Answer:
Neither technician
Explanation:
Neither technician is correct.
two bulbs are connected in series one bulb burn out
If one bulb in the series burns out then the circuit will break.
In a series circuit same current passes each resistor.
so, both the technician is incorrect bulb B will not work and current will not increase in the other bulb.
After the collision the magnitude of the momentum of the system is Mv
Given:
mass of 1st object = M
speed of 1st object = v
mass of 2nd object = M
speed of 2nd object = 0
To Find:
magnitude of the momentum after collision
Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.
Applying conservation of linear momentum
Mv + M(0) = 2MV
Mv = 2MV
V = v/2
So, after collision momentum is
p = 2MV = 2xMxv/2 = Mv
So, after collision momentum is Mv
Learn more about Momentum here:
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All matter is made of particles; these can be single atoms or atoms chemically joined to make molecules. Using this fact, matter can be classified into three broad groups: elements, compounds and mixtures. In an element, all the atoms are of the same type. If more than one type of atom is chemically joined, then a compound has been formed. If more than one type of atom or molecule is contained in the same substance, and the particles aren't chemically joined, this is a mixture.
Answer: The frequency heard will be f = 275.675Hz
Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.
The formula to calculate the Doppler effect is as follows
f = (
) · f₀
f is the observed frequency;
c is the speed of sound;
Vs is velocity of the source;
f₀ is the emitted frequency of source;
Substituting and calculating,
f = 
· 300
f = 275.675 Hz
Thus, the frequency heard by the police officer is 275.675Hz.
