Approximately 15 m/s is the speed of the car.
<u>Explanation:</u>
<u>Given:</u>
speed of sound - 343 m/s
You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,


If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,

Where,
- Observed frequency
f – Actual frequency
v – Velocity of sound waves
– Velocity of observer
- velocity of source
When source moves away from an observer at rest (
), the equation would be



By substituting the known values, we get






Approximately 15 m/s is the speed of the car.
Answer: 178.25*10^-6 T
Explanation: In order to solve this problem we have to take into account the equilibrium between the electric and magnetic forces in the electron, so it is given by:
Fm=evB
Fe=eE so
evB=eE the we have
v=E/B
Firsly we calculate the velocity of the electron before to get the parallel plates at 100V
eΔV=1/2*m*v^2 then
v=(2*eΔV/m)^1/2
v=(2*1.6*10^-19*3.1*10^3/9.1*10^-31)^1/2=33 *10^6 m/s
Then we can calculate B
B=E/v E.d=V where d is the separation between the plates and V is equal a 100V
B=V/(d*v)=100/(17*10^-3*33 *10^6)=178.25*10^-6 T
If they both start from the same height, then they both hit the ground at the
same time. It makes no difference if their horizontal speeds aren't equal.
The cannon ball still accelerates downward at the same rate as the baseball.
Answer:11.7 meters
Explanation: Gravitational acceleration (g)
9.8
m/s²
Initial velocity (v₀)
0
ft/s
Height (h)
11.77225
m
Time of fall (t)
1.55
sec
Velocity (v)
15.19
m/s
Answer:
The distance traveled by the ball is 8.5 m
Explanation:
Initial height of the ball, h₁ = 1.5 m above the ground
final height of the ball, h₂ = 5m
Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m
Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m
Total distance traveled = upward distance + downward distance
Total distance traveled = 3.5 m + 5m = 8.5 m
Therefore, the distance traveled by the ball is 8.5 m