1. During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the tre
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Answer:
(i) See attached image for the drawing
(ii) net force given in component form: (20, 20)N with magnitude:
Explanation:
First try to write all forces in vector component form:
The force F1 acting at 45 degrees would have multiplication factors of on both axes, to take care of the sine and cosine projections. Therefore, the:
x-component of F1 is
y-component of F1 is
As far as force F2, it is given already in x and y components, then:
x-component of F2 = 8 N
y-component of F2 = -6 N (negative meaning pointing down the y-axis)
Force F3 has only component (upwards) in the y-direction
x-component of F3 = 0 N
y-component of F3 =14 N
The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:
x-component of R = 12 + 8 = 20 N
y-component of R = 12 + 14 - 6 = 20 N
Therefore, the acceleration that the mass receives due to this force is given in component form as:
x-component of acceleration: 20 N / 5 kg =
y-component of acceleration: 20 N / 5 kg =
Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:
x-component of the velocity is:
y-component of the velocity is:
Answer:
λ = 5.85 x 10⁻⁷ m = 585 nm
f = 5.13 x 10¹⁴ Hz
Explanation:
We will use Young's Double Slit Experiment's Formula here:
where,
λ = wavelength = ?
Y = Fringe Spacing = 6.5 cm = 0.065 m
d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m
L = screen distance = 5 m
Therefore,
<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>
Now, the frequency can be given as:
where,
f = frequency = ?
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>f = 5.13 x 10¹⁴ Hz</u>
Explanation:
Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.
Hence, = 0. And, as
= 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.
Therefore,
Q = -(19000 J + 2000 J)
= -21000 J
or, |Q| = 21000 J
Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.