Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
Answer:
Explanation:
The lift is going down with acceleration
Initial speed u = 0
Final speed v = 6 m/s
distance s = 15.25 m
acceleration a = ?
v² = u² + 2 a s
6² = 0 + 2 x a x 15.25
a = 1.18 m /s²
Elevator is going down with acceleration .
mg - T = ma where T is tension in the cable .
722 x 9.8 - T = 722 x 1.18
7075.6 - T = 851.96
T = 6223.64 N .
Answer:
![d=691.71km](https://tex.z-dn.net/?f=d%3D691.71km)
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
![t=\frac{d}{v_t}-\frac{d}{v_l}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv_t%7D-%5Cfrac%7Bd%7D%7Bv_l%7D)
Here d is the distance at which the earthquake take place and
is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
![t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km](https://tex.z-dn.net/?f=t%3Dd%28%5Cfrac%7B1%7D%7Bv_t%7D-%5Cfrac%7B1%7D%7Bv_l%7D%29%5C%5Cd%3D%5Cfrac%7Bt%7D%7B%5Cfrac%7B1%7D%7Bv_t%7D-%5Cfrac%7B1%7D%7Bv_l%7D%7D%5C%5Cd%3D%5Cfrac%7B56.4s%7D%7B%5Cfrac%7B1%7D%7B5.05%5Cfrac%7Bkm%7D%7Bs%7D%7D-%5Cfrac%7B1%7D%7B8.585%5Cfrac%7Bkm%7D%7Bs%7D%7D%7D%5C%5Cd%3D691.71km)
Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by
![T = \frac{2u Sin\theta }{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2u%20Sin%5Ctheta%20%7D%7Bg%7D)
Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT
![v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}](https://tex.z-dn.net/?f=v_%7By%7D%3Du_%7By%7D-%20g%20%5Ctimes%20%5Cfrac%7B2u%20Sin%5Ctheta%20%7D%7Bg%7D)
![v_{y}=u Sin 30 - 2u Sin 30](https://tex.z-dn.net/?f=v_%7By%7D%3Du%20Sin%2030%20-%202u%20Sin%2030)
vy = - u Sin 30
The magnitude of final velocity is given by
![v = \sqrt{v_{x}^{2}+v_{y}^{2}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bv_%7Bx%7D%5E%7B2%7D%2Bv_%7By%7D%5E%7B2%7D%7D)
![v = \sqrt{\left (uCos 30 \right )^{2}+\left (uSin 30 \right )^{2}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cleft%20%28uCos%2030%20%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%28uSin%2030%20%20%5Cright%20%29%5E%7B2%7D%7D)
v = u
Thus, the velocity is same as it just reaches the ground.
<span>Strong nuclear forces hold the nucleus of an atom together. Weak nuclear forces are involved when certain types of atoms break down.</span>