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3 years ago

What does Newton’s law of universal gravitation say about the distance between objects?

1 answer:

3 0

Newton law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

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Why is the moon's umbra much smaller during a solar eclipse?

Stolb23 [73]

Answer:

the sun is larger than the moon so it creates a shadow

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2 years ago

7. When will an object's displacement and distance traveled be different?

mariarad [96]

If an object changes direction while travelling will an object's displacement and distance travelled be different.

Some people believe that distance and displacement are simply different names for the same quantity. However, distance and displacement are not the same thing. If an object changes direction while travelling, the total distance travelled is greater than the displacement between those two points.

The magnitude of the displacement is always less than or equal to the distance because it is measured along the shortest path between two points.

When the direction of displacement does not change, the magnitude of the displacement and distance are the same. When a body travels in a straight line, for example, its displacement and distance are the same.

Learn more about displacement and distance brainly.com/question/3243551

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8 0

2 years ago

The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?

Monica [59]

Hi there!

$\large\boxed{ -6.36m/s}$

Use the equation:

$v_{2} = -\frac{m_{1}}{m_{2}} v_{1}$

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

$v_{2} = -\frac{0.048}{2.95} *391$

$v_{2} = -6.36m/s$

8 0

3 years ago

You jump off a truck and accelerate toward the surface of the Earth. Does the Earth accelerate toward you?

Triss [41]

When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>

Gravitational attraction force is GMm/r² between the earth and a person.
M= mass of the earth

m= mass of the person

r= separation between them.

<h3>What's the acceleration of the earth towards the person when he jumps from a truck?</h3>

According to Newton's second law, Force = M×acceleration
Acceleration= Force / M
Here, Force = GMm/r²,

so acceleration of earth= Gm/r²

As this acceleration is very small, so we can't notice it.

Thus, we can conclude that the earth also accelerates towards us.

Learn more about the gravitational force here:

brainly.com/question/72250

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7 0

2 years ago

A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no

Rzqust [24]

Answer:

a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

Cos 20 = Aₓ / A

sin 20 = $A_{y}$ / A

Aₓ = A cos 160 = 2 cos 160

$A_{y}$ = A sin160 = 2 sin160

Aₓ = -1,879 mi

$A_{y}$ = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

cos 2600 = Bₓ / B

sin 260 = $B_{y}$ / B

Bₓ = B cos 260

$B_{y}$ = B sin 260

Bₓ = 4 cos 260

$B_{y}$ = 4 sin 260

Bₓ = -0.6946mi

$B_{y}$ = - 3,939 mi

Third vector C = 3 mi to 15 north east

cos 15 = Cₓ / C

sin15 = $C_{y}$ / C

Cₓ = C cos 15

$C_{y}$ = C sin15

Cₓ = 3 cos 15

$C_{y}$ = 3 sin 15

Cₓ = 2,898 mi

$C_{y}$ = 0.7765 mi

Now we can find the final position of the person

X = Aₓ + Bₓ + Cₓ

X = -1.879 -0.6949 + 2.898

X = 0.3241 mi

Y = $A_{y}$ + $B_{y}$ + $C_{y}$

Y = 0.684 - 3.939 +0.7765

Y = -2.4785 mi

a) We use Pythagoras' theorem

R = √ (x2 + y2)

R = √ (0.3241 2 + (-2.4785) 2)

R = 2.4996 mi

R = 2.5 mi

b) let's use trigonometry

Tan θ = y / x

Tanθ = -2.4785 / 0.3241

θ = tan⁻¹ (-7,647)

θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

3 0

3 years ago