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Sholpan [36]
3 years ago
8

An apple in a tree has a gravitational store of 8J. As it falls, it accelerates constantly until it hits the ground. What is the

apple's maximum kinetic energy store?
Physics
1 answer:
natita [175]3 years ago
5 0

Answer:

Explanation:

Given that on the tree the gravitational energy stored is 8J

Then, mgh = 8J.

The apple begins to fall and hit the ground, what is the maximum kinetic energy?

Using conservation of energy, as the above is about to hit the ground, the apple is at is maximum speed, and the height then is 0m, so the potential energy at the ground is zero, so all the potential of the apple at the too of the tree is converted to kinetic energy as it is about to hits the ground. Along the way to the ground, both the Kinetic energy and potential energy is conserved, it is notice that at the top of the tree, the apple has only potential energy since velocity is zero at top, and at the bottom of the tree the apple has only kinetic energy since potential energy is zero(height=0)

So,

K.E(max) = 8J

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A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc
Umnica [9.8K]
Frictional force always opposes applied force, so the net force on the cart would have to be 19N - 1.7N. The acceleration can then be solved by using the relation: F = ma. This is shown below:

Net force = 19 - 1.7 = 17.3 N

Acceleration = Force / mass
Acceleration = 17.3 / 2
Acceleration = 8.65 N/m
4 0
3 years ago
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The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden
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Answer:

m = 62.14 g

Explanation:

Energy used to melt the ice is the energy released by the condensation of the water forms on the glass

so here we have

energy for the condensation of water is given as

let mass of water condensed = m

E = m_1 L_f

now the energy of vaporization is given as

E = m_2 L_v

here we know that

L_f = 79.8 kCal/kg

L_v = 580 k Cal/kg

Now we have

8.55 \times 580 = m \times 79.8

m = 62.14 g

3 0
3 years ago
A man does 500 j work pushing a car a distance of 2m how much force does he apply
olasank [31]

Answer: 250 N

Explanation:

Use equation for work

W=F*d

d=2m

W=500J

F=?

-----------------------

W=Fd

F=W/d

F=500J/2m

F=250N

6 0
3 years ago
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Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M
stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity M_1 g (downward) and the tension in the string T (upward). So we can write

M_1 g - T = M_1 a

where

M_1 is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, M_2 g sin \theta. So the equation of motion is

T-M_2 g sin \theta = M_2 a

where

M_2 = 13.5 kg is the mass of the 2nd block

\theta=35.5^{\circ} is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

a=0

So the two equations become:

M_1 g - T=0\\T-M_2 g sin \theta = 0

Isolating T from the 1st equation,

T=M_1 g

And substituting into the 2nd equation, we can find the value of the mass M_1:

M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

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7 0
3 years ago
How to change this to mega and giga prefixes?urgent .thankyou
Akimi4 [234]
3*10^{8} m = 3*10^{5} km = 3*10^{2} Mm = 3*10^{-1} Gm

Each increase in the prefix is a division by 1000.
6 0
3 years ago
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