efficiency=work output/work input×100
since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input
efficiency=1500/3000×100
=50%
Energy consumed in doing the work = 300 Joules
Force applied on the object = 75 N
Let the distance moved by the object be d.
Work done by the force is determined by the force applied and the displacement happened in the direction of the force applied.
Work done = Force x displacement
300 = 75 x d
d = 4 m
Hence, the maximum distance moved by the object = 4 meters
Answer:
The tunnel probability for 0.5 nm and 1.00 nm are and respectively.
Explanation:
Given that,
Energy E = 2 eV
Barrier V₀= 5.0 eV
Width = 1.00 nm
We need to calculate the value of
Using formula of
Put the value into the formula
(a). We need to calculate the tunnel probability for width 0.5 nm
Using formula of tunnel barrier
Put the value into the formula
(b). We need to calculate the tunnel probability for width 1.00 nm
Hence, The tunnel probability for 0.5 nm and 1.00 nm are and respectively.
Answer:
The final velocity of the car is 2.02 m/s
Explanation:
Hi there!
The kinetic energy of the car as it runs along the first flat horizontal segment can be calculated using the following equation:
KE = 1/2 · m · v²
Where:
KE = kinetic energy
m = mass
v = velocity
Then, the initial kinetic energy will be:
KE = 1/2 · 0.100 kg · (2.77 m/s)²
KE = 0.384 J
When the car gains altitude, it gains potential energy. The amount of gained potential energy will be equal to the loss of kinetic energy. So let´s calculate the potential energy of the car as it reaches the top:
PE = m · g · h
Where:
PE = potential energy.
m = mass
g = acceleration due to gravity.
h = height.
PE = 0.100 kg · 9.8 m/s² · 0.184 m
PE = 0.180 J
Then, the final kinetic energy will be (0.384 J - 0.180 J) 0.204 J
Using the equation of kinetice energy, we can obtain the velocity of the car:
KE = 1/2 · m · v²
0.204 J = 1/2 · 0.100 kg · v²
2 · 0.204 J / 0.100 kg = v²
v = 2.02 m/s
The final velocity of the car is 2.02 m/s