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olasank [31]
3 years ago
11

If you climb to the top of Mt. Everest, you will be 8850 m (about 5.50 mi) above sea level.

Physics
1 answer:
Umnica [9.8K]3 years ago
5 0

Answer:

9.773m/s2

Explanation:

Given,

h=8848m

The value of sea level is 9.08m/s2

So,

Let g′ be the acceleration due to the gravity on the Mount Everest.

g′=g(1−h2h)

=9.8(1−640000017696)

=9.8(1−0.00276)

9.8×0.99724

=9.773m/s2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s2

Hope it helped!!!

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Jupiter is made of gas(like Saturn, Uranus and Neptune). What would happen to the strength of gravity if you
garik1379 [7]

Answer:

a) The strength of gravity decreases if one moved away from Jupiter

b) The strength of gravity increases if one fell into Jupiter

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Force \ (strength) \ of \ gravity = \dfrac{G \times M \times m}{R^2}

Where;

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7 0
3 years ago
Scenario
zepelin [54]

Answer:

 t = 23.255 s,   x = 2298.98 m,    v_y = - 227.90 m / s

Explanation:

After reading your extensive writing, we are going to solve the approach.

The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.

As there is a mixture of units in different systems we are going to reduce everything to the SI system.

         v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s

         y₀ = 2650 m

Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement

Y axis  

       y = y₀ + v₀ t - ½ g t²

the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero

       0 = y₀ + 0 - ½ g t²

       t = \sqrt{  \frac{2 y_o}{g} }

       t = √(2 2650/ 9.8)

       t = 23.255 s

Therefore, for the cargo to reach the desired point, it must be launched from a distance of

       x = v₀ₓ t

       x = 111.76 23.255

       x = 2298.98 m

at the point and arrival the speed is

        vₓ = v₀ₓ = 111.76

     

vertical speed is

         v_y = v_{oy} - gt

          v_y = 0 - gt

          v_y = - 9.8 23.25 555

         v_y = - 227.90 m / s

the negative sign indicates that the speed is down

in the attachment we have a diagram of the movement

7 0
3 years ago
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