Answer : The unbalanced force necessary to produce the movement is, 117.6 N
Solution :
Force : It is defined as the force is equal to the product of the mass of an object and the acceleration of an object.
Formula used :
where,
F = force
m = mass = 98 Kg
a = acceleration =
Now put all the given values in the above formula, we get the unbalanced force necessary to produce the movement.
Therefore, the unbalanced force necessary to produce the movement is, 117.6 N
Answer:
rate of fission =5.89*10^3 1\Year
Explanation:
we know that
rate of fission is given as
mass = 1.0 g
avogardo number = 6.02*10^23
molar mass of isotopes 235U =235
Putting all value to get rate of emission
rate of fission
rate of fission =5.89*10^3 1\Year
Answer:
1) Charge chord resistance is 75 Ω
2) Charge chord resistance is 6.33 Ω
Explanation:
1) To answer the question, we note that the the formula voltage is found as follows;
V = IR
Therefore,
2) Where the voltage, V = 19.5 V and the current, I = 3.33 A, we have;
Initial resistance R₁ = 19.5 V/(3.33 A) = 5.86 Ω
However, to reduce the current to 1.6 A, we have;
Therefore, where the resistance is found by the sum of the total resistance we have;
= R₁ + Charge chord resistance
∴ 12.1875 = 5.86 + Charge chord resistance
Hence, charge chord resistance = 6.33 Ω
Answer:
U = 1794.005 × 10⁶ J
Explanation:
Data provided;
Capacitance of the original capacitor, C = 1.27 F
Potential difference applied to the original capacitor, V = 59.9 kV
= 59.9 × 10³ V
Now,
The Potential energy (U) for the capacitor is calculated as:
Potential energy of the original capacitor, U = × C × V²
on substituting the respective values, we get
U = × 1.27 × ( 59.9 × 10³ )²
or
U = 1794.005 × 10⁶ J
Answer: The spring of the spring is 25 N/m.
Explanation:
Mass of the body = 25 g= 0.025 kg (1 kg = 1000 g)
Oscillation is 4 sec = 20
Oscillation in 1 sec =
Frequency of the vibration of the spring =
Force constant can be calculated bu using the relation between the frequency and, mass and spring constant 'k'
The spring of the spring is 25 N/m.