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Sergio [31]
3 years ago
5

57. Estimate Potential Energy A boulder with a

Physics
1 answer:
Oksanka [162]3 years ago
3 0

Answer: 4.9 x 10^6 joules

Explanation:

Given that:

mass of boulder (m) = 2,500 kg

Height of ledge above canyon floor (h) = 200 m

Gravita-tional potential energy of the boulder (GPE) = ?

Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).

GPE = mgh

GPE = 2500kg x 9.8m/s2 x 200m

GPE = 4900000J

Place result in standard form

GPE = 4.9 x 10^6J

Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules

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A motor exerts a force of 12,00N to lift an elevator 8.0m in 7.0secs what is the power produced by the motor?
SashulF [63]

Answer:

1371.4watt

Explanation:

from power=energy/time

BUT energy=force times distance

6 0
3 years ago
Which vector should be negative?
pentagon [3]

Answer:

 The velocity of a falling object

Explanation:

  The positive X axis is towards right and positive Y axis is towards up, so North direction is positive

  A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.

  Any vector whose magnitude is greater than 1 is never be a negative vector.

  The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.

7 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
3 years ago
Power is measured in unit of Joules per second or
Rasek [7]
D watts yw .yea yww
5 0
3 years ago
What is the change in internal energy if 60 J of heat is released from a system and 30 J of work is done on the system? Use U =
k0ka [10]

The change in internal energy of the system is +30 J

Explanation:

We can solve this problem by using the first law of thermodynamics, which states that the change in internal energy of a system is given by the equation:

\Delta U = Q -W

where

\Delta U is the change in internal energy

Q is the heat absorbed by the system (positive if it is absorbed, negative if it is released)

W is the work done by the system (positive if it is done by the system, negative if it is done by the surroundings on the system)

Therefore, in this problem, we have

Q=-60 J (heat released by the system)

W=-30 J (work done on the system)

Therefore, the change in internal energy is

\Delta U = -60 - (-30) = +30 J

Learn more about thermodynamics:

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brainly.com/question/3564634

#LearnwithBrainly

8 0
3 years ago
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