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2 years ago

5

I am pushing in a large box with 200 N of force. Clint is pushing on the box with 200 N of force in the opposite direction. Why

won’t the box move?

1 answer:

andreev551 [17]2 years ago

4 0

Answer:

Balanced forces

Explanation:

Balanced forces are where two forces of equal size act on an object in opposite directions. It means that in each direction, any pushes and pulls are balanced by another force in the opposite direction.

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Calculate the magnitude of the average gravitational force between earth and the moon

zepelin [54]

F = G mM / r^2, where
<span>F = gravitational force between the earth and the moon, </span>
<span>G = Universal gravitational constant = 6.67 x 10^(-11) Nm^2/(kg)^2, </span>
<span>m = mass of the moon = 7.36 × 10^(22) kg </span>
<span>M = mass of the earth = 5.9742 × 10^(24) and </span>
<span>r = distance between the earth and the moon = 384,402 km </span>

<span>F </span>
<span>= 6.67 x 10^(-11) * (7.36 × 10^(22) * 5.9742 × 10^(24) / (384,402 )^2 </span>
<span>= 1.985 x 10^(26) N</span>

5 0

2 years ago

If two stars have the same absolute magnitude, what can be a reason for the difference in their brightness?

AysviL [449]

B. their distances from the sun.

Explanation:

Absolute Magnitude:

Astronomers defines the absolute magnitude of a stars brightness in terms of how bright a star appears from a standard distance of 10 parsecs. Parsec is a unit of distance in astronomy. 10 parsecs is equal to 32.6 light years.

Apparent Magnitude:

Apparent magnitude of a star refers to how bright the star appears at its distance from the Earth.

If two stars have the same absolute magnitude but their apparent magnitude differs, the reason is that the distance of both the stars from the Earth varies. Hence their brightness differs when measured from Earth. The farther a star is from the Earth, the fainter its brightness.

Keywords: star, brightness, parsec, light years, apparent magnitude, absolute magnitude

Learn more about stars and absolute magnitude from:

brainly.com/question/13002384

brainly.com/question/1384449

#learnwithBrainly

5 0

2 years ago

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:

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

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

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

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

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

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

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

A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature

prisoha [69]

Answer:

$L = 2.746 cm$

Explanation:

As we know that thermal expansion coefficient of aluminium is given as

$\alpha = 24 \times 10^{-6} per ^oC$

now we also know that after thermal expansion the final length is given as

$L = L_o(1 + \alpha \Delta T)$

here we know that

$L_o = 2.739 cm$

$\alpha = 24 \times 10^{-6}$

$\Delta T = 108 - 0= 108^oC$

now we will have

$L = 2.739(1 + 24 \times 10^{-6} (108))$

$L = 2.746 cm$

3 0

3 years ago

What is the kinetic energy of a 0.5 kg puppy that is running 1.5m/s

bija089 [108]

Answer: Well the answer is KE = 5.625E-7 i just don't know the units for it...

Hope this helps....... Stay safe and have a Merry Christmas!!!!!!!!!! :D

3 0

2 years ago