Answer:
8.9
Explanation:
We can start by calculating the initial elastic potential energy of the spring. This is given by:
(1)
where
k = 35.0 N/m is the initial spring constant
x = 0.375 m is the compression of the spring
Solving the equation,
Later, the professor told the student that he needs an elastic potential energy of
U' = 22.0 J
to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:
Therefore, Tom needs to increase the spring constant by a factor:
Initial velocity Vi = 5.03m/s
Distance difference D = 4.80 m
Kinetic Friction coefficient u = 0.18
Static Friction coefficient Uf = 3.14
We know g = 9.81 m/s
Calculating th edistance at which it is stopped, d = Vi^2 / 2 ug =>
d = 5.03^2 / 2 x 0.18 x 9.81 => d = 25.3 / 3.53 = 7.167 m
Calculating the acceleration, a = (Vf^2 - Vi^2) / [2d] =>
Vf is 0 as it is at max distence, a = 5.03^2 / (2 x 7.167) = 1.765 m/ s^2
Vf = Square root of Vo^2 + (2aD) = > Square root of [0.00 + 2 x 1.765 x 4.80]
= Square root of [16.944]
Ss his speed = 4.12 m/s
Answer:
5
Explanation:
54.34+45.66=100.00 (you have to use the .00 because when you add to numbers you keep the number of decimals)
so you get 100.00
all numbers that are not 0 are sig figs so 1 is a sig fig
If a number ends with a 0 after a decimal place that 0 is a sig fig
all numbers between two sig figs are sig figs so that would make all of the numbers sig figs
Answer:
The kinetic energy dissipated is 3286.5 J
Explanation:
K.E before collision = 1/2m1v1^2 = 1/2×313×6^2 = 5634 J
K.E after collision = 1/2(m1+m2)v2^2
From the law of conservation of momentum:
m1+m2 = m1v1/v2 = 313×6/2.5 = 751.2 kg
K.E after collision = 1/2×751.2×2.5^2 = 2347.5 J
K.E dissipated = 5634 J - 2347.5 J = 3286.5 J
Answer:
The diagram shows a heater above a thermometer. The thermometer bulb is in the position shown. How the heat
