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Karo-lina-s [1.5K]
3 years ago
10

In which of the following cases would you feel weightless?

Physics
2 answers:
IrinaVladis [17]3 years ago
8 0

Answer:

A

Explanation:

Because of the absence of gravitational force in outer space.

Dafna11 [192]3 years ago
6 0
Falling from an airplane.
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When responding to sound, the human eardrum vibrates about its equilibrium position. suppose an eardrum is vibrating with an amp
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745.92 hz (b) 
this is hte answer only because i seen it on the question lol
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3 years ago
Which layer of the sun is responsible for producing the light shown in the picture above?
Eva8 [605]

Answer:

C. Photosphere

Explanation:

The lights shown in the figure comes from the outermost layer of the Sun. This layer is called photosphere.

This is the layer from where the light of the Sun is radiated, before travelling through space and reaching us.

The photosphere is the coldest layer of the Sun: its surface temperature is between 4500 and 6000 K. Its width is approximately 100 km.

A characteristic of the photosphere is the presence of the sunspots, which appear as darker spots, and are regions of lower temperature caused by a concentration of magnetic flux.

3 0
2 years ago
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For the sun, which is the final step in core fusion?
Marina86 [1]
The answer would be stage of<span> conversion of Hydrogen to Helium.
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3 0
3 years ago
If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. Th
jonny [76]

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency  

6 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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