Answer:
a)52.58 m/s
b)56.13°
Explanation:
assume the upward direction as positive
x-component of the velocity = 29.3×cos33.6°=24.40 m/s (remain constant)
y-component of the velocity which is -29.3sin33.6°= -16.21 m/s
time of flight = 68.3/24.40= 2.7991 seconds
now, we can obtain final velocity in y-direction
=43.66 m/s
=52.58 m/s
for direction
56.13° from the horizontal
I'd say b, precise, here.
If there's an error somewhere in the experiment or project, then it is consistently .... wrong. So, just 'cos you measure something precisely, it doesn't mean that you've measured it accurately. Maybe an example would be a measurement of length. If you used a metal ruler at zero degrees C, you can measure to say half a millimetre. A series of measurements of the same object would give very similar readings. But, if you used same metal ruler at, say 100 celsius (implausible) then you'd probably get a different set of readings. 'cos of the expansion of the metal ruler.
Brass is an alloy, a combination of two metals, and aluminum is a pure metal
Answer: option d. Explanation:1) The
direction of the
field lines inform about the
sign of the charges.
The field lines <span>
extend from the positive charges to the negative charges, so you can conclude that the charge C is positve and both charge A and charge B are negative:
</span><span>
</span><span>
</span><span>Charge C: positive
</span><span>
</span><span>Charge A: negative
</span><span>
</span><span>Charge B: netative
</span>
2) The
density of the lines (number of lines in a region) inform about the
magnitude of the electric field.
Since the charges are at the same distance, the magnitude of the electric field informs directly about the magnitude of the force and that about the magnitude of the charges.
Since, there are the
double of lines between C and B than between C and A, the magnitude of
charge B is the double than the magnitud of charge A.
From the five options given (a throug e) the only that is consistent with that charges A and B have the same sign, that charge C has different sign, and that charge B is the double of charge A is: