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Sav [38]
3 years ago
14

With what magnitude of force does a ball of mass 0.75 kilograms need to be hit so that it accelerates at the rate of 25 meters/s

econd2? Assume that the ball undergoes motion along a straight line.
Physics
2 answers:
SSSSS [86.1K]3 years ago
8 0

Answer:

Assume that the ball undergoes motion along a straight line. ... F = m A Force = (mass) x (acceleration) The question tells you the mass and the acceleration. All YOU have to do is take the numbers and pluggum into Newton's 2nd law. F = m A = (0.75 kg) (25 m/s²) = (0.75 x 25) kg-m/s² = 18.75 Newtons .

Explanation:

i looked it up ok

Zielflug [23.3K]3 years ago
8 0

<u>Answer: </u>

A ball of mass 0.75 kilograms needs to be hit so that it accelerates at the rate of 25 meters/second^2 assuming that the ball undergoes motion along a straight line and the mount of Force required is 18.75 newton.

<u>Explanation:</u>

According to Newton`s 2nd  law F=Ma,

where M mass of the object, a- Acceleration due to gravity.

Given-

Mass (m) = 0.75 kg

Acceleration (a) = 25 m/s^2

We know That, Force = mass × acceleration

Substituting the given values , we get

  F = 0.75 × 25

   F= 18.75 N  

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suter [353]

Answer: 0.223 m/s^{2}

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity g of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

F=G\frac{m_{E}m}{r^2}  (1)

Where:

F is the module of the force exerted between both bodies

G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}} is the gravitational constant

m_{E}=5.98(10)^{24} kg is the mass of the Earth

m are the mass of each communications satellite

r=R_{E}+h is the distance between the center of the Earth and the satellite

R_{E}=6.38(10)^{6} m is the radius of the Earth

h=3.59(10)^{7} m is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

F=mg  (2)

Combining (1) and (2):

G\frac{m_{E}m}{r^2}=mg  (3)

Isolating g:

g=\frac{G M_{E}}{r^2}  (4)

Remembering r=R_{E}+h:

g=\frac{G M_{E}}{(R_{E}+h)^2}  (5)

g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}  

Finally:

g=0.223 m/s^{2}  

5 0
3 years ago
Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest. Which best describes h
meriva

Answer:

8.1km/h Northwest

Explanation:

The 8.1km/h Northwest gives the best description of her distance from start to finish. This distance can be represented in a right angle triangle , this is the hypotenuse which is the longest side of the triangle. If we add 5.7 and 5.8 this gives 11.5km/h compared to 8.1km/h which is a smaller distance and the best.

6 0
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Beginning at the NW corner of the intersection of Pine &amp; 675, thence north 950 feet, thence west 380 feet, thence south 950
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Answer:

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The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.

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Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.

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6 0
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Current is produced when charges are accelerated by an electric field to move to a position of lower
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Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
motikmotik

Answer:

n_cladding = 1.4764

Explanation:

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Thus;

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θ_c = 85°

Now, critical angle is given by;

θ_c = sin^(-1) (n_cladding/n_core)

sin θ_c = (n_cladding/n_core)

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Plugging in the relevant values, we have;

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7 0
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