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3 years ago

A 5 kg rock is raised 28 m above the ground level. What is the change in its potential energy?

1 answer:

3 0

Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
=mg(h-y)
=50(28-0)
=1400 J

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3 years ago

Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the tot

guapka [62]

Answer:

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

$E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}$

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

$R = \frac{d}{\sqrt2}$

now we have

$E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}$

so the force on the charge is given as

$F = QE$

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

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3 years ago

A particle has 37.5 J of kinetic energy and 12.5 J of gravitational potential energy at one point during its fall from a tree to

Varvara68 [4.7K]

Answer: 50J

Explanation:

Mechanical energy follows the same principles of kinetic energy and potential energy, it is conserved. So Ei = Ef.

Mechanical energy is the sum of ALL energy's. There is no friction, so its just kinetic plus potential.

37.5 + 12.5 = 50J

Since the particle has not touched the ground, it has not transferred any energy to the ground yet, therefore the mechanical energy must still be 50J; mostly in kinetic energy with a very small amount of potential because of the low height relative to the ground.

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3 years ago

A ‘thermal tap’ used in a certain apparatus consists of a silica rod which

abruzzese [7]

Correct question is;

A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).

Answer:

ΔT = 268.67K

Explanation:

We are given;

d1 = 8mm

d2 = 1mm

At standard temperature and pressure conditions, the temperature is 273K.

Thus; Initial temperature; T1 = 273K,

Using the combined gas law, we have;

P1×V1/T1 = P2×V2/T2

The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:

V1/T1 = V2/T2

Now, volume of the tube is given by the formula;V = Area × height = Ah

Thus;

V1 = (πd1²/4)h

V2 = (π(d2)²/4)h

Thus;

(πd1²/4)h/T1 = (π(d2)²/4)h/T2

π, h and 4 will cancel out to give;

d1²/T1 = (d2)²/T2

T2 = ((d2)² × T1)/d1²

T2 = (1² × T1)/8²

T2 = 273/64

T2 = 4.23K

Therefore, Change in temperature is; ΔT = T2 - T1

ΔT = 273 - 4.23

ΔT = 268.67K

Thus, the temperature decreased to 268.67K

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3 years ago

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zalisa [80]

One plants don't orbit the sun, the earth does.
two true if its saying plants orbit the sun

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3 years ago

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