Ask question

Ask question

All categories

zloy xaker [14]

3 years ago

6

Kinetic vs potential energy

1 answer:

antiseptic1488 [7]3 years ago

5 0

Answer:

Kinetic energy is the energy obtained from movement while potential energy is the energy in something when it is at a height or has been manipulated in some way; it has work done to it.

Explanation:

Kinetic energy is, like I said, energy gained through movement or action. When an object stops moving, it converts into a different type of energy. Energy cannot be destroyed or created. An object with potential energy has a reference. A reference determines how high it is. The higher and heavier an object is, the more potential energy it has.

Hope this helps!

You might be interested in

The smallest known galaxy, Segue 2, has an approximate radius of 1.05 × 1015 kilometers. Use the conversion factors 1 light-year

scoundrel [369]

( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =

(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =

(0.03388) x (psc) x (10³) =

33.88 parsecs

5 0

3 years ago

When light in vacuum is incident at the polarizing angle (Brewster's angle) on a certain glass slab, the angle of refraction is

marysya [2.9K]

Answer:

Explanation:

θ( p ) + θ( r ) = 90

θ (r) = angle of refraction and θ ( p ) is polarising angle.

given θ ( r ) = 31.8

θ ( p ) = 90 - 31.8 = 58.2 degree

ii ) Tanθ ( p ) = n ( refractive index ) = Tan 58.2 = 1.6

3 0

3 years ago

A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha

lord [1]

Answer:

The value is $F_f = 46.935 \ N$

Explanation:

From the question we are told that

The magnitude of the horizontal force is $F = 92.7 \ N$

The mass of the crate is $m = 40.5 \ kg$

The acceleration of the crate is $a = 1.13 \ m/s$

Generally the net force acting on the crate is mathematically represented as

$F_{net} = F - F_f = ma$

Here $F_f$ is force of kinetic friction (in N) acting on the crate

So

$92.7 - F_f = 40.5 * 1.13$

=> $F_f = 46.935 \ N$

5 0

3 years ago

A spacecraft left Earth to collect soil samples from Mars. Which statement is true about the strength of Earth’s gravity on the

ryzh [129]

Answer:

: It Decreases.

As the spacecraft gets farther and farther from Earth, the gravitational

forces between the spacecraft and the Earth decrease.

Explanation:

5 0

3 years ago

Please help answer question

nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

4 0

3 years ago