Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, = 39 kHz
velocity of the bat, 
= 8.32 m/s
speed of sound in air, v = 340 m/s
The apparent frequency of sound striking the wall is calculated as;
The frequency reflected by the stationary wall to the bat is calculated as;
<span>When temperature is increased,
the rate of dissolving increases. The kinetic energy of the molecules of the
solute and solvent molecules is high thereby increasing their contact. An example
is mixing powdered sugar to the water. When you add water to the sugar, the
dissolving process is slow. However, when you increase the temperature of the
water by boiling it, the sugar dissolves immediately. </span>
Answer:
C) 0m
Explanation:
Since at the end of the day, it was not displaced
Displacement ti's a vector quantity
Answer:
The length of rod A will be <u>greater than </u>the length of rod B
Explanation:
We, know that the formula for final length in linear thermal expansion of a rod is:
L' = L(1 + ∝ΔT)
where,
L' = Final Length
L = Initial Length
∝ = Co-efficient of linear expansion
ΔT = Change in temperature
Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.
For Rod A:
∝₁ = 12 x 10⁻⁶ °C⁻¹
For Rod B:
∝₂ = β₂/3
where,
β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹
Therefore,
∝₂ = 24 x 10⁻⁶ °C⁻¹/3
∝₂ = 8 x 10⁻⁶ °C⁻¹
Since,
∝₁ > ∝₂
Therefore,
L₁ > L₂
So, the length of rod A will be <u>greater than </u>the length of rod B
15x 9.8 x .3 = 44.1
15 x 9.8 x 1 = 147
147 - 44.1 = 102.9J
102.9 J
