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3 years ago

Which of the substances listed in the table on the first page would heat up more quickly if an equal amount

1 answer:

4 0

My answer is: It's not possible to decide. Explanation: The inability to decide on an answer lies in the words "listed in the table on the first page". No first page. No table. No list. No table. No answer.

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A man (mass=68 kg) on a parachute is falling at terminal velocity (v=59 m/s)

raketka [301]

Explanation:

It is given that,

Mass of the man, m = 68 kg

Terminal velocity of the man, v = 59 m/s

We need to find the rate at which the internal energy of the man and of the air around him increase. The gravitational potential energy of the man is given by :

$E=mgh$

Differentiating equation (1) wrt t as :

$\dfrac{dE}{dt}=mg\dfrac{dh}{dt}$

Since, $v=\dfrac{dh}{dt}=59\ m/s$

$\dfrac{dE}{dt}=68\times 9.8\times 59$

$\dfrac{dE}{dt}=39317.6\ J/s$

So, the internal energy of the man and the air around him is increasing at the rate of 39317.6 J/s. Hence, this is the required solution.

3 0

3 years ago

A boat with a horizontal tow rope pulls a water skier. She skis off to the side, so the rope makes an angle of 12.0 ∘ with the f

Sladkaya [172]

Answer:

62.601kJ

Explanation:

We have a force with a horizontal component given by the angle of 15°. So wue can take this component to calculate the total work done, that is,

$W_T = Fd(cos\theta)$

Our values are,

$d=320m$

$Angle = 12\°$

$Force=200N$ ,

Substituting in Work equation,

$W_T = 200(320)(cos12)$

$W_T = 62601.4J$

$W_T = 62.601kJ$

4 0

3 years ago

If one oscillation has 5. 8 times the energy of a second one of equal frequency and mass, what is the ratio of their amplitudes?

quester [9]

The ratio of their amplitudes will be 2.41.

To find the answer, we have to know more about the simple harmonic motion.

<h3>How to find the ratio of their amplitudes?</h3>

It is given that,

$E_1=5.8E_2\\m_1=m_2 \\\omega_1=\omega_2$

We have the relation between these quantities and amplitudes as,

$A=\sqrt\frac{2E}{k} \\k=m\omega^2$

Here, for both the oscillation, k will be same.

Thus, the ratio of amplitude will be,

$\frac{A_1}{A_2} =\sqrt{\frac{E_1}{E_2} } =\sqrt{\frac{5.8E_2}{E_2} } =2.41$

Thus, we can conclude that, the ratio of their amplitudes will be 2.41.

Learn more about simple harmonic motion here:

brainly.com/question/22422926

#SPJ4

7 0

1 year ago

Someone is trying to balance a (110cm) plank with certain forces.

Mashutka [201]

Answer:

a. i. 30 Nm ii. This moment is a clockwise positive moment.

b. i. 15 Nm ii, This moment is a counter-clockwise negative moment.

c. i. The plank will not balance. ii. The plank would tip up.

d. 150 N

Explanation:

a) Calculate the moment of the 60N force (about O), then name its type.

i. Calculate the moment of the 60N force (about O)

Since moment = Force × perpendicular distance from point of moment ,

M = Fd

Since F = 60 N and d = 50 cm = 0.5 m

M = 60 N × 0.5 m = 30 Nm

ii. Then name its type.

This moment is a clockwise positive moment.

b) Calculate the moment of the 30N force (about O), then name its type.

i. Calculate the moment of the 30N force (about O),

Since moment = Force × perpendicular distance from point of moment ,

M' = F'd'

Since F' = 30 N and d' = 50 cm = 0.5 m

M' = 30 N × 0.5 m = 15 Nm

ii. Then name its type.

This moment is a counter-clockwise negative moment.

c) Will the plank balance? If not, which way will it tip?

i. Will the plank balance?

The plank will balance if the net moment on it is zero

So net moment, M' = positive moment - negative moment = M - M' = 30 Nm - 15 Nm = 15 Nm

Since the net moment on the plank is M" = 15 Nm ≠ 0,<u>the plank will not balance.</u>

ii Which way will it tip?

The plank would tip in the direction of the greater moment since the net moment is positive. <u>This moment tilts the plank in a clockwise direction, so the plank would tip up.</u>

d) What extra force would be needed at (B) to balance the plank?

The extra force must balance the net moment,

So M" = F"d" where F" = force and d" = distance of force from O = 10 cm = 0.10 m

F" = M"/d"

= 15 Nm/0.10 m

= 150 N

4 0

2 years ago

Waves with higher amplitude transfer ____energy.

natali 33 [55]

Answer:

- Waves with higher amplitude transfer HIGHER energy.

- Waves with higher frequency transfer HIGHER energy.

3 0

2 years ago