Answer: option D
Explanation: let us first define an heat engine.
An heat engine is that device that converts partly heat energy into mechanical energy.
Inside the heat engine is a substance that undergoes compression and expansion, intake and outtake of heat, this is known as a working substance.
For heat engines to work perfectly, the working substance has to take a substantial large amount of heat energy from a source, this source is the hot reservoir, the amount of heat energy given out by this reservoir is Qh and the temperature at this region is Th. The heat from the hot reservoir is accepted by the working substances and sent to a region that will discard it out, this region is known as the cold reservoir and the heat energy at this point is Qc and temperature is Tc.
Judging by the direction of heat flow, heat energy moves from hot reservoir to the cold reservoir for the heat engine to work perfectly fine, hence Qh must be greater than Qc and Th must be greater than Tc.
Hence Qh>Qc and Th>Tc
Answer:
L = 0.99 m = 99 cm
Explanation:
The period is the reciprocal of the frequency.
T = 1/0.5 = 2.0 s
T = 2π√(L/g)
L = g(T/2π)²
L = 9.8(2.0/2π)² = 0.99 m
If the system accelerates upward, it will cause the apparent gravity to increase. This will require a longer pendulum to keep the same period, or shorten the period if the length remains the same. This shows up in the equation where the product of gravity and the square of the period must remain constant for the length to remain constant.
Answer:
When the electrons jump to a higher energy state, they release energy as electromagnetic radiation, light.
Explanation:
When the solar wind gets past the magnetic field and travels towards the Earth, it runs into the atmosphere. As the protons and electrons from the solar wind hit the particles in the Earth's atmosphere, they release energy – and this is what causes the northern lights.
Answer:
(a) a= 0.139 m/s²
(b) d= 4.45 m
(c) vf= 1.1 m/s
Explanation:
a) We apply Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass (kg)
a : acceleration (m/s²)
Data
F₁= +2.05 * 10³ N : forward push by a motor
F₂= -1.87* 10³ N : resistive force due to the water.
m= 1300 kg
Calculation of the acceleration of the boat
We replace data in the formula (1):
∑F = m*a
F₁+F₂= m*a
a= 0.139 m/s²
b) Kinematics of the boat
Because the boat moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+at Formula (3)
Where:
d:displacement in meters (m)
t : time interval (s)
v₀: initial speed (m/s)
vf: final speed (m/s)
a: acceleration (m/s²
)
Data
v₀ = 0
a= 0.139 m/s²
t = 8 s
Calculation of the distance traveled by the boat in 8 s
We replace data in the formula (2)
d= v₀t+ (1/2)*a*t²
d= 0+ (1/2)*(0.139)*(8)²
d= 4.45 m
c) Calculation of the speed of the boat in 8 s
We replace data in the formula (3):
vf= v₀+at
vf= 0+( 0.139)*(8)
vf= 1.1 m/s
