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3 years ago

8

What property of a wave increases as the wave's energy increases?

1 answer:

Mariulka [41]3 years ago

5 0

Answer:

Frequency.

Explanation:

More energy = shorter wavelength = greater frequency

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ONLY 4 QUESTIONS PLEASE HELP!!!!!

rodikova [14]

1. B
2.D
3.C
4.B

Hope this helped. Good luck!

3 0

3 years ago

How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa

Anna [14]

Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature $T_{1}$ = 68 °F = 20° c

Final temperature $T_{2}$ = 212 °F = 100° c

Specific heat of water $C = 4.186 \frac{KJ}{kg c}$

Now heat transfer Q = m × C × ( $T_{2}$ - $T_{1}$ )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.

4 0

3 years ago

the state of Utah covers 82,144 square miles. the state of Montana covers 145,552 square miles. what is the total area of the tw

Fed [463]

The answer is...

82,144
+
145,552
-------------
227,696

3 0

3 years ago

Read 2 more answers

Which of these is an example of energy transfer by radiation?

andrezito [222]

Answer:

Hands rubbed together to become warm

6 0

3 years ago

Read 2 more answers

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

3 years ago