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tangare [24]
3 years ago
11

How much gravitational potential energy does a 1.0 kg hammer have on a shelf 1.5 m above the ground?

Physics
1 answer:
jok3333 [9.3K]3 years ago
8 0

Answer:

14.7 J

Explanation:

PE=MGH

PE= 1.0 x 9.8 x 1.5 = 14.7 J

You might be interested in
Problems with solar energy include _____.
german
First choice:  the inability of current technology to capture
large amounts of the Sun's energy

Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low.  But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.

Second choice:  the inability of current technology to store
captured solar energy

No.  We're pretty good at building batteries to store small amounts, or raising water to store large amounts.  Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.

Third choice:  inconsistencies in the availability of the resource

I think this is it.  If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
of luck during long periods of overcast skies.

Fourth choice:  lack of demand for solar energy

If there is a lack of demand, it's purely a result of willful manipulation
of the market by those whose interests are hurt by solar energy.
4 0
3 years ago
Read 2 more answers
How does increasing the distance between charged objects affect the electric force between them? the electric force increases be
konstantin123 [22]

the electric force decreases because the distance has an indirect relationship to the force

Explanation:

The electric force between two objects is given by

F=k \frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the two objects

As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.

3 0
3 years ago
Read 2 more answers
Describe the motion of a swing that requires 6 seconds to complete one cycle. What is its period and the frequency? Round to the
shutvik [7]

Period = 6 seconds and frequency = 0.167Hz .

<u>Explanation:</u>

We have , the motion of a swing that requires 6 seconds to complete one cycle. Period is the amount of time needed to complete one oscillation . And in question it's given that 6 seconds is needed to complete one cycle. Hence ,Period of the motion of a swing is 6 seconds . Frequency is the number of vibrations produced per second and is calculated with the formula of  \frac{1}{t} . SI unit of frequency is Hertz or Hz. We know that time period is 6 seconds so frequency =   \frac{1}{t}

⇒ frequency = \frac{1}{time}

⇒ frequency = \frac{1}{6}

⇒ frequency = 0.167Hz

Therefore , Period = 6 seconds and frequency = 0.167Hz .

7 0
3 years ago
Where would a car traveling on a roller coaster have the most kinetic energy ? and why?
goldenfox [79]

Answer:

As the car travels up the coaster it is gaining potential energy.

Explanation:

Because It has the greatest in amount of potential energy at the top of the coaster. when the car travels down the roller coaster it obtains speed and kinetic energy.

4 0
3 years ago
Read 2 more answers
A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an
Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V

ΔV=0.484mV  

3 0
3 years ago
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