A) <u>Weight = mass × acceleration (due to gravity) </u>
= 60×9.8
= 588 N
<u>B) Potential energy = mass x gravity x change in height
</u>
1,000 = 60.0 x 9.8 x h
h = 1.7 m
<u>C) Kinetic energyF = potential energyI
</u>
KEF = 1/2mv2
PEI = mgh = 1,000 J
1/2mv2 = 1,000
1/2(60.0)v2 = 1,000
v2 = 33.33
v = 5.77 m/s
Underline the words then eliminate the ones that arent part of the problem!
Answer:
The average acceleration of the bearings is 
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval 
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration

Put the value into the formula



Hence,The average acceleration of the bearings is 
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Via half-life equation we have:

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

There is 6.25 grams left of Ra-229 after 12 minutes.