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2 years ago

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A block is being pulled upward along an inclined surface at a constant speed. Which of the following statements is correct? Grou

p of answer choices The net force on the block points upward along the inclined surface. The net force on the block is zero. The net force on the block is larger than the magnitude of the weight of the block. The net force on the block is smaller than the magnitude of the weight of the block but greater than zero.

1 answer:

Alex17521 [72]2 years ago

5 0

Answer:The net force on the block is zero.

Explanation:

Given

Block is being pulled upward along an inclined surface at a constant speed

As speed is constant and moved in a straight line along the plane therefore its velocity is also constant .

and change in velocity is equal to acceleration therefore acceleration is zero here i.e. net force is zero acting on the body.

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Both technicians A and B

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Both trailer sway control, hill assist and hill descent controls are additional featires that enhance stability of electronics and their control systems. Majorly, these features track and reduce skidding in electronics, therfore, enhancing electronic system stability. During the process, these newly added features help to automatically apply brakes and direct the sytem where the controller wants to take it.

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A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

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Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

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It starts from rest, then after 15s it would achieve a speed of

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(b) The distance angle swept by it is:

$\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad$

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The air that flows over the top part of an airplane's wing moves faster than the air that flows across the bottom. This faster m

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

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Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

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$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

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<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

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$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

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Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

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