Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Answer:
Maharashtra - mashru or himroo / dhoti and lugda
Gujarat - patola / ghagra choli
Punjab - pat / kurta and pajama
Odisha - ikat / Sadi
West Bengal - tossa / kurta
Karnataka - Mysore silk / mundu
The light that makes up the image of the fish is refracted when it leaves the water and enters the air. Your brain thinks the light traveled straight, so the fish is not located in the direction from you that your brain thinks it is.
Answer:
a) T ’= 0.999 s
, b) t = 3596.4 s
Explanation:
The angular velocity of a simple pendulum is
w = √g / L
The angular velocity, frequency and period are related
w = 2π f = 2π / T
2π / T = √ g / L
T = 2π √ L / g
L = T² g / 4π²
L = 1² 9.8 / 4π²
L = 0.248 m
To know the effect of the temperature change let's use the thermal expansion ratios
ΔL = α L ΔT
ΔL = 24 10⁻⁶ 0.248 (-4 - 20)
ΔL = 142.8 10⁻⁶ m
Lf - L = -142. 8 10⁻⁶
Lf = 142.8 10⁻⁶ + 0.248
Lf = 0.2479 m
Let's calculate new period
T ’= 2π √ L / g
T ’= 2π √ (0.2479 / 9.8)
T ’= 0.999 s
We can see that the value of the period is reduced so that the clock is delayed
b) change of time in 1 hour
When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is
t = 3600 0.999
t = 3596.4 s
Therefore the clock is delayed almost 4 s
