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MakcuM [25]
3 years ago
13

the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i

t, how much gravitational potential energy does the rock have?
Physics
1 answer:
Elanso [62]3 years ago
7 0
Gravitational potential energy can be calculated using the formula:

PE_{grav} =mgh

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know. 
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it. 
PE_{grav} =mgh
PE_{grav} =(10kg)(3.7m/s^{2})(1m)
PE_{grav} =37J


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A force of 4 kg weight acts on a body of mass 9.8 kg calculate the acceleration
White raven [17]

Answer:

here given is a weight

then force becomes mg

that is F=Mg

=4*9.8

then by using the formula

F=Ma

a=F/M

=4*9.8/9.8

=4

Explanation:

3 0
3 years ago
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) It is necessary to maintain a constant fluid volume in the body. Therefore, the body adjusts the water outputs to be equal to
loris [4]

Answer:

Evaporative Water Loss = 2 kg

Explanation:

According to the given condition, the water entering the body must be equal to the water leaving the body. Therefore,

Water Entering the Body = Water Leaving the Body

Feed Water + Drinking Water + Metabolic Water = Urine Water + Evaporative Water Loss

using the given values:

1 kg + 5 kg + 0.5 kg = 4.5 kg + Evaporative Water Loss

Evaporative Water Loss = 1 kg + 5 kg + 0.5 kg - 4.5 kg

<u>Evaporative Water Loss = 2 kg</u>

8 0
2 years ago
After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average
lara [203]

Answer:

compression of spring is x = 0.12 m

Assumed k = 160,000 N/m ........ Truck's suspension system

Explanation:

Given:

- The mass of average person m_p = 69 kg

- Total number of persons n_p = 27

- The mass of each goat m_g = 15 kg

- The total number of goats n_g = 3

- The mass of each chicken m_c = 3 kg

- The total number of goats n_c = 5

- The total mass of bananas m_b = 25 kg

Find:

How much are the springs compressed?

Solution:

- Using equilibrium equation on the taptap in vertical direction:

                                 F_net = F_spring - F_weight = 0

- Compute the force due to all the weights on the taptap:

                                F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b)*9.81

                                F_weight = (69*27 + 3*15 + 5*3 + 25)*9.81  

                                F_weight = 19109.88 N

- The restoring force of a spring is given by:

                                F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

                                 F_weight = F_spring

                                 19109.88 = k*x

                                 x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

                                 x = 19109.88 / 160,000

                                 x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.

8 0
3 years ago
If a certain silver wire has a resistance of 3.00 Ï at 11.0°c, what resistance will it have at 25.0°c?
Romashka [77]
11.30      times the nuber of toys
7 0
3 years ago
The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?
RoseWind [281]

Answer:

E = 2.9775\times10^5 J

Explanation:

Given:  The latent heat of fusion for Aluminum is L = 3.97\times10^5  J/Kg

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J

Therefore, energy required to melt 0.75 Kg aluminum

E = 2.9775\times10^5 J

5 0
3 years ago
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