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2 years ago

This diagram shows a stationary box sitting on the floor. Which statements about the other force exerted on the box are

1 answer:

5 0

It's a normal force exerted by the floor. It balances the gravitational force. Option A is correct. The normal force is balanced by the gravitational force.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). Mathematically it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and accelertaion in the different components and balancing the equation gets components in the y-direction.

The normal force is balanced by the gravitational force. It's a normal force exerted by the floor. It balances the gravitational force.

Hence, option A is correct.

To learn more about the friction force refer to the link;

brainly.com/question/1714663

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3 years ago

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3 years ago

Energy is conserved. This means that in any system, _________. a) energy is constantly recycled b) total energy input equals tot

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Answer:

b) total energy input equals total energy output

Explanation:

The first law of thermodynamics is a generalization of the conservation of energy in thermal processes. It is based on Joule's conclusion that heat and energy are equivalent. But to get there you have to get around some traps along the way.

From Joule's conclusion we might be tempted to call heat "internal" energy associated with temperature. We could then add heat to the potential and kinetic energies of a system, and call this sum the total energy, which is what it would conserve. In fact, this solution works well for a wide variety of phenomena, including Joule's experiments. Problems arise with the idea of heat "content" of a system. For example, when a solid is heated to its melting point, an additional "heat input" causes the melting but without increasing the temperature. With this simple experiment we see that simply considering the thermal energy measured only by a temperature increase as part of the total energy of a system will not give a complete general law.

Instead of "heat," we can use the concept of internal energy, that is, an energy in the system that can take forms not directly related to temperature. We can then use the word "heat" to refer only to a transfer of energy between a system and its environment. Similarly, the term work will not be used to describe something contained in the system, but describes a transfer of energy from one system to another. Heat and work are, therefore, two ways in which energy is transferred, not energies.

In an isolated system, that is, a system that does not exchange matter or energy with its surroundings, the total energy must remain constant. If the system exchanges energy with its environment but not matter (what is called a closed system), it can do so only in two ways: a transfer of energy either in the form of work done on or by the system, either in the form of heat to or from the system. In the event that there is energy transfer, the change in the energy of the system must be equal to the net energy gained or lost by the environment.

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3 years ago

Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

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3 years ago