Given Information:
Frequency of horn = f₀ = 440 Hz
Speed of sound = v = 330 m/s
Speed of bus = v₀ = 20 m/s
Answer:
Case 1. When the bus is crossing the student = 440 Hz
Case 2. When the bus is approaching the student = 414.9 Hz
Case 3. When the bus is moving away from the student = 468.4 Hz
Explanation:
There are 3 cases in this scenario:
Case 1. When the bus is crossing the student
Case 2. When the bus is approaching the student
Case 3. When the bus is moving away from the student
Let us explore each case:
Case 1. When the bus is crossing the student:
Student will hear the same frequency emitted by the horn that is 440 Hz.
f = 440 Hz
Case 2. When the bus is approaching the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330+20 )
f = 440 ( 330/ 350 )
f = 440 ( 0.943 )
f = 414.9 Hz
Case 3. When the bus is moving away from the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330-20 )
f = 440 ( 330/ 310 )
f = 440 ( 1.0645 )
f = 468.4 Hz
Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.
From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.
Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.
Learn more about length on:
brainly.com/question/25292087
Answer:
41.8m/s^2
Explanation:
Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s
From the equations of motion, v = u + at
a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2
Answer:
Explanation:
DetaM=4 x 1.02875 - 4.002603
DetaM= 0.028697u
Using E= mc²
= 0.028697 x 1.49x*10^-10
= 4.2x10^-12J
Answer:
here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location
Explanation:
As we know that gravitational field is defined as the force experienced by the satellite per unit of mass
so we will have
now in order to find the acceleration of the satellite we know by Newton's II law
so we will have
so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location
