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borishaifa [10]
3 years ago
10

Two identical boxes (equal dimensions and mass) are at rest, the first on a 30-degree incline and the second on a 45-degree incl

ine. Which of the following statements is FALSE?
(A) The frictional forces acting on the two boxes have the same magnitude.
(B) The normal contact force acting on the first box is larger than the normal contact force acting on the second
(C) The gravitational forces acting on the two boxes have the same magnitude.
(D) The net forces acting on the boxes have equal magnitude.
Physics
1 answer:
klemol [59]3 years ago
3 0

Answer:

A)

Explanation:

If both boxes are at rest, this means that the friction force must be equal to the component of the gravity force, trying to accelerate the box down the incline.

If the angle that the incline forms with the horizontal is θ, the component of the weight along the incline is as follows:

Fgp = m*g*sinθ

As sin θ is different for the 30-degree incline than for the 45-degree one, the components of  the weight along the incline, that must be equal to the friction forces (due to the boxes are at rest), can't be equal each other, as masses are the same.

So, the frictional forces acting on the two boxes have NOT the same magnitude, so A) is FALSE.

All the other choices are true.

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A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te
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Answer:

Te =  23.4 °C

Explanation:

Given:-

- The mass of iron horseshoe, m = 1.50 kg

- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

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What is the final temperature of the water–horseshoe system?

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- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

                             m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )

- Solve for (Te):

                             m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )

                             Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

                             Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

                             Te = 2462600 / 105322

                             Te =  23.4 °C    

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