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Setler79 [48]
3 years ago
5

A 170-lb man carries a 10-lb can of paint up a helical staircase that encircles a silo with radius 15 ft. if the silo is 40 ft h

igh and the man makes exactly two complete revolutions, how much work is done by the man against gravity in climbing to the top? ft-lbs
Physics
1 answer:
goldenfox [79]3 years ago
4 0
Work done against the gravity = Total weight*Vertical distance covered.

Total weight = 170+10 = 180 lb
Vertical distance covered = 40 ft

Therefore,

Work done = 180*40 = 7200 lb-ft
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If the radius of an object in circular motion is doubled, what change will occur in the centripetal force?
cestrela7 [59]

Answer:

The centripetal force will be 1/2 as big as it was. (option c)

Explanation:

Recall that centripetal force (F_c) is defined as: F_c=m\,* \frac{v^2}{r} where "v" is the tangential velocity of the object in circular motion, "r" is the radius of rotation and "m" is the object's mass.

So if we start with such formula with a given mass, radius, and tangential velocity, and then we move to a situation where everything stays the same except for the radius which doubles, then the new centripetal force (F'_c) will be given by: F'_c=m\,* \frac{v^2}{2r}

and this is half (1/2) of the original force:

F'_c=m\,* \frac{v^2}{2r}\\F'_c=m\,* \frac{v^2}{r}*\frac{1}{2} \\F'_c=F_c\,*\,\frac{1}{2}

which is expressed by option "c" of the provided list.

8 0
3 years ago
Read 2 more answers
The moon has a mass of 7.35x1022 kg and is a lot farther away than is shown in textbooks. The mass of the
anyanavicka [17]

Answer:

F=1.98\times 10^{20}\ N

Explanation:

Given that,

The mass of a Moon, M_m=7.35\times 10^{22}\ kg

The mass of the Earth, M_e=5.98\times 10^{24}\ kg

The moon's mean orbit distance around the earth is, r=3.84\times 10^8\ m

We need to find the gravitational force exerted on the moon by the Earth.

The formula of gravitational force is given by :

F=G\dfrac{M_mM_e}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.35\times 10^{22}\times 5.98\times 10^{24}}{(3.84\times 10^8)^2}\\\\F=1.98\times 10^{20}\ N

So, the required force is 1.98\times 10^{20}\ N.

3 0
3 years ago
oxygen (o) is a gas found in the 16th column of the periodic table. What statement is true about oxygen and he other elements in
Tpy6a [65]
<h3>I'm pretty sure, all the elements in the column have similar chemical properties.</h3>
8 0
4 years ago
An open bed truck is moving at 20 m/s and comes to a stop. A 100 kg crate sitting in the back of the truck remains at rest in th
zzz [600]

The work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

<h3>Work done by friction to prevent the crate from moving</h3>

The work done is determined by applying the principle of conservation of energy.

W = K.E

W = ¹/₂mv²

where;

  • m is mass of crate
  • v is speed of the crate with respect to the truck = 20 m/s

W = ¹/₂(100)(20²)

W = 20,000 J

Thus, the work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

3 0
2 years ago
Newton's laws of motion work well for ordinary situations on earth. However, these laws of motion do not work for all cases. Und
soldier1979 [14.2K]
I believe the answer is C: For objects at extremely fast speeds.
Hope this helps!
5 0
3 years ago
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