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3 years ago

13

Which substance has a melting point greater than room temperature?

1 answer:

blagie [28]3 years ago

4 0

Answer:

I think the answer is D.)

Explanation:

If it means something needs to melt into liquid i would have gone with B.) or C.) but since it doesn't specifiy. I thought D.) since all you have to do is heat it and it melts or boils.

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HELPPPPPPPPPPPPPP 70PTS NOT A LIE

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2 years ago

What is the difference between a wedge and a screw?

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3 years ago

Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

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3 years ago

A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. The maximum depth in a lake to

Svetach [21]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

F = $5.20 \times 10^{5}$ N

g = 9.8 m/s

radius = $\frac{diameter}{2}$

= $\frac{30 cm}{2}$ = 15 cm = 0.15 m (as 1 m = 100 cm)

Formula to calculate depth is as follows.

F = $\rho \times g \times h \times A$

or, h = $\frac{F}{\rho \times g \times A}$

h = $\frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}$

= 751 m

Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.

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3 years ago