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3 years ago

A ball is thrown straight up. At the top of its path its acceleration is

1 answer:

6 0

C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)

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What is 6.9 × 107 written in standard form?

n200080 [17]

Answer: 6.9x 107 in standard form is 69,000,000

7 0

2 years ago

Read 2 more answers

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago

A passenger jet travels from Los angles to Bombay, India, in 22 h. The return flight takes 17 h. The difference in flight times

Hoochie [10]

Answer:

70.5 mph

Explanation:

A passenger jet travels from Los Angeles to Bombay, India, in 22h.

The return flight takes 17 h.

The difference in flight times is caused by winds over the Pacific Ocean that

blow primarily from west to east.

If the jet's average speed in still air is 550 mi/h what is the average speed

of the wind during the round trip flight? Round to the nearest mile per hour.

Is your answer reasonable?

Let w = speed of the wind

Write a distance equation (dist is the same both ways

17(550+w) = 22(550-w)

9350 + 17w = 12100 - 22w

17w + 22w = 12100 - 9350

39w = 2750

W = 2750/39

w = 70.5 mph seems very reasonable

Confirming if the solution by finding the distances using these value

17(550+70.5) = 10549 mi

22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph

6 0

3 years ago

If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi

Oduvanchick [21]

you can find it using the equation: potential energy=mass*gravitational acceleration*height.

energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ

Sometimes they use 10 instead of 9.8 for the g constant.

Rember to make me Brainliest!!!

3 0

3 years ago

A diver is is pushed upwards by a diving board. She weighs 739 N and accelerates from rest to a speed of 4.60 m/s while moving 0

gavmur [86]

Answer:

Answer A

Explanation:

It seems more likely

8 0

2 years ago