Answer:
a= 3.49 m/s^2
Explanation:
magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.
we know that tangential acceleration a_t= change in velocity /time taken
now 90 km/h = 25 m/s
a_t = 25/17 = 1.47 m/s^2.
radial acceleration a_r = v^2/r
v= a_t×t = 1.47×13 = 19.11 m/s
a_r = 19.11^2/115= 3.175
now,
a= 3.49 m/s^2
Answer:
4.4345× 10^-7V
Explanation:
The computation of the half voltage for a 1.2T magnetic field applied is shown below
The volume of one mole of copper is
v = m ÷p
= 63.5 ÷ 8.92
= 7.12cm
Now the density of free electrons in copper is
n = Na ÷ V
= 6.02 × 10^23 ÷ 7.12
= 8.456× 10^28/m^3
Now the half voltage is
= IB ÷ nqt
= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)
= 4.4345× 10^-7V
A) 4.7 cm
The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is
where
n is the order of the maximum
is the wavelength
is the distance between the slits
In this problem,
n = 5
So we find
And given the distance of the screen from the slits,
The distance of the 5th bright fringe from the central bright fringe will be given by
B) 8.1 cm
The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:
To find the angle corresponding to the 8th dark fringe, we substitute n=8:
And the distance of the 8th dark fringe from the central bright fringe will be given by
The centripetal acceleration a is 4.32 10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4 10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4 10^-3)^2 / 0.037
a = 4.32 10^-4 m/s^2.