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3 years ago

When an orange rolls off a table, its potential energy is converted into?

1 answer:

7 0

Since the orange rolls off the table and is (falling) its potential energy turnes into kinetic energy.

Hope this helped
:D

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The more force you put on a material, the more it .......?

Brilliant_brown [7]

Answer:

the faster it moves(if it is free) . the less the inertia (potential) and the more the inertia (kinetic)

the more (at times x2) force and momentum it comes back at you

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3 years ago

Read 2 more answers

How would the earth move if the sun suddenly disappeared

Lerok [7]

Just a guess without asking Google, that without the sun's gravity holding us in orbit, we would end up drifting whatever direction we were headed as it disappeared?

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3 years ago

The most narrow region of the electromagnetic spectrum is

Lostsunrise [7]

The answer is A. X rays or it can be visible light but mostly A.

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3 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

bixtya [17]

Answer:

a. $P=18.61\ W$

b. $\eta_c=0.6976=69.76\%$

c. $\eta=0.4718=47.18\%$

Explanation:

Given:

temperature of the hotter reservoir, $T_H=1250\ K$

temperature of the colder reservoir, $T_C=378\ K$

heat absorbed by the engine, $Q_H=1.42\times 10^{5}\ J$

heat rejected to the cold reservoir, $Q_L=7.5\times 10^4\ J$

time duration of the energy transfer, $t=1\ hr=3600\ s$

<u>Now the work done by the engine:</u>

Using energy conservation,

$W=Q_H-Q_L$

$W=14.2\times 10^4-7.5\times 10^4$

$W=6.7\times 10^4\ J$

<u>Hence the power output:</u>

$P=\frac{W}{t}$

$P=\frac{6.7\times 10^4}{3600}$

$P=18.61\ W$

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{378}{1250}$

$\eta_c=0.6976=69.76\%$

now actual efficiency:

$\eta=\frac{W}{Q_H}$

$\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}$

$\eta=0.4718=47.18\%$

6 0

4 years ago

You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 690.0 kg and was tra

kirza4 [7]

Answer:

115 km/h

Explanation:

$m_1$ = Mass of car A = 690 kg

$m_2$ = Mass of car B = 520 kg

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration

u = Initial velocity

v = Final velocity

$a=\mu g\\\Rightarrow a=0.75\times 9.81\\\Rightarrow a=7.3575\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 7.3575\times 6+0^2}\\\Rightarrow v=9.396\ m/s$

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow u_1=\frac{(m_1 + m_2)v-m_2u_2}{m_1}\\\Rightarrow u_1=\frac{(690+520)9.396-520\times -\frac{74}{3.6}}{690}\\\Rightarrow u_1=31.96\ m/s$

Converting to km/h

$31.96\times 3.6=115.056\ km/h$

Initial velocity of car A = 115 km/h

7 0

3 years ago