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Bad White [126]
1 year ago
13

the force that gravitation exerts upon a body, equal to the mass of the body times the local acceleration of gravity

Physics
1 answer:
PIT_PIT [208]1 year ago
8 0

The force that gravitation exerts upon a body, equal to the mass of the body times the local acceleration of gravity is known as weight.

Weight is the force of gravity acting on a body.

The formula is :

                                W =mg

Here,

W is the weight or force acting on the body. m is the mass of the body, and g is the gravitational acceleration.

Since weight is also a force, so its SI unit is also newton. The value of weight varies from place to place depending on the gravity. Its value can also be equal to zero.

If you need to learn about the difference between mass and weight, click here

brainly.com/question/23876249?referrer=searchResults

#SPJ4

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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
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Answer:

C. the area of the rectangle plus the area of the triangle under the line

Explanation:

Based on the information provided, the velocity vs. time graph is a line with a positive slope and a y-intercept of (0, 3).  The displacement is the area under this line.  This area can be divided into a triangle and a rectangle.  So of the options available, C is the correct one.

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3 years ago
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What is the key characteristic for double displacement
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It exchanges the cations or the anions of two ionic compounds.
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3 years ago
You have a ball with a mass 1.3 kg tied to a rope, and you spin it in a circle of radius 1.8m. If the ball is moving at a speed
BlackZzzverrR [31]

Answer:

6.5\; {\rm N}.

Explanation:

When an object travel at a speed of v in a circle of radius r, the (centripetal) acceleration of that object would be a = (v^{2} / r).

In this question, the ball is travelling at v = 3\; {\rm m\cdot s^{-1}} in a circle of radius r = 1.8\; {\rm m}. The (centripetal) acceleration of this ball would be:

\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}.

By Newton's Laws of Motion, for an object of mass m, if the acceleration of that object is a, the net force on that object would be m\, a. Since the acceleration of this ball is a = 5\; {\rm m\cdot s^{-2}}, the net force on this ball would be:

\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}.

7 0
2 years ago
If a questions says "at what point does the ball stop?" what is it asking for?
Bad White [126]

Explanation:

If a question says "at what point does the ball stop?", it means we need to find the position of the ball when its final velocity is equal to 0. It can be calculated using the equation of kinematics as follows :

d = ut + (1/2) at²

and

v²-u²=2ad

Where, u is initial velocity, v is final velocity, a is acceleration, t is time and d is displacement.

5 0
3 years ago
A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total
worty [1.4K]

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

6 0
3 years ago
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