Michael is doing an experiment. He wants to drop a bowling ball and a stuffed bear out the window of a tall building. Under what
1 answer:
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Answer:
a)Amplitude ,A = 2 mm
b)f=95.49 Hz
c)V= 30 m/s ( + x direction )
d) λ = 0.31 m
e)Umax= 1.2 m/s
Explanation:
Given that
As we know that standard form of wave equation given as
A= Amplitude
ω=Frequency (rad /s)
t=Time
Φ = Phase difference
So from above equation we can say that
Amplitude ,A = 2 mm
Frequency ,ω= 600 rad/s (2πf=ω)
ω= 2πf
f= ω /2π
f= 300/π = 95.49 Hz
K= 20 rad/m
So velocity,V
V= ω /K
V= 600 /20 = 30 m/s ( + x direction )
V = f λ
30 = 95.49 x λ
λ = 0.31 m
We know that speed is the rate of displacement
The maximum velocity
Umax = 1200 mm/s
Umax= 1.2 m/s
Answer:
B. +m
Explanation:
The magnification of an image is defined as the ratio between the size of the image and of the object:
where we have
y' = size of the image
y = size of the object
There are two possible situations:
- When m is positive, y' has same sign as y: this means that the image image is upright
- When m is negative, y' has opposite sign to y: this means that the image is upside down
Therefore, the correct option representing an upright image is
B. +m
Answer:
A. The work done during the process is W = 0
B. The value of heat transfer during the process Q = 442.83
Explanation:
Given Data
Initial pressure = 100 k pa
Initial temperature = 25 degree Celsius = 298 Kelvin
Final pressure = 300 k pa
Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.
⇒ =
------------- (1)
Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.
⇒ P ∝ T
⇒ =
⇒ Put all the values in the above formula we get the final temperature
⇒ =
× 298
⇒ = 894 Kelvin
(A). Work done during the process is given by W = P × ()
From equation (1), =
so work done W = P × 0 = 0
⇒ W = 0
Therefore the work done during the process is zero.
Heat transfer during the process is given by the formula Q = m (
-
)
Where m = mass of the gas = 1 kg
= specific heat at constant volume of nitrogen = 0.743
Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )
⇒ Q = 442.83
Therefore the value of heat transfer during the process Q = 442.83