Answer:
r = 0.114 m
Explanation:
To find the speed of the proton, from conservation of energy, we know that
KE = PE
Thus, we have;
(1/2)mv² = qV
Where;
V is potential difference = 1kv = 1000V
q is charge on proton which has a value of 1.6 x 10^(-19) C
m is mass of proton with a constant value of 1.67 x 10^(-27) kg
Let's make the velocity v the subject;
v =√(2qV/m)
v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))
v = 4.377 x 10^(5) m/s
Now to calculate the radius of the circular motion of charge we know that;
F = mv²/r = qvB
Thus, mv²/r = qvB
Divide both sides by v;
mv/r = qB
Thus, r = mv/qB
Value of B from question is 0.04T
Thus,
r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)
r = 0.114 m
r = 8.76 m
<span>v = 25.0 km/</span><span>h = 25*5/18 m/s = 6.94 m/s
</span><span>centripetal force = mv²/r = 1275*6.94²/40 = 1537.18 N </span>
Answer:
Time period of the motion will remain the same while the amplitude of the motion will change
Explanation:
As we know that time period of oscillation of spring block system is given as

now we know that
M = mass of the object
k = spring constant
So here we know that the time period is independent of the gravity
while the maximum displacement of the spring from its mean position will depends on the gravity as


so we can say that
Time period of the motion will remain the same while the amplitude of the motion will change
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series

In parallel


Solving the above quadratic equation


∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm