The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = ( - ) / ( + )
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
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Force on -7 uC charge due to charge placed at x = - 10cm
now we will have
towards left
similarly force due to -5 uC charge placed at x = 6 cm
now we will have
towards left
Now net force on 7 uC charge is given as
towards left
<span>When the difference between two results is larger than the estimates error, the result is</span>
A. and C. are both correct. Work is force x distance, and 1 J is 1 N-m .
Answer:
39.9 V
Explanation:
Given that
Capacitor are indentical that is why the value of C will be same for both the capacitor.
We know that energy stored in the capacitor is given as
V=Voltage
C=Capacitance
For A :
E₁= 2 x 10⁻³ J
Voltage = V₁
For B :
E₂=1.8 x 10⁻⁴ J
V₂ = 12 V
Now by using above two equation
Therefore the voltage on capacitor A will be 39.9 V