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kati45 [8]
3 years ago
11

The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. If a rock weighs 75.0 N on earth, compu

te the acceleration due to gravity on the surface of Venus in comparison to the earth's gravitational acceleration
Physics
1 answer:
IgorLugansk [536]3 years ago
7 0

Answer:

g = 0.905 gE

W  = 67.9 N

Explanation:

given data

mass of Venus mv =  81.5% = 0.815

radius Rv = 94.9% = 0.949

weighs W = 75.0 N

solution

we apply here acceleration due to gravity at earth surface that is

g = \frac{Gm}{R^2}   = 9.80 m/s²  ............1

so

g = \frac{G(0.815)}{0.949R^2}  

g = 0.905 gE

and

W = m gv

W = 0.905 m gE

W  = 0.905 × 75

W  = 67.9 N

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Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

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7 0
3 years ago
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A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

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          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

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Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

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The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

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          y = y₃ + y₂

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b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

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The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

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        t = t_{s} + t_{b}

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        t =63.84 s

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suter [353]

Answer:

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