Answer: V = 15 m/s
Explanation:
As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,
F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz
Using doppler effect formula
F = C/ ( C - V) × f
Where
F = observed frequency
f = source frequency
C = speed of light = 3×10^8
V = speed of the car
Substitute all the parameters into the formula
2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10
2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)
1.000000049 = 3×10^8/(3×10^8 - V)
Cross multiply
300000014.7 - 1.000000049V = 3×10^8
Collect the like terms
1.000000049V = 14.71429
Make V the subject of formula
V = 14.71429/1.000000049
V = 14.7 m/s
The speed of the car is 15 m/s approximately.
Answer:
<h2>
14.66secs</h2>
Explanation:
Given the formula for calculating the depth in metres expressed as
depth in meters = ½ (1500 m/sec × Echo travel time in seconds)
Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.
10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds
10,994 = 750 * Echo travel time in seconds
Dividing both sides by 750;
Echo travel time in seconds = 10,994 /750
Echo travel time in seconds ≈ 14.66secs (to two decimal places)
Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
You can use the displacement method or the eureka can so basically in the displacement can what you have to do is to put some water into a measuring cylinder and measure its volume before adding the irregular shaped object and then measuring the level of water which had been displaced and then eureka can you can check online
