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2 years ago

Which of the following is an example of chemical energy converted into mechanical energy?

Physics

1 answer:

EastWind [94]2 years ago

5 0

Answer:

Explanation:

A is wind energy converted to mechanical energy

B is chemical energy stored in food creating energy for movement.

C electric energy to light energy

D building a campfire is chemical energy from burning wood to light energy.

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If I push on the wall with 75 Newtons of force, the wall will push back with __

Scilla [17]

Answer: 0

Explanation:

75 newtons will push back canceling it out make it 0

7 0

2 years ago

A 920-kg compact car moving at 92 m/s has approximately 3,893,440 Joules of kinetic energy. What is the change in kinetic energy

tensa zangetsu [6.8K]

Answer:

Change in kinetic energy = 3297280 J

Explanation:

Given that,

Mass, m = 920 kg

Speed of a car, v = 92 m/s

Kinetic energy, K = 3,893,440 J

If the speed of a car, V = 36 m/s

Net kinetic energy is given by :

$K_n=\dfrac{1}{2}mV^2\\\\=\dfrac{1}{2}\times 920\times (36)^2\\\\K_n=596160\ J$

The change in kinetic energy = 3,893,440 - 596160

= 3297280 J

So, the change in kinetic energy of the car is 3297280 J.

3 0

2 years ago

11) We are past the year 1990:<br> A.) Observation<br> B.) Inference

Sliva [168]

A an observation hehdhhdhdhdhdhhdhdhdhdhdb

7 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

Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values

Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and 4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and 1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and 2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and 2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge Right Charge Resulting force(N)

1μC 4μC 90 N

4μC 1μC 90 N

2μC 2μC 90 N

1μC 2μC 45 N

1μC 8μC 180 N

2μC 8μC 360 N

5 0

3 years ago