Answer:
The speed of ejection is 
Solution:
As per the question:
Magnetic field density, B = 0.4 T
Density of the material in the sunspot, 
Now,
To calculate the speed of ejection of the material, v:
The magnetic field energy density is given by:

This energy density equals the kinetic energy supplied by the field.
Thus


where
m = mass of the sunspot in
= 


Answer: the correct answer is 7.8026035971 x 10^(-13) joule
Explanation:
Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get
m = 4.00260 u + 222.01757 u = 226.02017 u .
The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is
Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,
which is equivalent to an energy change of
Delta E = (0.00523 u)*(931.5MeV/1u)
Delta E = 4.87 MeV
Converting 4.87 MeV to Joules
1 joule [J] = 6241506363094 mega-electrón voltio [MeV]
4 mega-electrón voltio = 6.40870932 x 10^(-13) joule
4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule
Answer:
The force applied on the big piston is 1306.67 N
Explanation:
Given;
force applied on small piston, F₁ = 200 N
diameter of the small piston, d₁ = 4.37 cm
radius of the small piston, r₁ = d₁/2 = 2.185 cm
Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²
Area of the big piston, A₂ = 98 cm²
The pressure of the piston is given by;

Where;
F₂ is the force on big piston

Therefore, the force applied on the big piston is 1306.67 N
Answer:
1410 Hz
Explanation:
Capacitance is reduced by 2, so the angular frequency will increase by a factor of
.
Answer:
ball clears the net
Explanation:
= initial speed of launch of the ball = 20 ms^{-1}
= angle of launch = 5 deg
Consider the motion of the ball along the horizontal direction
= initial velocity = 
= time of travel
= horizontal displacement of the ball to reach the net = 7 m
Since there is no acceleration along the horizontal direction, we have
Eq-1
Consider the motion of the ball along the vertical direction
= initial velocity = 
= time of travel
= Initial position of the ball at the time of launch = 2 m
= Final position of the ball at time "t"
= acceleration in down direction = - 9.8 ms⁻²
Along the vertical direction , position at any time is given as

Since Y > 1 m
hence the ball clears the net