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3 years ago

A dump truck contains a load of soil. Which action will leave the dump truck's

Physics

2 answers:

antoniya [11.8K]3 years ago

5 0

D ) increase the force applied by the engine

soldier1979 [14.2K]3 years ago

3 0

Answer:

Explanation:

This will not change the weight and therefore not change the inertia

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An electrical heater is used to supply 245.0 J of energy to a 15.0 g sample of aluminum, originally at 310.0 K. Determine the fi

Softa [21]

Answer:

$T_f=328.1 K$

Explanation:

To determine the final temperature of the sample, we use the specific heat formular as follows:

$Q=mCp(T_f-T_0)\\T_f=\frac{Q}{mCp} +T_{0}\\mol Al= \frac{15g}{26.98gmol^{-1}}=0.555 mol\\T_f=\frac{245.0J}{0.555mol\times24.35Jmol^{-1}K^{-1}} +310.0K\\T_f=328.1 K$

Finally, the temperature of the aluminium sample has raised 18 K.

8 0

4 years ago

A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 33 °. A 1.2 kg block is projected up th

Serga [27]

Answer:

Explanation:

Initial kinetic energy = 29 J

work done against gravity = mgsin33 x d , m is mass of the block

= 1.2 x 9.8 sin 33 x .9

= 5.76 J

potential energy stored in compressed spring

= 1/2 k x², k is spring constant and x is compression

= .5 x 200 x .3²

= 9

energy left = 29 - ( 5.76 + 9 )

= 14.24 J

b )

energy stored in spring when compression is .4 m

= 1/2 x 200 x .4²

= 16 J

required kinetic energy = 16 + 5.76

= 21.76 J

Block must be projected with energy of 21.76 J .

7 0

3 years ago

If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the

koban [17]

Answer:

$\displaystyle \frac{480}{7}\approx 68.6\; \rm mph$ .

Explanation:

The average speed of an object is equal to total distance over total time.

Distance traveled: $\rm 50 \; mi$ .

How much time is taken? This trip is divided into two halves, each of distance $\displaystyle \frac{50}{2} = 25\;\rm mi$ .

Time spent on the first half of the trip:

$\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours$ .

Similarly, time spent on the second half of the trip:

$\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours$ .

In total:

$\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours$ .

Average speed:

$\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}$ .

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to $\rm 60\; mph$ rather than $\rm 80\; mph$ .