K = 1/2 m x v^2
m = mass on the cart
V = velocity imparted to the cart
KA = 1/2 mA x vA^2.......................(1)
KB = 1/2 mB x vB^2........................(2)
Diving equation 1 by equation 2, we get -
KA/KB = mA/mB
= 2
KA = 2 x KB
Option A is correct
Answer:
Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>
Explanation:
Hope it helps..
Just correct me if I'm wrong, okay?
But ur welcome!!
(;ŏ﹏ŏ)(◕ᴗ◕✿)
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>
Answer: Yes.
Explanation:
Assuming Earth and Moon are isolated is space, it is possible to have a point where Earth and Moon will pull at an object with equal force.
That point will be closer to the Moon than the Earth because Moon's gravitational field strength is weaker than Earth's gravitational field strength.