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olya-2409 [2.1K]
4 years ago
15

Is orange juice in an orange potential energy or kinetic energy?

Physics
2 answers:
joja [24]4 years ago
6 0

Answer:

Potential energy.

Explanation:

The potential energy is the energy which is possessed by the body by the virtue of its position with respect to the others, electric charges and with other factors.

And the kinetic energy is defined as the energy of an object which is produced by the virtue of the motion of an object.

Therefore, in the given situation orange juice does not possess any velocity it is in rest. So orange juice is an orange potential energy.

kicyunya [14]4 years ago
3 0
It would be potential energy because it is not moving. <span />
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Grace [21]
Initial velocity u = 40 
Angle at launch = 55 degrees
 At maximum height v = 0, velocity equation v^2 = u^2 - 2gh,
 0 = (40 x sin55)^2 - 2 x 9.81 x h => (40 x 0.819)^2 = 19.62h => h = 32.76^2 /
19.62
 Maximum height = 54.7 m 
 We have v = u - gt => 0 = (40 x sin55) - 9.81 x t => t = 32.76 / 9.81 => t =
3.34 s
 Time taken to hit the ground is 2t = 2 x 3.34 = 6.68 s 
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8 0
4 years ago
Please help with these physics problems 1. A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to
Tatiana [17]

Answer:

1.) h = 164.8 m

2.) U = 49.1 m/s

3.) t = 1.43 seconds

Explanation:

1.) A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to the ground. The height of the building is...? 

Since the soccer ball is dropped from the building, the initial velocity U will be equal to zero

Using second equation of motion

h = Ut + 1/2gt^2

Substitutes the time into the formula

h = 1/2 × 9.8 × 5.8^2

h = 164.8 m

2. The Falcon 9 launches to a height of 123 meters. What is its vertical initial velocity?

At maximum height final velocity = 0

Using the third law of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8 × 123

U^2 = 2410.8

U = 49.1 m/s

3. An apple falls from rest off a 10.m m tree. How long will it take before it hits the ground?

Since the apple fall from rest, the initial velocity U will be equal to zero

Using the second equation of motion,

h = Ut + 1/2gt^2

substitute all the parameters into the formula

10 = 1/2 × 9.8 × t^2

10 = 4.9t^2

t^2 = 10/4.9

t^2 = 2.04

t = 1.43 seconds

8 0
3 years ago
A square conducting plate 52.0 cm on a side and with no net charge is placed in a region, where there is a uniform electric fiel
Mrac [35]

Explanation:

Given:  uniform electric field E= 82.0 kN/C.

a) charge density σ =ε_0 E.

therefore, \sigma =82\times10^3\times3.85\times10^{-12}\\=0.0000003157= 315.7 nC/m^2

b)Total charge on each face = σA

q=σA

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5 0
4 years ago
"A problem involves a car of mass m going down a track from a height H, and round a loop of radius r. The loop is frictionless.
neonofarm [45]

Answer:

Explanation:

At the topmost position,  the car does not have zero velocity but it has velocity of v so that

v² /r = g or centripetal acceleration should be equal to g ( 9.8 )

Considering that,  the car must fall from a height of 2r + h where

mgh = 1/2 mv²

= 1/2 m gr

So h = r/2

Hence the ball must fall from a height of

2r + r /2

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v² / r = g .

4 0
3 years ago
A 30 kg box sits on the floor it requires 275N of force to get it moving once it is moving it only takes 225N of force.what are
gavmur [86]

1) The coefficient of static friction is 0.935

2) The coefficient of kinetic friction is 0.765

Explanation:

1)

When a force is applied on a box sitting on the floor, the force that must be applied in order to make the box moving is equal to the maximum force of static friction between the floor and the box, which is:

F = \mu_s mg

where

\mu_s is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

g=9.8 m/s^2

F = 275 N

Therefore, the coefficient of static friction is

\mu_s = \frac{F}{mg}=\frac{275}{(30)(9.8)}=0.935

2)

Once the box is in motion, the force that must be applied in order to make the box moving at constant velocity is equal to the force of kinetic friction between the floor and the box, which is:

F = \mu_k mg

where

\mu_k is the coefficient of kinetic friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

g=9.8 m/s^2

F = 225 N

Therefore, the coefficient of kinetic friction is

\mu_k = \frac{F}{mg}=\frac{225}{(30)(9.8)}=0.765

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

8 0
3 years ago
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