When calculated energy transferred between objects use the definition of heat as
Answer:
a ) 1.267 radian
b ) 1.084 10⁻³ mm
Explanation:
Distance of screen D = 1.65 m
Width of slit d = ?
Wave length of light λ = 687 nm.
Distance of second minimum fro centre y = 2.09 cm
Angle of diffraction = y / D
= 2.09 /1.65
= 1.267. radian
Angle of diffraction of second minimum
= 2 λ / d
so 2 λ / d = 1.267
d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm
=1084.45 nm = 1.084 x 10⁻³ mm.
Answer:
wavelength decreases and frequency increase
Explanation:
the higher the wavelength the smaller the frequency , the smaller the wavelength the higher the frequency
C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
