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Reil [10]
4 years ago
5

Which is an example of a superconductor?

Physics
1 answer:
astra-53 [7]4 years ago
3 0
Diamond,Aluminium,<span>Gallium</span>
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A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
What makes these image a represent of what research <br>​
ycow [4]
The economic down fall
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3 years ago
The process of heat radiation is the ONLY method of heat transfer that can occur
lina2011 [118]
Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. so...<span> electromagnetic waves ?</span>
5 0
4 years ago
Read 2 more answers
a 1.05 kg water bottle is sitting on the teacher's desk it is .20 m tall and has a radius of .03 m find the force that the water
arsen [322]
The force applied would be 1.05*9.8 = 10.3 N
the pressure is equal to F/a
area will be πr^2 = 0.002826
thus pressure will be = 10.3/0.002826= 3644.72 N/m^2
4 0
3 years ago
To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
8 0
3 years ago
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