Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Answer:
v = 0.489 m/s
Explanation:
It is given that,
Mass of a box, m = 1.5 kg
The compression in the spring, x = 6.5 cm = 0.065 m
Let the spring constant of the spring is 85 N/m
We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.
So, the speed of the box is 0.489 m/s.
Answer:
Explanation:
F = mω²R
F = 15(2π/8.5)²(7.8)
F = 63.93044788...
F = 63.9 N
answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.
Answer:
1 = 5.4 J
2 = 0.1979 C
3 = 5
Explanation:
Energy in a capacitor, E is
E = 1/2 * C * V²
E = 1/2 * 3000*10^-6 * 60²
E = 1/2 * 3000*10^-6 * 3600
E = 1/2 * 10.8
E = 5.4 J
E = Q²/2C = 6.53 J
E * 2C = Q²
Q² = 6.53 * 2 * 3000*10^-6
Q² = 13.06 * 3000*10^-6
Q² = 0.03918
Q = √0.03918
Q = 0.1979 C
The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.
As you crank the variable resistor to higher resistance, the total current in the loop decreases. The power dissipated by the light bulb ... the heat and brightness it produces ... depend directly on the current through it, so they decrease too. Your circuit is a perfect incandescent light dimmer circuit.
