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fiasKO [112]
1 year ago
5

How does the increasing mass effect the force of an object in motion?​

Physics
1 answer:
irina [24]1 year ago
7 0

Answer:

<u>According </u><u>to </u><u>second </u><u>law </u><u>of </u><u>motion</u><u>,</u><u>t</u><u>he acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.</u>

<em>So </em><em>simply</em><em>,</em><em> </em><em>it </em><em>can </em><em>be </em><em>affected </em><em>due </em><em>to </em><em>increasing </em><em>force </em><em>as </em><em>there </em><em>is </em><em>close </em><em>relationship </em><em>between </em><em>momentum.</em>

Explanation:

<em>The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.</em>

<em>I </em><em>hope </em><em>it </em><em>was </em><em>helpful </em><em>for </em><em>you </em><em>:</em><em>)</em>

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chubhunter [2.5K]

Answer:

Pressure = 9.94 x 10⁶ Pascals

Explanation:

given data

mass = 51 kg

radius = 0.400 cm

solution

we know Pressure that is express as here

Pressure = total force on an area ÷ the area of the area   .................1

and

Force is the woman's weight so weight will be

Weight = mass × gravity  .................2

put here value

Weight = 51 × 9.8 m/s²

Weight = 499.8 Newtons

and

Area of a circle of bottom of the heel = (π) × (radius)²    ...................3

put here value

Area = (π) × (0.40 cm)²

Area = 0.502654 cm²

Area = 0.0000502654 m²

and

now we put value in equation 1 we get

Pressure = force ÷ area

Pressure = 499.8 ÷ 0.0000502654

Pressure = 9943221.381 N/m²

Pressure = 9.94 x 10⁶ Pascals

6 0
3 years ago
Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is
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}  

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