As you mentioned, we will use <span>Equipartition Theorem.
</span><span>H2 has 5 degrees of freedom; 3 translations and 2 rotation
</span>Therefore:
Internal energy = (5/2) nRT
You just substitute in the equation with the values of R and T and calculate the internal energy as follows:
Internal energy = (5/2) x 2 x <span>8.314 x 308 = 32.0089 x 10^3 J</span>
(a) Fx = 1.464 N
(b) Fy = 1.952 N
(c) F(x, y) = 1.464 i + 1.952 j
Given
Mass = 1kg
Acceleration = 2.44 m/s2
Angle with positive X axis = 53°
As we know
F = ma
By substituting value
F= 1×2.44 N
F= 2.44 N
(a) Component of force in X direction
Fx = F Cosθ
Fx = 2.44 Cos(53°)
Fx = 2.44 × 0.60 = 1.464 N
(b) Component of force in Y direction
Fy = F Sinθ
Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N
(c) Net force in vector notation
F(x, y) = 1.464 i + 1.952 j
Thus we got net force.
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Option c) 1.5 V
Explanation:
<em>As the circuit is build in series first we will find the current passing through the complete circuit. Current stays the same in each element is the series cirucuit, however, the voltage is different.</em>
Voltage is given by the following formula:
V = IR
<em>Because we have to find current through whole circuit, we will first find resistance of the whole circuit.</em>
Equivalent Resistance R(eq): R1 + R2 = 60 + 60 = 120 ohm
Current passing through whole circuit be:
= 0.025
Now we will find out the voltage between C and D:
Current stays the same in series circuit: I = 0.025 c
Resistance between C and D is, R = 60 ohm
Voltage becomes, V = IR = 0.025 * 60 = 1.5 V
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force
by the worker must be equal to the friction force
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2

b. The work is done on the crate by this force is the product of its force
and the distance traveled s = 4.35

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

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