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2 years ago

Pls guys what is displacement

Physics

2 answers:

insens350 [35]2 years ago

7 0

Explanation:

Displacement is a difference between an initial value and a final value, not factoring in any changes that happen in between. For example you let an object of a certain volume sink to the bottom of a container of water, the water level will be displaced, or changed, by exactly the volume of that object.

Example: A marble with volume of 3 cm³ is dropped into a graduated cyllinder with 10 mL of water. The water level rises to the 13 mL mark, so the displacement of the water by the marble is exactly 3 mL.

This doesn't just apply to a change in volume: If I move a box 3 meters along an axis in the positive direction, we say its displacement along that axis is 3 meters. If I moved it in the negative direction the same distance, the displacement would be -3 meters.

KengaRu [80]2 years ago

4 0

Answer:

Displacement is the shortest distance from your initial position to your final position. Therefore, displacement is a measurement of distance and not an "end-point."

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Don't text while driving
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A girl rides her scooter to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the

Whitepunk [10]

Answer:

6.93 km/h

Explanation:

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3 years ago

Read 2 more answers

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$