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2 years ago

Sometimes you can see a faint reflection in the surface of a shiny plate or cup.Why ?

Physics

1 answer:

bazaltina [42]2 years ago

4 0

Answer:

The degree of reflection whether faint or bright you see on the surface of an object is an indication that light particles had hit the surface. Since light is a wave and as part of its characteristics can get reflected. However, the amount of light reflected by a surface is dependent on the smoothness of the surface which can be shiny or dull, it can also be dependent on the nature of the surface which can be glass, water, and so on. So, from the question, you can see a faint reflection on the surface of a shiny plate or cup because of the smoothness of the surface which reflects the lights that hit it from a particular direction at the same angle.

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A 4000kg truck has a head-on inelastic collision with a 2500kg truck.

iogann1982 [59]

Answer:it could be B

Explanation:

im not sure

7 0

3 years ago

Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms

lakkis [162]

Answer:

(a) 333.77 J

(b) 237.85 J

(d) 4667.85 J

Explanation:

Temperature of source, TH = 314 K

Temperature of A, Tc = 292 K

Temperature of B, Tc' = 298 K

heat taken out, Qc = 4430 J

Let the heat deposited outside is QH and QH' by A and B respectively.

$\frac{Q_H}{Q_c}=\frac{T_H}{T_c}\\\\Q_H = \frac{4430\times314}{292}=4763.77 J$

Now

$\frac{Q_H'}{Q_c}=\frac{T_H}{T_c'}\\\\Q_H' = \frac{4430\times314}{298}=4667.85 J$

(a) Work done for A

W = QH - QC = 4763.77 - 4430 = 333.77 J

(b) Work done for B

W' = QH' - Qc = 4667.85 - 4430 = 237.85 J

(d) QH' = 4667.85 J

4 0

3 years ago

An automatic switch that opens when the current reaches a set value is called:

marishachu [46]

I believe the correct answer would be letter c) circuit breaker

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3 years ago

A 1459 kg car is traveling WEST at 43 m/s. A 9755 kg truck is traveling EAST at 11 m/s. They collide head-on, and stick together

otez555 [7]

Answer:

<em>Both vehicles move east at 3.97 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is:

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

Assume both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Assuming east direction to be positive, we have an m1=1459 kg car traveling west at v1=-43 m/s. An m2=9755 kg truck is traveling east at v2=11 m/s. They collide head-on and stick together after that.

Computing the resultant velocity after the collision:

$\displaystyle v'=\frac{1459*(-43)+9755*11}{1459+9755}$