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3 years ago

1d. Conservation of energy is demonstrated in this roller coaster example.

Physics

1 answer:

Dvinal [7]3 years ago

6 0

Answer:

can you clear your question I can't understand

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Waves can transfer energy through

lina2011 [118]

Answer:

electromagnetic waves

Explanation:

"wave" is a common term for a number different ways in which energy is transferred

3 0

3 years ago

A student needs to travel a total of 400 miles to reach his vacation destination . if he drives at an average speed of 50 mph ho

Arte-miy333 [17]

Answer:

The answer is 8 hours.

Explanation:

M : H

50 : 1

400 : X

400 * 1 = 50 x

400 / 50 = 8

8 0

3 years ago

Read 2 more answers

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate w

grigory [225]

A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2 to the east , how far did it travel during the accelration? Give your answer in units of kilometers per hour/sec. To find the acceleration of the car we have to 
1. First determine the suitable formula for this word problem.
Which is a. A=vf-vi/t which will be
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s
Solution: 
x = v0t + ½at2
x = 16.645375 m

7 0

3 years ago

where would information on the chemical and physical properties of a specific chemical be located in a laboratory or in the work

goldenfox [79]

Answer:

Both

Explanation:

8 0

3 years ago