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2 years ago

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

Physics

1 answer:

lisabon 2012 [21]2 years ago

6 0

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

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An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar

Iteru [2.4K]

The direction of the electric field would be south.

qE/m = 115
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 = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) 
 = 762.87 * 10^(-12) 
 = 6.27 x 10^-10 N/C

Hope this answers the question. Have a nice day. Feel free to ask more questions.

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2 years ago

Can i get help for the parts for the two questions? MATHPHYSSSS

Anestetic [448]

Explanation:

003 (part 1 of 2)

Pressure is force divided by area.

P = F / A

P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)

P = 229,320 Pa

003 (part 2 of 2)

There are approximately 6895 Pa in 1 psi.

P = 229,320 Pa × (1 psi / 6895 Pa)

P = 33.3 psi

004 (part 1 of 2)

Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).

Impulse = change in momentum

F Δt = m Δv

F (36 s) = (300 × 0.003 kg) (5.2 sin 57° m/s − (-5.2 sin 57° m/s))

F = 0.218 N

004 (part 2 of 2)

Pressure is force over area.

P = F / A

P = 0.218 N / 0.712 m²

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7 0

2 years ago

4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest

yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}=v_{i}-gt$ (1)

Where:

v(i) is the initial velocity
v(f) is the final velocity
g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

$0=30-9.81t$

Solve it for t:

$t=3.06\: s$

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}^{2}=v_{i}^{2}-2gh$ (2)

$0=v_{i}^{2}-2gh$

We know that the initial velocity is 30 m/s.

$0=30^{2}-2gh$

Solving it for h we have:

$h=\frac{30^{2}}{2*9.81}$

$h=45.87 \: m$

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

$10=30-(9.81t)$

So, the time elapsed to get 10 m/s is:

$t_{upward}=2.04\: s$

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

$t_{upward}=2.04+t$

$t=1.02\: s$

This is the time when the ball has 10 m/s downward.

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

$t=2t_{upward}$

$t=2*3.06$

$t=6.12\: s$

The displacement will be zero after 6.12 s.

e) Here we need to find the time when v(f) is 15 m/s

$15=30-gt$

$t=\frac{15}{9.81}$

$t=1.53\: s$

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0

2 years ago

How many position coordinates are required to identify the position of a system undergoing one-dimensional motion? What symbol s

gregori [183]

One-dimensional motion can be plotted through the Cartesian plane which has a coordinates of (x,y). These coordinates are the abscissa and ordinates. Since, there are two coordinates, the answer to the second item is two.

The symbol that can be used to identify systems position is (x,y). Since this is one dimensional motion, it is possible that one of the two coordinates becomes zero.

6 0

3 years ago

Select to show the energy of pendulum 1. Be sure that friction is set to none. Drag the pendulum to an angle (with respect to th

iragen [17]

Answer:

it have Potential energy

Explanation:

given data

Drag the pendulum to an angle 30∘

to find out

what form of energy does it have

solution

we know that pendulum start no kinetic energy when it release from any rest position then in starting it have potential energy only so that when pendulum is angle 30∘ at some height from ground so when it start it have potential energy same as in starting.

we know that the total energy is always conserve

so it have potential energy

3 0

2 years ago