Ask question

Ask question

All categories

3 years ago

15

A 0.75 kilogram apple is thrown upward from the ground. The apple reaches a height of 5.5m, what is the beginning gravitational

potential energy of the Apple

1 answer:

MaRussiya [10]3 years ago

6 0

Answer:

U(0.0m) = 0J

U(5.5m) = 40.42 J

Explanation:

The gravitational potential energy is given by the following formula:

$U=mgh$

h: height

m: mass of the apple = 0.75kg

g: gravitational acceleration = 9.8m/s^2

When the apple is at 5.5m from the ground the gravitational potential energy is:

$U=(0.75kg)(9.8m/s^2)(5.5m)=40.42\ J$

when the apple is on the ground you have:

$U=mg(0m)=0\ J$

You might be interested in

2. In your own words, what is direct plagiarism?

alekssr [168]

Answer:

copying another writer's work with no attempt to acknowledge that the material was found in external source is considered as a direct plagiarism.

7 0

3 years ago

Read 2 more answers

Suppose the force of gravity on the electron was comparable to the electric force created by the deflection plates. how would th

Alexandra [31]

"F=Vector Sum Of The Two Forces" Is the answer.

3 0

3 years ago

A falling skydiver has a mass of 105 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resis

LenKa [72]

F-free = m*g - F_air = m*a

F_air = 1.2 * m

a= (105 kg * 9.8 m.s^2 - 5*105) / 105 kg

a = 9.3 m/s

Hope this helps

7 0

3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

3 years ago

A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of

miskamm [114]

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= $\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}$

= 1.72 x 10³ N.

8 0

3 years ago