Answer:
Both technicians are right, to be able to make a threaded joint you need to use the external thread on one part of the rod using the tap and die set, and on the other side of the rod you need to have an internal thread using the thread repair insert kit
Answer:
Total number of lamps will be 4
Explanation:
We have given power of the lamp W = 400 watt
Potential difference across the lamp V=110 volt
We know that power is equal to 
So 
Total current is given 15 A
As it is given that lamps are connected in parallel so total current is the sum of current through each lamp
So number of lamp will be 
As the lamp can not be in negative
So total number of lamps will be 4
Answer:
You have a displacement of 5 units to the right.
Explanation:
First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.
The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let 
be the masses of the three fragments.
Let 
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,
So the x- component of the velocity of the m2 fragment after the explosion is,
∴ 
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
