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sertanlavr [38]
3 years ago
8

if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use

w=mgh since work done and mass of a person is same everywhere​
Physics
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

h_{moon} = 11.8\ m

Explanation:

Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:

W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} =  11.8\ m

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2 years ago
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

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3 years ago
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Explanation:

5 0
3 years ago
A 6.8 kg bowling ball and 7.4 kg bowling ball rest on a rack 0.74 m apart. What is the force of gravity pulling each ball toward
zimovet [89]

The gravitational force between the two balls is 6.13\cdot 10^{-9} N

Explanation:

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For the balls in this problem,  we have

m_1 = 6.8 kg

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Substituting into the equation, we find the gravitational force between the two balls:

F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0
3 years ago
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