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4 years ago

An object's weight is dependent upon its location in the universe. Why is this true?

Physics

2 answers:

solmaris [256]4 years ago

8 0

Explanation :

We know that the weight of an object is given by :

$W=mg$

Where,

m is the mass

g is the acceleration due to gravity.

The value of g changes from one location to another.

An object's weight is dependent upon its location in the universe. This statement is true. This is because g varies in different places so the weight also varies.

Hence, the correct option is (D) " This is true because weight is the magnitude of gravitational force acting on an object, and since gravitational force varies in different places in the universe, weight also varies ".

Bezzdna [24]4 years ago

6 0

Because an object's weight is the measure of the gravitational
forces between the object and other things that have mass.

In its travels around the universe, the object can be in the
neighborhood of other things with huge mass or tiny mass,
and it can be nearer to them or farther away from them. So
the gravitational forces between the object and other things
can have widely different values in different places.

Read the choices very carefully.
I think 'D' is the one that says this.

You might be interested in

A non uniform rod has mass

Doss [256]

Answer:

$r_{cm} = L/3$

Explanation:

Mass: M, Length: L.

$\sigma (x) = b(L-x)$

The formula that gives center of mass is

$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ...}{m_1 + m_2 + ...} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}$

In the case of a non-uniform mass density, this formula converts to

$\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx }$

where the denominator is the total mass and the nominator is the mass times position of each point on the rod.

We have to integrate the mass density over the total rod in order to find the total mass. Likewise, we have to integrate the center of mass of each point (xσ(x)) over the total rod. And if we divide the integrated center of mass to the total mass, we find the center of mass of the rod:

$\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx } = \frac{\int\limits^L_0 {xb(L-x)} \, dx }{\int\limits^L_0 {b(L-x)} \, dx } = \frac{b\int\limits^L_0{(xL - x^2)} \, dx }{b\int\limits^L_0 {(L-x)} \, dx } = \frac{\frac{x^2L}{2} - \frac{x^3}{3}}{Lx - \frac{x^2}{2}}\left \{ {{x=L} \atop {x=0}} \right.$

Here x's are cancelled. Otherwise, the denominator would be zero.

$r_{cm} = \frac{\frac{xL}{2}-\frac{x^2}{3}}{L-\frac{x}{2}}\left \{ {{x=L} \atop {x=0}} \right. = \frac{\frac{L^2}{2}-\frac{L^2}{3}}{L-\frac{L}{2}} = \frac{\frac{L^2}{6}}{\frac{L}{2}} = \frac{L}{3}$

8 0

3 years ago

Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a

bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>

This is a single force that balances other given forces.

The given parameters:

First force, F₁ = 10 N
Second force, F₂ = 10 N
Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

$F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N$

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

5 0

3 years ago

Any two importance of measurement in points (i am giving you 30 points)

erma4kov [3.2K]

Answer:

1.It helps in trade and business.

2. It helps to perform scientific calculations.

5 0

3 years ago

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the elect

RideAnS [48]

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o r}$

It is clear that the electric field is inversely proportional to the distance. So,

$\dfrac{E}{E'}=\dfrac{r'}{r}$

$E'=\dfrac{Er}{r'}$

$E'=\dfrac{125\times 3.5}{1.5}$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

5 0

4 years ago

As a baseball is being caught, its speed goes from 43 to 0 m/s in about 0.003 s. Its mass is 0.145 kg. (Take the direction the b

juin [17]

The question is incomplete, these are the missing parts of the question.

b) What is the baseball's acceleration in g's? (Indicate the direction with the sign of your answer.)

(b) What is the size of the force acting on it?

Answer:

a) -14333.3ms-2

b) -1462.58g

c) 2078N

Explanation:

The acceleration (deceleration) is easily found from from the information provided. Then the acceleration is obtained in terms of g as shown in the solution attached. The force is obtained from F=ma as shown in the image attached. Note the acceleration is negative because the ball was coming to rest. (The ball was caught).

6 0

3 years ago