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2 years ago

5

A particular lightbulb is designed to consume 40 W when operating on a car's 12-V DC electric power. If you supply that bulb wit

h 12-V AC power from a transformer, how much power will it consume

1 answer:

Andrew [12]2 years ago

4 0

40V because it will provide the same amount of power.

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The breaking in (blank) bonds in food releases energy for your body to use

kicyunya [14]

Answer:

The breaking in <em>molecular</em> bonds in food releases energy for your body to use.

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2 years ago

Why are frogs said to have two lives?

Stells [14]

They live half their lives in water and most of their life on land

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3 years ago

Which object has the most gravitational potential energy?

Kipish [7]

Answer: An 8 kg book at a height of 3 m has the most gravitational potential energy.

Explanation:

Gravitational potential energy is the product of mass of object, height of object and gravitational field.

So, formula to calculate gravitational potential energy is as follows.

U = mgh

where,

m = mass of object

g = gravitational field = $9.81 m/s^{2}$

h = height of object

(A) m = 5 kg and h = 2m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 2 m\\= 98.1 J (1 J = kg m^{2}/s^{2})$

(B) m = 8 kg and h = 2 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 2 m\\= 156.96 J (1 J = kg m^{2}/s^{2})$

(C) m = 8 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 3 m\\= 235.44 J (1 J = kg m^{2}/s^{2})$

(D) m = 5 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 3 m\\= 147.15 J (1 J = kg m^{2}/s^{2})$

Thus, we can conclude that an 8 kg book at a height of 3 m has the most gravitational potential energy.

3 0

3 years ago

A ruined in a race decides to accelerate right up to the moment he crosses the line He is the initially travelling at 5m/s and a

docker41 [41]

Answer:

S = V t + 1/2 a t^2 = 5 m/s * 5 s + .2 m/s^2 * 25 s^2 =

25 m + 5 m = 30 m distance traveled

Vf = V + a t = 5 m/s + .4 m/s^2 * 5 s = (5 + 2) m/s = 7 m/s final velocity

6 0

2 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago