Ask question

Ask question

All categories

2 years ago

13

Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is

different and your example
Please help due today!!

1 answer:

Andreas93 [3]2 years ago

5 0

Answer:

The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.

Examples of Gravitational Energy

A raised weight.

Water that is behind a dam.

A car that is parked at the top of a hill.

A yoyo before it is released.

You might be interested in

Q5: An ice skater moving at 12 m/s coasts

ExtremeBDS [4]

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

8 0

3 years ago

The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 12t + 6 for t ≥ 0

avanturin [10]

Answer:

(a) $v(t) = 6t^2 - 6t - 12$ , $a(t) = 12t - 6$

(b) When $0 \leq t < 0.5$ , object is slowing down, when $t > 0.5$ object is speeding up.

Explanation:

(a) To get the velocity function, we need to take the derivative of the position function.

$v(t) = \frac{ds(t)}{dt} = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12$

To get the acceleration function, we need to take the derivative of the velocity function.

$a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6$

(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0

$12t - 6 < 0 \\12t < 6 \\t < 0.5$

On the other hand, object is speeding up when a > 0

$12t - 6 > 0 \\12t > 6 \\t > 0.5$

Therefore, when $0 \leq t < 0.5$ , object is slowing down, when $t > 0.5$ object is speeding up.

6 0

3 years ago

During which phase of mitosis do the chromosomes line up across the center of the cell?

nirvana33 [79]

Metaphase; the centromeres of duplicated chromosomes line up in the middle of the cell. (It's also the shortest phase of mitosis).

5 0

3 years ago

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0

3 years ago

A current I flows down a wire of radius a.

Helga [31]

Answer:

(a) $K = \frac{I}{2\pi a}$

(b) $J = \frac{I}{2\pi as}$

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

$K = \frac{dI}{dL}$

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

$dL = 2\pi r$

The radius of the wire = a

$dL = 2\pi a$

The surface current density $K = \frac{I}{2\pi a}$

(b) The current density is inversely proportional

$J \alpha s^{-1}$

$J = \frac{k}{s}$ ......(1)

k is the constant of proportionality

$I = \int\limits {J} \, dS$

$I = J \int\limits \, dS$ ........(2)

substituting (1) into (2)

$I = \frac{k}{s} \int\limits\, dS$

$I = k \int\limits^a_0 \frac{1}{s} {s} \, dS$

$I = 2\pi k\int\limits\, dS$

$I = 2\pi ka$

$k = \frac{I}{2\pi a}$

substitute $J = \frac{k}{s}$

$J = \frac{I}{2\pi as}$

7 0

2 years ago