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ryzh [129]
4 years ago
15

Explain why an object’s weight is dependent upon where in the universe it is located.

Physics
1 answer:
strojnjashka [21]4 years ago
6 0

Answer:

The weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

Explanation:

Everything obeys the law of universal gravity proposed by Newton. Where the attraction force with which one body attracts another depends on the mass of the body that attracts the body with less mass, the radius of the body with the greatest mass and the universal gravitational constant.

For a better understanding of this concept let us use examples with numeric values. In the first example we will determine with what force planet Earth attracts a person of 70 [kg].

And in the second example we will perform the same exercise but on a planet like Jupiter

Example 1:

A person with a mass of 70 [kg] is located on planet Earth which has a mass of 5.97x10^24 [kg] and a terrestrial radius of 6371 [km]. Find the force exerted by the planet upon the person.

We have the  law of universal gravity

F=G*\frac{m_{1} *m_{2} }{r^{2} } \\where\\G = 6.67*10^-11[\frac{N*m^{2} }{kg^{2}} ]\\m_{1}=70[kg]\\m_{2}=5.97*10^{24} [kg]\\r= 6371[km]

Now replacing the values in the equation we have:

F=6.67*10^{-11} *\frac{70*5.97*10^{24} }{(6371*10^3)^{2} } \\F=687[N]

We can appreciate that if we only use the terms of mass of the earth, the gravitational constant and the radius of the earth, we will have the value of the gravity accelaration of the earth. Let's check

g_{earth} =6.67*10^{-11} *\frac{5.97*10^{24} }{(6371*10^3)^{2} } \\g_{earth} = 9.81 [m/s^{2} ]

Example 2:

A person with a mass of 70 [kg] is located on planet Jupiter which has a mass of 1.899x10^27 [kg] and a  radius of 71492 [km]. Find the force exerted by the planet upon the person.

F=6.67*10^{-11}*\frac{70*1.899*10^{27} }{(71492*10^3)^{2} }  \\F= 1734.73[N]

We will calculate the value of the gravity accelaration of Jupiter. Let's check

g_{jupiter} = 6.67*10^{-11} *\frac{1.899*10^{27} }{(71492*10^3)^{2} } \\g_{jupiter}= 24.8 [m/s^2]\\

Therefore an object in jupiter will weigh more than 2.5 times its weight than on planet Earth.

We found that the force of attraction or the weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

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alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
What crisis is occurring in California?
nlexa [21]
Hey there!

Option A

A climate crisis is occurring in California where sudden forest fires occur due to this .
3 0
2 years ago
A water tank is filled with up to 3.5 m height.Calculate the pressure given by the tank at its bottom​
anyanavicka [17]

Answer:

96.04Pa

Explanation:

height=3.5m

gravity=9.8%

density=9.8/3.5

=2.8

Preassure=h×g×d

=3.5×9.8×2.8

=96.04Pa

6 0
3 years ago
A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o
maxonik [38]
For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
v = y*f = 1.5*220 = 330 m/s
4 0
3 years ago
For which one of the following situations will the path length equal the magnitude of the displacement? a A ball on the end of a
PSYCHO15rus [73]

Answer:

d A ball is rolling down an inclined plane.

Explanation:

When path length is equal to the displacement

then we can say that the motion of the object must be in straight line so that the distance and displacement must be same

SO here we can say

a A ball on the end of a string is moving in a vertical circle.

In circular path distance and displacement is not same

b A toy train is traveling around a circular track.

In circular path distance and displacement is not same

c A train travels 5 miles east before it stops. It then travels 2 miles west.

Net displacement is 3 miles East while distance is 7 miles

d A ball is rolling down an inclined plane.

Here its motion is in straight line so we can say that path length and displacement will be same

e A ball rises and falls after being thrown straight up from the earth's surface.

In this type of to and fro motion path length is not same as displacement

7 0
3 years ago
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