Yes, but it's a null vector .So,no vector can have a component at right angle to itself unless it is a zero vector.
Explanation:
The given data is as follows.
mass (m) = 0.160 kg, spring constant (k) = 8 n/m,
Maximum speed (
) = 0.350 m/s
Formula for angular frequency is as follows.
= 7.07 rad/sec
(a) Formula to calculate the amplitude is as follows.
A = 
= 
= 0.05 m
Hence, value of amplitude is 0.05 m.
(b) Displacement = 0.030 m
Formula for mechanical energy is as follows.
M.E = 
Putting the values into the above formula as follows.
M.E =
= 
= 
Joule
For x = 0.03,
As, P.E = 
= 
= 
Hence, calculate the kinetic energy as follows.
K.E = M.E - P.E
= ( - ) J
= 
J
Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is J.
In projectile motion horizontal motion has zero acceleration and when there is no air resistance acting on the object the motion will be uniform. Motion in the vertical direction is accelerated downwards to to the acceleration due to gravity
We are asked to compare two series circuits having equal number of light bulbs.
1st circuit is powered by 6 batteries each having a voltage of 1.5V
2nd circuit is powered by a single battery having a voltage of 9V.
The six batteries in the 1st circuit can be connected together in series or in parallel.
When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means
Voltage of pack = number of batteries*voltage of each battery
Voltage of pack = 6*1.5
Voltage of pack = 9 volts
But the current remains same in the series connection since there is only path for the current to flow.
On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.
Circuit 1:
In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Circuit 2:
In this circuit, we have 1 battery which provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit just like in circuit 1.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Conclusion:
Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.
