for acceleration we can define that rate of change in velocity is know as acceleration
So whenever velocity of train is changing with time we can say train is accelerating
Now here if initially train is standstill then after some time its speed is 5 m/s
so here the train is accelerated first time
Then on straight path its speed changed from 5 m/s to 10 m/s so here train gets accelerated second time
After this train chugged around a curve with same speed 10 m/s
SO here since train is moving in curve so here its direction of velocity is continuously changing and this type of acceleration is known as centripetal acceleration
SO this is accelerated Third time
Then its speed decreases and it comes to speed of 5 m/s from 10 m/s
So here it is acceleration of train for Fourth time
Then finally train comes to stop so again its speed changed from 5 m/s to 0
so this is acceleration of train Fifth time
So total train will accelerate 5 times in whole path
Answer:
a. 13.7 s b. 6913.5 m
Explanation:
a. How much time before being directly overhead should the box be dropped?
Since the box falls under gravity we use the equation
y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.
So,
y = ut - 1/2gt²
y = 0 × t - 1/2gt²
y = 0 - 1/2gt²
y = - 1/2gt²
t² = -2y/g
t = √(-2y/g)
So, t = √(-2 × 919 m/-9.8 m/s²)
t = √(-1838 m/-9.8 m/s²)
t = √(187.551 m²/s²)
t = 13.69 s
t ≅ 13.7 s
So, the box should be dropped 13.69 s before being directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m
Answer:
E_Phase = 560V
Explanation:
The computation of the voltage i.e. dropped across each phase is shown below:
Given that
The delta connection line voltage is
E_line = 560 V
And, in the case of delta connection, the line voltage would be equivalent to the phase voltage
That means
E_Phase = E_Line
= 560 V
Hence, the voltage i.e. dropped across each phase is
E_Phase = 560V
The diameter of the sphere is 7.5 cm, therefore its volume is
V = [(4π)/3]*(7.5/2 cm)³ = 220.8932 cm³
The density of the lead ball is 11.34 g/cm³, therefore its mass is
m = (220.8932 cm³)*(11.34 g/cm³) = 2.5049 x 10³ g = 2.5048 kg
Answer: 2.5 kg (nearest tenth)
Im pretty sure its neptune. Its mot pluto because its not a planet. Good luck.
