roller coaster

Moreover, according to the law of conservation of energy, the total mechanical energy of the roller coaster (the sum of potential+kinetic energy) is constant during the motion:

$E=PE+KE=const.$

This implies that:

- When PE increases (because h increases), KE decreases (because v decreases)

- When PE decreases (because h decreases), KE increases (because v increases)

Now we can apply these conclusions to the motion of the roller coaster:

- When it moves from A to B, potential energy is converted into kinetic energy, so PE decreases and KE increases

- When it moves from B to C, kinetic energy is converted into potential energy, so PE increases and KE decreases

- When it moves from C to D, potential energy is converted into kinetic energy, so PE decreases and KE increases

- When it moves from D to E, kinetic energy is converted into potential energy, so PE increases and KE decreases

8 0

3 years ago

Is a burning wax candle nuclear energy?

sasho [114]

no because nuclear energy come from kinetic not potential energy. burning a wax candle is an example of heat/thermal energy .

6 0

3 years ago

A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

By substituting the values, we get

$\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}$

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0

3 years ago