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3 years ago

In the figure above, a box of mass 2.5kg is initially at rest on a rough inclined plane. Assuming the

Physics

1 answer:

neonofarm [45]3 years ago

5 0

Answer:

Explanation:

By work - energy theorem on the box

Wmg + wn + wf = kf - ki

=> mgh + 0 + wf = 1/2 mv^2f - 1/2mv^2i

=> wf = 1/2mv^2f - 1/2mv^2i - mgh

:. Wf = 1/2×2.5×4^2 - 0- 2.5×9.8×1.5

= 20 - 36.75

= -16.75 J

Work done comes out negative here which is expected as the friction always act opposite to the velocity of an object. Friction is a non conservative force and it causes loss of mechanical energy.

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If the current through a 20-ω resistor is 8.0 a , how much energy is dissipated by the resistor in 1.0 h ?

fiasKO [112]

1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J

4 0

3 years ago

A typical meteor that hits the earth's upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at

attashe74 [19]

Answer:

Answer:u=66.67 m/s

Explanation:

Given

mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg

velocity of meteor v=40km/s \approx 40000 m/s

Kinetic Energy of Meteor

K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}

K.E.=2\times 10^6 J

Kinetic Energy of Car

=\frac{1}{2}\times Mu^2

=\frac{1}{2}\times 900\times u^2

\frac{1}{2}\times 900\times u^2=2\times 10^6

900\times u^2=4\times 10^6

u^2=\frac{4}{9}\times 10^4

u=\frac{2}{3}\times 10^2

u=66.67 m/s

8 0

3 years ago

If an object is projected upward from ground level with an initial velocity of 144144 ft per sec, then its height in feet after

Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

$v_0 = 144 ft/s$

The equation that describes its height at time t is

$s(t) = -16t^2 + 144 t$ (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

$v(t) = s'(t) = -32t +144$ (2)