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3 years ago

A man weighs 90 N on the surface of earth of radius R at what height above the surface of earth is weight will be 40 N.

Physics

1 answer:

lapo4ka [179]3 years ago

4 0

Answer:

3/2 R

Explanation:

Let the Mass of man be m, Mass of earth be M.

By Newtons, law of gravitation,

F = GMm/R^2, Where G is gravitational constant

90 = GMm/R^2

When man weighs 40N,

GMm/(New R)^2 = 40

= 4/9 (90)

GMm/(New R)^2 = 4/9 (GMm/R^2)

(New R)^2 = 9/4 R^2

New R = 3/2 R

Therefore, height is 3/2 R

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A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

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What's the time it takes a car to attain the speed of 30 m/s when accelerating from rest at 2 m/s (squared)

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A jet has amass of 40,000 kg. The thrust for each of the four engines is 20,000 N. What is the jet's acceleration?

BartSMP [9]

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