Answer:
Explanation:
A)
W = work done by the freezer = 800 J
Q = heat removed from the freezer = 1735 J
Q' = Heat expelled into the room
Coefficient of performance is given as
inserting the values
B)
Heat expelled is given as
Q' = W + Q
Q' = 800 + 1735
Q' = 2535 J
Answer:
The answer is explained below.
Explanation:
The energy emitted during the de-excitation of an electron from a higher energy level to a lower energy level is directly proportional to the frequency of the emitted light.
Here, the total sum of the energies of 2 frequencies of light emitted in different stages is equal to the energy of a single frequency of light during the de-excitation of fourth level to ground level directly.
Hence the total sum of of the frequencies of 2 lights emitted in different stages is equal to the frequency of single frequency of light emitted during the de-excitation from fourth level to ground level directly.
The some of the energies of 2 frequencies emitted by one electron is equal to the energy of a single frequency when electron jumps directly.
Answer:
potential energy = mgh
= 400÷1000 × 10× 4÷100
= 0.4 × 10 × 0.04
=4/10 ×10×4/100
= 4/10 × 4/10
=16/100
= 0.16 joules
m1 (400) stretches 4cm
m1 (100g) stretches 1cm
so, m2(800g) stretches 8 cm
potential energy of m2 = mgh
= 800/1000 ×10×8/100
= 0.8 × 0.8
=8/10 ×8/10
= 64/100
=0.64 joules
Ratio of s1 to s2
16/100 ÷ 64/100
= 1:4 ( answer)
Answer:
0.594 m/s
Explanation:
First, find the time it takes to land.
Given, in the y direction:
Δy = 2.225 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(2.225 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.674 s
Next, find the horizontal velocity.
Given, in the x direction:
Δx = 0.400 m
a = 0 m/s²
t = 0.674 s
Find: v₀
Δx = v₀ t + ½ at²
(0.400 m) = v₀ (0.674 s) + ½ (0 m/s²) (0.674 s)²
v₀ = 0.594 m/s
Answer:
you cant caculate this because there is no question
Explanation:
