The work done by the force is 47.1 J
Explanation:
The work done by a force in moving an object is given by
(1)
where
F is the magnitude of the force
d is the distance covered by the object
is the angle between the direction of the force and the motion of the object
In this problem, the force applied to the object is
F = 3.0 N
This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

And the distance covered is equal to the circumference of the circle, which is:

where r = 2.5 m is the radius.
Now we can substitute into eq.(1) to find the work done:

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Answer:
200 km\h
or 0.621 mp\h its the same speed just different mesuarements
The force needed to stretch the steel wire by 1% is 25,140 N.
The given parameters include;
- diameter of the steel, d = 4 mm
- the radius of the wire, r = 2mm = 0.002 m
- original length of the wire, L₁
- final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
- extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
- the Youngs modulus of steel, E = 200 Gpa
The area of the steel wire is calculated as follows;

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;


Thus, the force needed to stretch the steel wire by 1% is 25,140 N.
Learn more here: brainly.com/question/21413915
Answer:5000000000hertz
Explanation:
Wavelength=6cm=6/100 m=0.06m
Frequency=velocity/wavelength
Frequency=(3×10^8)÷0.06
Frequency=5000000000 hertz
Answer:
See Explanation
Explanation:
a) We know that;
v = λf
Where;
λ = wavelength of the wave
f = frequency of the wave
v = velocity of the wave
So;
T = 2 * 2.10 s = 4.2 s
Hence f = 1/4.2 s
f = 0.24 Hz
The wavelength = 6.5 m
Hence;
v = 6.5 m * 0.24 Hz
v = 1.56 m/s
b)The amplitude of the wave is;
A = 0.600 m/2 = 0.300 m
c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same
Where d = 0.30 m
A = 0.30 m/2 = 0.15 m