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a_sh-v [17]
3 years ago
12

What is air pressure?

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0
Air Pressure is a force that is exerted onto your body by air molecules.
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Why do the passengers on a high-flying airplane not appear weightless, similar to the astronauts on the space station?
sergey [27]

<span>Even in space, there is still presence of gravity. The cause of weightlessness is not how far above the earth the space shuttle is but rather how fast it is travelling. The shuttle is in free fall causing weightlessness, but it is travelling fast enough to miss the earth as it falls. Similarly, the airplane could also provide weightlessness if it went free fall as well. However, that ends as the plane hits the ground. </span>

4 0
3 years ago
When hiking at altitude, the air pressure drops to 825,000Pa. Solve the force that site exerts against the side of a rectangular
OLga [1]

Answer:

The force is 1237500N

Explanation:

We can apply the following expression

pressure=\frac{Force}{Area}

To find the area of rectangular tent wall:

Area=length *width\\Area=2.00m*0.750m\\Area=1.5m^{2}

Now to find the force:

pressure=825,000pa

pressure=\frac{Force}{Area}

Force=pressure*Area\\Force=825000pa*1.5m^{2} \\Force=1237500N

7 0
3 years ago
g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron
Butoxors [25]

Answer:

The value is v  =  2.999 *10^{8} \  m/s

Explanation:

From the question we are told that

   The time taken to travel to the planet from earth is t = 189 \ light-years

    The  time to be spent on the ship is  t_{s} =  12 \  years

Generally speed can be obtained using the mathematical relation represented below

       t_s  =  2 * t *  \sqrt{1 -  \frac{v^2}{c^2 } }

The 2 in the equation show that the trip is a round trip i.e going and coming back

=>    12 =  2 * 189 *  \sqrt{1 -  \frac{v^2}{(3.0*10^{8})^2 } }

=>     v  =  2.999 *10^{8} \  m/s

5 0
3 years ago
A bar on a hinge starts from rest and rotates with an angular acceleration α = 12 + 5t, where α is in rad/s2 and t is in seconds
Rama09 [41]

Answer:

Ф = 239.73 rad

Explanation:

α = 12 + 15×t

W = ∫α×dt

   = ∫(12 + 5×t)×dt

   = 12×t + 2.5×t^2

then:

Ф = ∫W×dt

   = ∫(12×t + 2.5×t^2)dt

   = 6×t^2 + 5/6×t^3

therefore the angle at t = 4.88s is:

Ф = 6×(4.88)^2 + 5/6×(4.88)^3

   = 239.73 rad    

5 0
3 years ago
A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a
sleet_krkn [62]

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N   ⇒   k = (68.6 N) / (0.105 m) ≈ 653 N/m

5 0
2 years ago
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