PE=30J, m=?, g=10m/s2, h=10m
1 answer:
Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:
PEgrav = mgh
Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)
With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.
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Do operations that you can with what is given first.
Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.
m = 0.30kg
PEgrav = mgh
Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)
With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.
Do operations that you can with what is given first.
Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.
m = 0.30kg
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Answer:
B. 2 m/s
B. Acceleration = 4.05 m/s² and Tension = 297.5 N.
Explanation:
A force is applied on a mass m whose acceleration is 4 m/s
Force = mass × acceleration
a = F/m = 4 m/s
4 m/s = F/m
F = 4 m/s (m)
If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:
2F/4 m = F/ 2m
4m/s (m) / 2m = 2 m/s
a = 2 m/s
2.
Given that
mass = 30 kg
mass = 50 kg
= 0.1
From the question; we can arrive at two cases;
That :
----- equation (1)
---- equation (2)
50 a = 50 g - T
30 a = T - 30 g sin 30 - 4 × 30 g cos 30
By summation
80 a =g
80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)
80 a = 324 m/s ²
a = 324/80
a = 4.05 m/s²
From equation , replace a with 4.05
50 × 4.05 = 50 × 10 - T
T = 500 -202.5
T =297.5 N
Answer:
<em>The internal energy change is 330.01 J</em>
Explanation:
Given
the initial volume = 5.75 L
the final volume = 1.23 L
is the external pressure = 1.00 atm
q the heat energy removed = -128 J (since is removed from the system)
expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;
W = -ΔV
W = -1.00 x(1.23 - 5.75)
W = -1.00 x -4.52
W = 4.52 L atm
converting to joules we have
W = 4.52 L atm x 101.33 J/ L atm = 458.01 J
The internal energy change during compression can be calculated thus;
ΔU = q + W
ΔU = -128 J + 458.01 J
ΔU = 330.01 J
Therefore the internal energy change is 330.01 J