Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J
Finding out the acceleration 12/3 = 4m/s^2
thus it is descending so the actual acceleration would be 9.8-4 = 5.8 m/s^2
the weight will be 90*5.8 = 522 N
522/9.8 = 53.2 kg
The amount left of a given substance can be calculated through the equation,
A = (A0) x 0.5^n/h
From the given scenario,
A/A0 = 0.75 = 0.5*(60/h)
The value of h from the equation is 144.565 minutes.
Answer:
288.2 K
Explanation:
= Volumetric expansion coefficient = 
= Initial temperature = 273.1 K
= Final temperature
= Original volume = 150 mL
Change in volume is given by
The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL
Answer:
Explanation:
Given that,
Mass of student 1, m₁ = 65 kg
Mass of student 2, m₂ = 70 kg
The distance between the students, d = 3 m
We need to find the magnitude of the gravitational force between the two students. The formula for the magnitude of the gravitational force between two masses is given by :
So, the gravitational force between the two students is 
.
