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
If the acceleration constant..
you can use the formula s = ut + 1/2at²
Known that :
s = ?
u = 0
t = 2s
a = 10ms-²
Then you can apply the formula
s = ut + 1/2at²
s = 0 + 1/2(10)(2)²
s = 5 × 4
s = 20m
Answer : 20m
Explanation :
The gravity can be 9,8 or 10. Also im not sure how people teach you but in my school, if the ball goes down the gravity is positive and not negative thats why i put 10ms-² and not -10ms-²
s = displacement/distance
u = initial speed
a = acceleration
t = time
sorry if im wrong
To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as
The potential energy, 
The volume, 
The potential energy per unit volume is defined as the energy density.
The energy density related with electric field is given by
Here, the permitivity of the free space is
Therefore, rerranging to find the electric field strength we have,
Therefore the electric field is 2.21V/m
Answer:
the answer is ture
Explanation:
because the energy substances must absorb in order to change from liquid to gas
