Answer:
C
Explanation:
Answer is C.
Answer:
(a): should be pumped out of the room 18 L/min to keep the water level constant.
(b): should be pumped out of the room 78 L/min to reduce the water level by 4 cm/hr.
Explanation:
S= 90 m²
rate= 1.2 cm/hr = 0.012 m/hr = 0.0002 m/min
Water leak= S*rate= 90 m² * 0.0002 m/min
Water leak= 0.018 m³/min * 1000 L/m³
Water leak= 18 L/min (a) Water should be pumped out to keep the level constant.
By the rule of 3:
1.2 cm/hr ------------- 18 L/min
(4+1.2) cm/hr -------- x= 78 L/min (b) Water should be pumped out to reduce the level by 4 cm/hr.
The new length of aluminum rod is 17.435 cm.
The linear expansion coefficient is given as,
Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.
and linear expansion coefficient is
Substitute,
Hence, The new length of aluminum rod is 17.435 cm.
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Answer:
the new length is 17.435cm
Explanation:
the new length is 17.435cm
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Answer:
Explanation:
A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of 1 = 0.0128(kg/m). It’s displacement function is D(x,t) = Acos(kx - t). It’s amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with 2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and , for the original wave, as well as k and the reflected JJJJJJave and the transmitted wave.
track with a radius of 30 meters. What
is the car's rate of centripetal
acceleration?
The car's rate of centripetal acceleration in the circular path is 4.8 m/s².
The given parameters;
The centripetal acceleration of the car is calculated as follows;
where;
Substitute the given parameters and solve for the centripetal acceleration;
Thus, the car's rate of centripetal acceleration is 4.8 m/s².
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Answer:
https://youtu.be/stW-C7F7QOg
Answer:
f3 = 102 Hz
Explanation:
To find the frequency of the sound produced by the pipe you use the following formula:
n: number of the harmonic = 3
vs: speed of sound = 340 m/s
L: length of the pipe = 2.5 m
You replace the values of n, L and vs in order to calculate the frequency:
hence, the frequency of the third overtone is 102 Hz