increased with an increased current flow
No, it will only melt if the temperature is lowered. If you compress it, it will change the shape, but it will not change the state it is in (i.e. solid).
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:
a = 1.764m/s^2
Explanation:
By Newton's second law, the net force is F = ma.
The equation for friction is F(k) = F(n) * μ.
In this case, the normal force is simply F(n) = mg due to no other external forces being specified
F(n) = mg = 15kg * 9.8 m/s^2 = 147N.
F(k) = F(n) * μ = 147N * 0.18 = 26.46N.
Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.
Thus, F(net) = F(k) = ma
26.46N = 15kg * a
a = 1.764m/s^2
