physics 10 points to first person to give me a detailed answer step by step?

Question

1.A ball of mass 0.5 kg is rolling across a table top with a speed of 5 m/s. When the ball reaches the edge of the table it rolls down an incline onto the floor 1.0 meter below (without bouncing) What is the speed of the ball when it reaches the floor?

2. A hot whells car of mass 0.025 kg is traveling on a horizontal track with a velocity of 5.0 m/s. If the track suddenly turns upward, how high up the track can the car travel?

Answer 1

1. Use conservation of energy: PE(initial) + KE(initial) = PE(final) + KE(final
so:
mg(h1) + .5m(v1)^2 = mg(h2) + .5m(v2)^2 ["^2" means squared]
plugging in (solving for v2, which we'll call v from here on, and saying that the final potential energy is 0):
.5(9.8)1 + .5*.5(25) = 0 + .5*.5v^2
4.9 + 6.25 = .25v^2
v^2 = 11.15/.25=44.6
v=6.68m/s

2. Again, conservation of energy. The initial potential energy is 0, and the final kinetic energy is zero, so we have
.5m(v1)^2 = mgh
(mass cancels out:)
.5(5)^2 = gh
h = 12.5/g
h = 1.28

Answer 2

Both of these questions can be answered using the concept of conservation of energy.
By tracking the ball/car's Kinetic Energy (KE) and gravitational Potential Energy (PE), we can learn information about the ball/car's speed and height above our reference point.

KE = 1/2 mv^2
where m is the ball/car's mass and v is the speed of the ball/car.

PE = mgh
where m is the mass of the ball/car, g is the gravitational acceleration the ball/car experiences (9.81 m/s^2), and h is the height of the ball/car above our reference point.

Question #1)
m = .5 kg
v = 5 m/s
h = 1.0 m

The ball posses some initial amount of Kinetic Energy due to its motion as well as some gravitational Potential Energy due to its height above the floor. As the ball rolls down the ramp, it looses its PE in exchange for an increase in its KE. The change (drop) in PE equals the change (increase). in KE. The final KE = the initial KE + the change in KE, which is equal to the ball's initial PE.
KE_final = KE_initial + PE_initial

KE_initial = 1/2 (.5 kg) (5 m/s)^2
KE_initial = 6.25 Joules

PE = mgh
PE = (.5 kg)(9.81 m/s^2)(1.0 m)
PE = 4.91 Joules

KE_final = 6.25 J + 4.91 J = 11.16 Joules

Now we use this value to calculate the ball's final speed.
KE_final = 11.16 Joules = 1/2 (.5 kg) v^2
v^2 = 44.64 m^2/s^2
v = 6.8 m/s

Question #2)
In this question, all of the car's initial KE is converted into PE as the car rolls up the ramp. At the end, the car has no KE and its final PE value equals its initial KE value.

KE_initial = 1/2 (.025 kg) (5.0 m/s)^2
KE_initial = .3125 Joules

PE_final = .3125 Joules = (.025 kg)(9.81 m/s^2) * h
solving for the height, h,
h = .3125 Joules / (.025 kg * 9.81 m/s^2)
h = 1.27 meters

Answer 3

1)energy remains the same from the table top to the floor 1.0m below. At the table top, energy consider gravity potential energy and kinetic energy, while on the the floor , enery consider only kinetic energy.
For table top, Energy,E= 1/2 (m)(v)(v) + mgh
= 1/2(0.5)(5)(5) + (0.5)(9.81)(1.0)
=11.155J
On the floor, E= 1/2 (m)(v)(v) = 11.155
1/2(0.5)(v)(v) =11.155
v= sqrt(44.62)
= 6.68 m/s
2) The is no force acting up, so the car cant even move up.

Answer 4

Fg = G(m1)(m2) / r^2 The on the unique radius, r, Fg = 5.0*(10^4) N the recent radius will be (a million/4)r as we've moved 3/4th alongside a thanks to middle of the ordinary body. As all different variables will be held consistent, Fg at this new radius will only be: Fg = G(m1)(m2) / [(a million/4)(r)]^2 [(a million/4)(r)]^2 = (a million/16)r^2 So we've: G(m1)(m2) / (a million/16)r^2 = 16*G(m1)(m2) / r^2 as we've a million/(a million/16) = 16 So we've at a million/4 the radius, the Fg is 16 cases better. So only mulitiply the unique Fg by 16 to get inspite of it truly is i visit't discover my calculator