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Damm [24]
3 years ago
14

Find the weight of a 80 kg man on the surface of the moon? What should be his mass on the moon and the earth? (ge =9.8 m/s2, gm

=1.63 m/s2 )
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

If a man's mass is 80 kg, then ...

-- He weighs 258 newtons on Mars.

-- He weighs 785 newtons (about 176.4 pounds) on Earth.

-- He weighs 130 newtons on the Moon.

-- He weighs different amounts in other different places.

His mass doesn't change.  It's 80 kg no matter where he goes.

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6 0
4 years ago
Read 2 more answers
If the net force on a block is zero
amm1812

If the net force on a block is zero, the block will move at constant velocity

Explanation:

We can answer this question by applying Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:

\sum F = ma (1)

where

\sum F is the net force on the object

m is its mass

a is its acceleration

In this problem, we have a block, and the net force on it is zero:

\sum F = 0

According to eq.(1), this also implies that

a=0

So, the acceleration of the block is zero.

However, acceleration is the rate of change of velocity of a body:

a=\frac{\Delta v}{\Delta t}

where \Delta v is the change in velocity in a time of \Delta t. Since the acceleration is zero, this means that \Delta v=0, and therefore the velocity of the object is constant.

Learn more about Newton's second law:

brainly.com/question/3820012

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8 0
3 years ago
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

\delta = 0.001932m

\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
3 years ago
Need help ASAP please help me please
lbvjy [14]

Answer:

145

Explanation:

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Answer:

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