Which would cause the largest increase in a mountain climber’s gravitational potential energy?

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

Bad White [126]

A) Agreed.
b) Value agreed but units should be W (watts). 

c) Here's one method... 

15 miles = 24140 m 

1 gallon of gasoline contains 1.4×10⁸ J. 

So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J 

Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J 

Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J 

So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. 

Efficiency = useful work/total energy supplied 
= 1000/5800 
= 0.17 (=17%)

8 0

3 years ago

Which sentence describes an example of sublimation?

k0ka [10]

answer C is the correct one

7 0

3 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago

Help on physics problem (picture above)

natali 33 [55]

The two letters are B and A, in that order.

7 0

3 years ago

When bouncing a ball, the bouncing motion results in the ball ____________.

yan [13]

Answer:

B) changing position

Explanation:

When a ball bounces to the ground it hits the ground with some energy. The amount of energy with which it hits the ground is kinetic energy. When it comes in the contact with the ground kinetic energy gets converted into potential energy. This potential energy again gets converted into kinetic energy and balls moves again from the ground and bounces multiple times. So, due to multiple bounce the position of the ball changes.

Thus, When bouncing a ball, the bouncing motion results in the ball changing position.

6 0

3 years ago

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