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3 years ago

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

Physics

2 answers:

beks73 [17]3 years ago

5 0

Potential energy. Kinetic energy is movment

Anuta_ua [19.1K]3 years ago

4 0

It is potential because it has the POTENTIAL to move. <span />

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Define newton's first law and provide an example

Lynna [10]

Answer:

An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

Explanation:

When an athlete tries to stop his career, it takes several meters to stop completely, due to the inertia produced.

When trying to push a car, at first it is very difficult, because, due to inertia, the car tends to remain still. But once it is put into motion, the effort is much less to be done, since then inertia causes it to keep moving.

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3 years ago

Read 2 more answers

Two samples of dirt are collected from a suspect's tread in his shoe and a crime scene. She notes very similar characteristics.

elena-14-01-66 [18.8K]

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4 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

$E=7*10^{9}N/C$

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field is mathematically given as

$E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\$

$E=7*10^{9}N/C$

For more information on this visit

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3 years ago

When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical

laila [671]

Answer:

I = 18 x 10⁻⁹ A = 18 nA

Explanation:

The current is defined as the flow of charge per unit time. Therefore,

I = q/t

where,

I = Average Current passing through nerve cell

q = Total flow of charges through nerve cell

t = time period of flow of charges

Here, in our case:

I = ?

q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C

t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s

Therefore,

I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)

6 0

3 years ago