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;


Answer:
<h3>The answer is 15 N</h3>
Explanation:
The force acting on an object can be found by using the formula
<h3>Force = mass × acceleration</h3>
From the question
mass = 50 g = 0.05 kg
acceleration = 300 m/s²
We have
force = 0.05 × 300
We have the final answer as
<h3>15 N</h3>
Hope this helps you
The force applied would be 1.05*9.8 = 10.3 N
the pressure is equal to F/a
area will be πr^2 = 0.002826
thus pressure will be = 10.3/0.002826= 3644.72 N/m^2
Answer:
The height of the cliff is 90.60 meters.
Explanation:
It is given that,
Initial horizontal speed of the stone, u = 10 m/s
Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)
The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s
Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :



h = 90.60 meters
So, the height of the cliff is 90.60 meters. Hence, this is the required solution.
Answer:
-0.55m/s
Explanation:
Given that: For the boy
Weight = 745N
Velocity = +0.35 m/s
Mass of the boy = ?
g = 9.81m/s^2
W = mg
745 = m×9.81
m = 75.94kg
For the girl
Given that:
Weight = 477 N
g = 9.81m/s^2
m = ?
W = mg
477 = m×9.81m/s^2
m = 48.62kg
To solve for the v of the girl, the two has to add up
48.62kg×v + 75.94kg×+0.35 m/s = 0
48.62v + 26.579 = 0
48.62v = - 26.579
v = -26.579/48.62
v = -0.5466
v = -0.55m/s
Hence, the velocity of the girl is -0.55m/s.
The negative sign is as a result of the two of them moving is opposite direction.