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3 years ago

Which of the following does not have a global cycle?

Physics

2 answers:

Darya [45]3 years ago

8 0

The answer is c hope it helps

Musya8 [376]3 years ago

3 0

It's Iron, I believe.

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Explain the energy transfer if a book on the edge of a table, falls to the floor. There should be 3 types of energy transfers.

gavmur [86]

Answer:

potential energy

kinetic energy

thermal energy

Explanation:

The book's potential energy can be released by knocking it off the table. As the book falls, its potential energy is converted to kinetic energy. When the book hits the floor this kinetic energy is converted into heat and sound by the impact.

6 0

3 years ago

Read 2 more answers

What is the gravitational potential energy of a 150 kg object suspended 5m above the Earth's surface

ella [17]

The gravitational potential energy referred to the ground level is given by
$U=mgh$
where m is the mass of the object, $g=9.81 m/s^2$ is the gravitational acceleration and h is the height of the object with respect to the ground.

Therefore in our problem the potential energy is
$U=(150 kg)(9.81 m/s^2)(5 m)=7357.5 J$

6 0

3 years ago

He distance between two consecutive crests is 2.5 meters. Which characteristic of the wave does this distance represent? amplitu

kupik [55]

Frequency is measured in units of reciprocal time.
Period is measured in units of time.
Phase is a number without units that represents a fraction of a wave.

None of these is measured in meters, so none of them can be the answer.
It must be either amplitude or wavelength.

Amplitude is a quantity that's measured at one or two points in the same wave.
The question is talking about points on consecutive waves.

<em>Wavelength is</em> the only choice left. That must be it.

3 0

3 years ago

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A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

3 years ago

The linear impulse delivered by the hit of a boxer is 202 N · s during the 0.244 s of contact. What is the magnitude of the aver

zlopas [31]

Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force $\times$ Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = $\frac{Impulse}{Time}$ = $\frac{202}{0.244}$ = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

7 0

3 years ago