A piston releases 18 J of heat into its surroundings while expanding from 0.0002 m^3 to 0.0006 m^3 at a constant pressure of 1.0
1 answer:
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Answer:
Explanation:
Given that,
Mass of a person, m = 84 kg
The person is standing at a top of Mt. Everest at an altitude of 8848 m
We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :
E = mgh, g is the acceleration due to gravity
So, the gravitational potential energy of the person is
Answer:
<em>Answer: Work equals force times distance. 3,000 J</em>
Explanation:
Work Done By A Force
When some force is applied and a displacement
is achieved, the work done by the force is given by
Note that the work is a scalar magnitude as the result of the dot-product of two vectors. If the force and the displacement are parallel, then the vectors can be replaced as its magnitudes F,x and the work is
The dot product becomes a simple arithmetic product, i.e force times distance.
Sara weighs 500 Nw and she climbs up a 6 meter set of stairs. She needs to lift her weight up, so the force is the weight and the distance is the height of the stairs, thus
Answer: Work equals force times distance. 3,000 J
Answer:
A. 110Hz,220Hz, 330 Hz
Explanation:
for organ open at open both ends;
the length of the organ for the fundamental frequency, L = A---->N + N----->A
A---->N = λ /4 and N----->A = λ /4
L = λ /4 + λ /4 = λ /2
λ = 2 x 1.5m = 3.0 m
Wave equation is given by;
V = Fλ
Where;
V is the speed of sound
F is the frequency of the wave
F = V/ λ
F₀ = V / 2L
Where;
F₀ is the fundamental frequency
F₀ = 330 / 2(1.5)
F₀ = 330 / 3
F₀ = 110 Hz
the length of the organ for the first overtone, L = A---->N + N----->A + A----->N + N----->A
L = 4λ /4
L = λ
λ = 1.5 m
F₁ = 330 / 1.5
F₁ = 220 Hz
Thus, F₁ = 2F₀
For open organ at one end
the length of the organ for the fundamental frequency, L = N------A
L = λ /4
λ = 4L
F₀ = V/4L
F₀ = 330 / (4 x 0.75)
F₀ = 110 Hz
the length of the organ for the first overtone, L = N-----N + N-----A
L = λ/2 + λ / 4
L = 3λ /4
F₁ = 3F₀
F₁ = 3 x 110
F₁ = 330 Hz
Thus the fundamental frequency for both organs is 110 Hz,
The first overtone for the organ open at both ends is 220 Hz
The first overtone for the organ open at one end is 330 Hz
The correct option is "A. 110Hz,220Hz, 330 Hz"