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3 years ago

I need help on the top part

Physics

2 answers:

zhuklara [117]3 years ago

8 0

2 goes with energy stored in food and 3 goes with energy stored in a bolder on a montainside

Talja [164]3 years ago

4 0

4 goes with energy stored in a stretched spring.

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A condition that affects the ability to sleep or the quality of sleep is referred to as a __________.

Anit [1.1K]

The answer is B) Sleep disorder. Notice that is says, "A condition that affects the ability to sleep or quaility of sleep." Meaning how much sleep you get. Also known as sleep disorder. Answer is B. Hope that helps :)

7 0

3 years ago

Read 2 more answers

What is the radius of the event horizon of a 10 solar mass black hole?

deff fn [24]

Answer:

The number take for the "radius" is normally taken to be the radius from the center to the event horizon, since nothing inside that can be observed, and since it is a point of no return. This radius is directly proportional to the mass of the black hole. From the Wikipedia article on event horizon: "For the mass of the Sun this radius is approximately 3 kilometers and for the Earth it is about 9 millimeters. In practice, however, neither the Earth nor the Sun has the necessary mass and therefore the necessary gravitational force, to overcome electron and neutron degeneracy pressure." The last sentence means that it is unlikely for the Sun to become a black hole, since it has too little mass to become one.

Explanation:

CAN I GET BRAINLIEST

6 0

3 years ago

A 2kg blob of putty moving at 4 m/s slams into a 6kg blob of putty at rest what is the speed of the to stick together blobs imme

andrew11 [14]

<h2>Answer</h2>

1m/s

<h2>Explanation</h2>

Given that:

<em>Mass of first blob = 2kg = m1</em>

<em>Velocity of blob = 4m/s = v1</em>

<em>Mass of second blob = 6kg = m2</em>

<em>Velocity of blob = 0m/s = v2</em>

<em />

To find:

<em>Final velocity = Vf</em>

<em />

<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>

<em />

<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>

(2*4) + (6*0) = Vf(2+6)

8 + 0 = Vf(8)

8 = Vf(8)

Vf = 1 m/s

So the speed of two blobs immediately after colliding = 1 m/s

3 0

3 years ago

A satellite in a nearly circular orbit is 2000 km above earth's surface. the radius of earth is approximately 6400 km. if the sa

Ann [662]

Velocity is computed using the formula:

$v= \frac{d}{t}$
Where:
V = speed
d = distance traveled
t = time/period

First you need to consider that the orbit is circular. To get the measurement or the distance going around Earth, you will need to get the circumference of the path.

$C = 2 \pi r$

Where:
C = circumference
π = 3.14
r = radius

The Earth has a radius of 6,400km, but you also need to consider that the satellite is orbiting above the surface of the Earth, so you add in the 2,000km to that radius.

r = 6,400Km + 2,000Km = 8,400Km

Next step is to insert that into our circumference formula:

$C = 2 \pi r$
$C = 2 \pi 8,400Km$
$C = 52,778.76 Km$

The distance traveled would then be 52,778.76Km

Now that we have the distance, we can then get the velocity:

$v= \frac{d}{t}$
$v= \frac{52,778.76Km}{12hrs}$
$v= 4,398.23km/hr$

The speed of the satellite is 4,398.23km/hr.

7 0

3 years ago

A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because

Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

$\tau = F*d$

Where

F= Force

d = Distance

Our values are given as,

$m_1 = 0.32Kg$

$m_2 = 0.4Kg$

$d = 8.7*10^{-2}m$

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

$\tau = T_2-T_1$

$\tau = F_2*d-F_1*d$

$\tau = m_2g*d-m_1*g*d$

$\tau = (m_2-m_1)g*d$

$\tau = (0.4-0.32)(9.8)(8.7*10^{-2})$

$\tau = 0.068N\cdot m$

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is $\tau = 0.068N\cdot m$

8 0

3 years ago