Answer:
Wnet, in, = 133.33J
Explanation:
Provided that
Pump heat QH = 1000J
Hot temperature TH= 300K
Cold temperature TL= 260K
Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:
According to the first law of thermodynamics,
COP(HP, rev) = 1/(1-TL/TH)
COP(HP, rev) = 1/(1-260/300)
COP(HP, rev) = 1/(1-0.867)
COP(HP, rev) = 1/0.133
COP(HP, rev) = 7.5
The power necessary to operate the heat pump is given by
Wnet, in = QH/COP(HP, rev)
Wnet, in = 1000/7.5
Wnet, in = 133.333J. QED
Thus, the 133.33J represents the initial work input during the heat transfer process.
<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">
QL=QH-Wnet, in
QL=1000-133.333
QL=866.67J
</pbased></padditionally...>
Answer:
Explanation:
The measurement of pressure is indicated as 
where p denotes the pressure, 
signifies density, and h represents height
Given values include pressure 
, gravity's acceleration 
, and height =1.163 m
Answer:
The final size is nearly the same as the initial size because the increase in size
is remarkably small
Solution:
According to the problem:
The proton beam energy is E = 250 GeV =
Distance traveled by the photon, d = 1 km = 1000 m
Proton mass, 
Initial size of the wave packet, 
Now,
This operates under relativistic principles
The rest mass energy for the proton is expressed as:
This proton energy is 
Thus, the speed of the proton, v
The time to cover 1 km = 1000 m of distance is calculated as:
T = 
T = 
According to the dispersion factor;
Thus, the widening of the wave packet is relatively minor.
Hence, we can conclude that:
where
= final width
Answer:
a. β = 8.23 K
b. β = 28.815 K
Explanation:
The performance of the heat pump can be calculated using the formula
β = TH / (TH - TC)
a.
TH = 15 ° C + 273.15 K = 288.15 K
TC = - 20 ° C + 273.15 K = 253.15 K
β = 288.15 K / (288.15 K - 253.15 K)
β = 8.23 K
b.
TH = 15 ° C + 273.15 K = 288.15 K
TC = 5 ° C + 273.15 K = 278.15 K
β = 288.15 K / (288.15 K - 278.15 K)
β = 28.815 K
Answer:
Explanation:
The data indicates that point A is located midway between two charges.
To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
This field points towards Q⁻.
A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.
To find the resultant field, we add these contributions:
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
For the force acting on an electron placed at A:
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
