The name of the brightest star is ursae majoris. it is a dwarf star and lie in the constellation of ursa major. that is why it has such similar name. around this star , we have three planets around this star. the distance of this star is around 46 light years. also it's mass is almost same as that of the sun. it rotates with a period of 24 hours
Answer:
ω=6684.51 rpm
Explanation:
r= 30cm= 0.3m
a= 15000gs (convert to m/s^{2}
1g = 9.8 m/s^{2}
a= 15000 *9.8 = 147000 m/s^{2}
a=\frac{v^{2} }{r}
147000 = \frac{v^{2} }{0.3}
147000*0.3 = v^{2}
44100 = v^{2}
√44100 = v
210m/s = v
v=210m/s (linear velocity)
we will convert this to angular velocity
ω=\frac{v}{r}
ω= 210/0.3
ω= 700 rads^{-1}
we will convert this to rev per minute
1rad per second = 9.5493 rev per minute
ω= 700*9.5493
ω=6684.51 rpm
Answer:
The ladder is moving at the rate of 0.65 ft/s
Explanation:
A 16-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second. We need to find the rate at which the top of the ladder moving down when the foot of the ladder is 5 feet from the wall.
The attached figure shows whole description such that,
.........(1)
We need to find, 
at x = 5 ft
Differentiating equation (1) wrt t as :
Since, 
At x = 5 ft,
So, the ladder is moving down at the rate of 0.65 ft/s. Hence, this is the required solution.
Answer:
The voltage is V = 37.5 [V]
Explanation:
To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.
V = I*R
where:
V = voltage [Volt]
I = current = 0.25[amp]
R = resistance = 150 [ohm]
V = 0.25*150 = 37.5 [V]
Explanation:
v= (f) x ( lambda)
1.7 ms^-1/12.05 m = f =o.14 hz
