Answer:
Explanation:
In magnetic field , charged particle will have circular path . Let the radius of their circular path be r₁ and r₂ . Let their velocity at the time of entering magnetic field be v₁ and v₂ .
The velocity with which they will come out of electric field can be measured from following equation
Eq = 1/2 m v² , E is electric field , q is charge on the particle , m is mass and v is velocity .
v² = 2Eq / m
radius of circular path can be measured by the following expression
m v² / r = Bqv
2Eq / r = Bqv
r = 2Eq / Bqv
= 2E / Bv
r² = 4E² / B²v²
= 4E²m / B²x 2Eq
since E , B and q are constant
r² = K . m
r₂² / r₁² = m₂ / m₁
1.5²
m₂ / m₁ = 1.5²
= 2.25
Answer:
The Required horizontal force is 230.04N
Explanation:
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
where:
F_{Hn} is the required Force
u is the friction coefficient
m is the mass
g is gravitational acceleration=9.8m/s^2
Eq (1)
Now, weight increases by 42% and friction coefficient decreases by 19%
New weight=(1.42*m*g) and new friction coefficient=0.81u
Eq (2)
Divide Eq(2) and Eq (1)
The Required horizontal force is 230.04N
<span>Lifting an object increases the gravitational potential energy of the system. If you release the object, that potential energy will be transformed into the energy of an object in motion which is termed as the kinetic energy as it falls toward earth. Hope this helps.</span>
