The acceleration of the box as it slides down the ramp is 4.9 m/s².
<h3>
What is the acceleration of the box down the incline?</h3>
The acceleration of the box down the incline is determined by applying Newton's second law of motion as shown below;
F - Ff = ma
where;
- F is the parallel force on the box
- Ff is the frictional force on box = 0
- m is the mass of the box
- a is the acceleration of the box
F - 0 = ma
F = ma
mg sinθ = ma
g sinθ = a
where;
- g is acceleration due to gravity
- θ is the angle of inclination of the incline
a = 9.8 x sin(30)
a = 4.9 m/s²
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Answer:
Explanation:
a ) wave length of waves in water
= velocity / frequency
= 1482 / (18 x 1000)
= .0823 m
= 8.23 cm
b ) Applying Doppler's effect relation
frequency of reflected wave
= 18000 [ 1482 / (1482 - 4.95 ) ]
= 18000 x 1.003385
= 18061
Difference in frequency= 61 Hz
Answer:
f = 614.28 Hz
Explanation:
Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :
f = 614.28 Hz
So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.
Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg
