Considering the magnetic force is providing the centripetal force,
we get, mv^2/r = qvB
so, r = mv/qB
as the Kinetic energy, K of the particle is given by.5mv^2
so, v = sqrt(2K/m)
substituting the in radius equation,
<span>r = m*sqrt(2K/m)/qB
r = sqrt(2Km)/qB</span>
Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:
where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by:
MA = (Effort Distance)/(Effort Resistance) = L/H
L = MA * H = 5 * 8" = 40"
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
Answer:
Explanation:
Given that,
An object starts from rest at the origin (x₀ = 0) and travels in a straight line with a constant acceleration (a = constant).
The relation between the position (x) and time (t) is given by :
Let v is the velocity of the object.
Let a is the acceleration of the object. It is given by :
So, the acceleration of the object is . Hence, this is the required solution.