Answer:
Explanation:
mass of object, m = 3 kg
spring constant, K = 750 n/m
compression, x = 8 cm = 0.08 m
angle of gun, θ = 30°
(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.
(b) Let v be th evelocity of ball at the tim eof launch.
by using the conservation of energy
1/2 Kx² = 1/2 mv²
750 x 0.08 x 0.08 = 3 x v²
v = 1.265 m/s
By use of the formula of maximum height
h = 0.02 m
h = 2 cm
Answer:
(a) W = 1329.5 J = 1.33 KJ
(b) ΔU = 24.27 KJ
Explanation:
(a)
Work done by the gas can be found by the following formula:
where,
W = Work = ?
P = constant pressure = (0.991 atm)() = 100413 Pa
ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)() = 0.01324 m³
Therefore,
W = (100413 Pa)(0.01324 m³)
<u>W = 1329.5 J = 1.33 KJ</u>
<u></u>
(b)
Using the first law of thermodynamics:
ΔU = ΔQ - W (negative W for the work done by the system)
where,
ΔU = change in internal energy of the gas = ?
ΔQ = heat added to the system = 25.6 KJ
Therefore,
ΔU = 25.6 KJ - 1.33 KJ
<u>ΔU = 24.27 KJ</u>
Answer:
Part a)
Part b)
Part c)
Part d)
Net force on a closed loop in uniform magnetic field is always ZERO
Explanation:
As we know that force on a current carrying wire is given as
now we have
Part a)
current in side 166 cm and magnetic field is parallel
so we have
here we know that L and B is parallel to each other so
Part b)
For 68.1 cm length wire we have
here we know that
so we have
Part c)
For 151 cm length wire we have
here we know that
so we have
Part d)
Net force on a closed loop in uniform magnetic field is always ZERO