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3 years ago

Which has the least potential energy

Physics

2 answers:

Inessa [10]3 years ago

7 0

An object with a very tiny mass, being at a very low height.

makvit [3.9K]3 years ago

4 0

Answer:

The plasma state would have the least potential energy and the greatest kinetic energy.

Explanation:

i just took the test and i go it right

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A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0

3 years ago

a spacecraft that seems to be motionless in deep space is given some type of quick push. which statement describes what will hap

bogdanovich [222]

An applied force applied causes a body such as the spacecraft to move. The magnitude of the force determines the change in the velocity

What will happen to the spacecraft is given by option C. from among the possible question options.

C. The spacecraft will begin to move and it will continue moving until it is stopped by an equal and opposite force

Reason:

The possible question options obtained from a similar question includes;

A. The spacecraft will move for some time then stop slowly

B. Air resistance will prevent the spacecraft from moving

C. The spacecraft will begin to move and the motion will continue until a force equal and opposite to the applied force stops it

D. The quick push will not cause the spacecraft to move, as a quick push works on Earth only

The state of the space craft = Motionless

Location of the spacecraft = Deep space

Type of force applied = Quick push

The statement that describes what will happen = Required

Solution;

Let F, represent the force applied, and let Δt be the duration of the applied force, we have;

The impulse of the force, F × Δt = m·(v₂ - v₁)

Where;

m = The mass of the spacecraft

v₁ = The initial velocity of the spacecraft = 0

v₂ = Final velocity of the spacecraft

Plugging in v₁ = 0, gives;

F × Δt = m·v₂

The space craft is given a velocity, v₂, and according to Newton's First Law of Motion, it continues moving in a straight line until another force acts on it

Therefore, the correct option is option C. The spacecraft will begin to move and the motion will continue until it is stopped by an equal and opposite force

Learn more about Newton's First Law of Motion here:

brainly.com/question/20841616

8 0

3 years ago

Formula for calculating work done in machines

Archy [21]

Answer:

force×distance

Explanation:

i think that's the answer and it could also be expressed as m×g×d

7 0

3 years ago

How do longitudinal waves transport energy?

marissa [1.9K]

A longitudinal wave transports energy through the medium without permanently transporting matter.

Hope this helps :D

6 0

3 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

3 years ago