How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 7.1 % increase in the emitted frequency?

Answer: Kinetic Energy to Electrical.

Explanation: The magnet is rotated as a result of the spinning wheels, and this results in a powerful stream of electrons, therefore converting kinetic to electrical.

Answer:

The minimum coefficient of static friction required, µ = 0.10

Note. The question is incomplete. The complete question is given below:

While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.

The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

Answer:

6.77m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the velocities before collision

v is the final collision

Given

m1 = 300g  = 0.3kg

u1 = 6.0m/s

m2 = 10g = 0.01kg

u2 = 30m/s

Required

The bird's speed immediately after swallowing v

Substitute the given values into the formula

m1u1 + m2u2 = (m1+m2)v

0.3(6) + 0.01(30) = (0.3+0.01)v

1.8+0.3 = 0.31v

2.1 = 0.31v

v = 2.1/0.31

v = 6.77m/s

Hence the bird's speed immediately after swallowing is 6.77m/s

Answer:

here u go

Explanation:

Earth's rotation is the rotation of planet Earth around its own axis. Earth rotates eastward, in prograde motion. As viewed from the north pole star Polaris, Earth turns counterclockwise.

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

[tex]specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3[/tex]

Fraction of the object's weight below the surface of water is calculated as;

[tex]= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}[/tex]

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

Answer:

880J/kelvin

Explanation:

Q =MC ×change in t

c =C/m

C=Q/change in t

c= Q/ m× change in t

c = 26400 / 2.0 × 15

c = 880 J/kelvin