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dolphi86 [110]
4 years ago
6

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.

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Explanation:

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What derived unit is used to measure the slope of the line in this graph?
vlabodo [156]

Answer:

C. g/cm³

Explanation:

The slope is measured by calculating the variation of the Y values over the X values between two points on a line.  

So, the formula is: Slope = Δy/Δx

That means that we also take the units.

In this case, the Y-axis unit is in g, while the X-axis unit is in cm³.

Dividing a Y-variation over an X-variation will give you g/cm³.

In this case, let's assume the line passes through (10,100) (not exactly, but close enough for the example), and it passes through (0,0)

So the slope would be: (100-0) g / (10-0) cm³ = 10 g/cm³

3 0
4 years ago
Which statement correctly describes mass-energy equivalence?
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A 415-kg container of food and water is dropped from an airplane at an altitude of 300m. First, consider the situation ignoring
frozen [14]

1) 7.8 s

Ignoring air resistance, the only force acting on the container is gravity. Therefore, we can use the following suvat equation for the free-fall case:

s=ut+\frac{1}{2}at^2

where

s = 300 m is the displacement of the container

u = 0 is the initial vertical velocity of the container, which starts from rest

a = g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have chosen downward as positive direction, so all quantities are positive

Solving the equation for t, we find:

s=\frac{1}{2}gt^2\\t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(300)}{9.8}}=7.8 s

2) 76.4 m/s

The speed of the container at any time t can be found by using another suvat equation:

v=u+at

where

v is the speed at time t

u = 0 is the initial speed

a = g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Substituting t = 7.8 s, the time of flight, we can immediately find the final speed of the container, just before hitting the ground:

v=0+(9.8)(7.8)=76.4 m/s

3) 4067 N

In this case, there is a parachute that produces a drag force acting upward.

When the container is falling at constant speed, 6 m/s down, it means that its acceleration is zero, so the net force acting on it is zero. This means that the air drag balances the weight of the container, therefore we can write:

F=mg

where

F is the air drag

m is the mass of the container

g is the acceleration of gravity

Substituting m = 415 kg, we find the magnitude of the drag force:

F=(415)(9.8)=4067 N

5 0
4 years ago
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