Oooo that ones hard. ummm... idk i think we should just leave it to the experts ya know.
Answer: C
X = Displacement of the spring
Hooke's law: It states that the applied force F is proportional to the displacement of spring .
F ∝ x
Where, x = displacement of spring in meters
F = force, measured in Newtons
In another words The force F is equal to the constant K times the disparagement.
F = k.x
Where k is constant and it depends on elastic material.
Spring has restorative force.
If the spring moves in opposite direction then,
F = - k.x
A negative sign indicates that the spring resists and force is to the left. The compression of the spring is greater than the restoring force.
Example: A mass 'm' stretches a spring at a displacement x.
Answer:
v=32.49 m/s
Explanation:
Given that
Distance ,d= 66 m
Initial speed of the car ,u = 0 m/s
Coefficient of friction ,μ = 0.8
Lets take the total mass of the car = m
The acceleration of the car is given as
a = μ g ( g= 10 m/s² )
Now by putting the values in the above equation we get
a= 0.8 x 10 m/s²
a= 8 m/s²
We know that ,final speed is given as
v²= u ²+ 2 a d
Now putting the value
v²=0² + 2 x 8 x 66
v²= 1056
v=32.49 m/s
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;
Therefore, the magnitude of the electric field is 0.1108 N/C
Answer:
Answer B
Explanation:
An increase in resistance makes it harder for the electric current to pass through
