Answer:
Mass of the box = 0.9433 kg
Explanation:
Mass of racket-ball 
= 0.00427 kg
Velocity of racket-ball before collision 
= 22.3 m/s
Velocity of racket-ball after collision with box 
= -11.5 m/s
[Since ball is bouncing back, so velocity is taken negative.]
Velocity of the box before collision 
= 0 m/s
<em>[Since the box is stationary, so velocity is taken zero]</em>
Velocity of box moving forward after collision 
= 1.53 m/s
To find the mas of the box 
.
By law of conservation of momentum we have:
Momentum before collision = Momentum after collision
This can be written as:
We can plugin the given value to find
Adding both sides by 0.4911
Dividing both sides by 1.53.
∴ 
kg
Mass of the box = 0.9433 kg (Answer)
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
Answer:
0.4 ohms.
Explanation:
From the circuit,
The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.
1/R' = 1/R1+1/R2
R' = (R1×R2)/(R1+R2)
R' = (2.4×1.2)/(2.4+1.2)
R' = 2.88/3.6
R' = 0.8 ohms.
Hence the current flowing through the circuit is
I = V'/R'................ Equation 1
Where V' = voltmeter reading
I = 6/0.8
I = 7.5 A
This is the same current that flows through the variable resistor.
Voltage drop across the variable resistor = 9-6 = 3 V
Therefore, the resistance of the variable resistor = 3/7.5
Resistance = 0.4 ohms.
Answer:
1)
2)displacement
3)
Explanation:
At equilibrium position the weight of the man should be balanced by force in the spring
thus we have at equilibrium
Applying values we get
2)
When we add another identical spring we get an equivalent spring with spring constant as
Applying values we get
Thus at equilibrium we have
3) Equivalent spring constant will be as calculated earlier
