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marusya05 [52]
3 years ago
13

Dams hold collected water back from flowing at their normal rate. What type of energy is in the collected water above the dam? c

hemical energy electrical energy potential energy kinetic energy
Physics
2 answers:
Andreyy893 years ago
8 0
It would be mechanical potential energy.
forsale [732]3 years ago
3 0
Potential energy because the moment the dam breaks, a lot of water will burst out. Obviously this isn't chemical energy because there is not chemical reaction taking place. It's not electrical energy because there is no electricity at the dam. This leaves us with potential and kinetic energy. A key word in here is holds back, aka stored water. Storing almost always leads to potential energy.
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Answer:

the answer to your question is 4 cm long

Explanation:

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A straight, vertical wire carries a current of 2.15 A downward in a region between the poles of a large superconducting electrom
enot [183]

Answer:

a) F = 0.01234N to the south direction

b) F = 0.01234N to the north direction

c) F = 0.01234N

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Explanation:

The magnetic force in a magnetic field is given by:

F = BIlsin \theta

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Current flowing through the wire, I = 2.15 A

a) If the magnetic field direction is east

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b) If the magnetic field direction is south

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That is , F = BIL sin 90

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c) If the magnetic field direction is 30 degrees south

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Therefore the magnitude of the force still remains 0.01234

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8 0
3 years ago
Why do you want the water to drip off the metal before it is placed in the calorimeter?
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Answer:

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An open beaker of pure water has a water potential of.
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A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
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Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

6 0
3 years ago
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