Answer:
The correct answer is B)
Explanation:
When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.
The formula for calculating the velocity of a point on the edge of the wheel is given as
= 2π r / T
Where
π is Pi which mathematically is approximately 3.14159
T is period of time
Vr is Velocity of the point on the edge of the wheel
The answer is left in Meters/Seconds so we will work with our information as is given in the question.
Vr = (2 x 3.14159 x 1.94m)/2.26
Vr = 12.1893692/2.26
Vr = 5.39352619469
Which is approximately 5.39
Cheers!
To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as

Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia


The value for our angular velocity is not in SI, then


Replacing our values we have that


The precession frequency is




Therefore the precession period is 5.4s
Answer:
The answer to the question is as follows
The acceleration due to gravity for low for orbit is 9.231 m/s²
Explanation:
The gravitational force is given as

Where
= Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by
and = 1 kg
rEarth = The distance between the two masses = 6.40 x 10⁶ m
therefore at a height of 400 km above the erth we have
r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m
and
=
= 9.231 N
Therefore the acceleration due to gravity =
/mass
9.231/1 or 9.231 m/s²
Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²
Speed uses distance and velocity uses displacement in its calculation.
For 100 m race, distance = displacement. Hence speed = velocity
For 400m race, distance ≠ displacement. distance = 400m whereas displacement = 0m. Hence speed ≠ velocity
Answer:
1. a
2. b
3. b
Explanation:
1.
Resistance is the property of a conductor to offer resistance to the flow of current. The lower the resistance better is the conductivity of wire.
We know that the resistance of a wire depends on several factor which are inter-connected by an equation as:
where:
R = resistance of the wire
length of the wire
cross sectional area of the wire
from the above relation we observe that

- Also when the temperature of the wire is significantly high then the lattice vibration cause obstruction in the path of the flowing charges and reduce the current flow.
2.
As the collision between the electrons and protons increases the speed of the flow of charges will decrease because the opposite charges attract each other and as we know that electrical current is the rate of flow of charge.
3.
Heating up of wire due to sunlight will cause lattice vibration in the conductor and will obstruct the movement of the charges which build up electric current, hence increasing the resistance of conductivity.