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2 years ago

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Sort the following forces as relevant or not relevant to this situation. The symbols are defined as follows: normal force = n⃗ ,

tension force = T⃗ , horizontal force of the road on the car = F⃗ rc, horizontal force of the road on the truck = F⃗ rt, weight = w⃗ , force of the car pushing on the truck = F⃗ ct, and force of the truck pushing on the car = F⃗ tc.

1 answer:

Irina18 [472]2 years ago

7 0

Answer:

it would be least to graetest

Explanation:

10-84

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An automatic coffee maker uses a resistive heating element to boil the 2.4 kg of water that was poured into it at 21 °C. The cur

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Answer:

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$

is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$

is the temperature change

Thus, given that:-

Mass of water = 2.4 kg

Specific heat = 4.18 J/g°C

$\Delta T=100-21\ ^0C=79\ ^0C$

So,

$\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ$

Heat Supplied $Q=VIt$

where $i=current$

$V=Voltage$

$8.5\times 120\times t=2.4\times 4186\times 79$

$t=778.10 s$

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2 years ago

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, $m=33.2\ kg$

Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

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2 years ago

The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f

tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

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2 years ago

Process in which permanent deformation of metals occurs due to applied stress and results in breaking of bonds and then reformin

luda_lava [24]

Answer:

This process involves the motion of dislocations and is termed slip (or glide in some textbooks)

Explanation:

Plastic deformation of metals (and other crystalline materials) usually occurs by slip, which is the sliding of planes of atoms over one another by dislocation movements.

On a microscopic scale, stress causes planes of crystalline objects to leave their original position and slide over other planes into new positions, these microscopic movements manifest as a slip on a macroscopic scale. And the planes do not return back to their original position after the removal of the dislocation-causing stress.

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2 years ago

A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c

Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

The plane wall is initially with constant properties is initially at a uniform temperature, To.
Suddenly the surface x=L is exposed to convection process such that T∞>To.
The other surface x=0 is maintained at To

Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

There is only conduction in 1-D.
The system bears constant properties.
The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

$T(x,0)=T_o$

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u /> $T(0,t)=T_o$ <u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u /> $-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]$ <u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0

2 years ago