PHYSICS HIGH SCHOOL

3. During a race, a sprinter increases from 5.0 m/s to 7.5 m/s over a period of 1.25s. What is the sprinter’s average acceleration during this period?Also please try and explain
Please and Thank you

Answers

Answer 1
Answer: (7.5)-(5)
----------- = 2 m/s^2
(1.25-0)

Average acceleration is calculated using the equation V(f)-V(i) / t(f)-t(i). (Final velocity minus initial velocity divided by final time minus initial time)

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How many g in 6 1/2 kg

Answers

there are 6,500 grams in 6 1/2 kilograms

What is an isobar?a. a unit of measurement for air pressure
b. a line along which air pressure is constant
c. a region in which the weather is not changing
d. a symbol on a weather map showing wind direction

Answers

Answer:

B.a line along which air pressure is constant

Explanation:

i took the quiz and got it right

A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter 1.0 m. The ball makes 2.0 revolutions every 1.0 s. What are the magnitude and direction of the acceleration of the ball? What's the tension in the string?

Answers

Answer :

Explanation :

It is given that:

mass of the ball, m=175\ g=0.175\ Kg

Radius of circle, r=(diameter)/(2)=0.5\ m

The ball makes 2.0 revolutions every 1.0 s. So, angular speed is \omega=4\pi\ radian/sec

Since, it is moving in circular path so centripetal acceleration will act here.

So, centripetal acceleration \alpha =m\omega^2r

\alpha=0.175\ Kg* (4\pi)^2* 0.5

So, \alpha=13.803\ m/s^2

Hence, the acceleration is 13.803\ m/s^2 and it is directed towards the center of rotation.

Tension is a force which is given by :

                                 F=ma

                        F=0.175\ Kg*13.803\ m/s^2

                                   F=2.415\ N

This is the required answer.                        

For circular motion.

Centripetal acceleration = mv²/r = mω²r

Where v = linear velocity, r = radius = diameter/2 = 1/2 = 0.5m

m = mass = 175g = 0.175kg.

Angular speed, ω = Angle covered / time

                         = 2 revolutions / 1 second

                         = 2 * 2π  radians / 1 second

                         = 4π  radians / second

Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5    Use a calculator

                                                         ≈13.817  m/s²

The magnitude of acceleration ≈13.817  m/s² and it is directed towards the center of rotation.

Tension in the string = m*a

                                   = 0.175*13.817

                                   = 2.418 N

What is Newton's third law motion answer part a and b

Answers

Answer:

hope it helps

Explanation:

Newtons third law is that objects exert equal and opposite forces on each other.

'every action has an equal and opposite reaction'.

Answer:

to every action there's equal and opposite reaction

A ball is equipped with a speedometer and launched straight upward. The speedometer reading after two seconds is 40 m/s; the ball is moving upward. At what approximate times would the ball display the following speedometer readings; 0 m/s and 30 m/s

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