Acceleration = (change in speed) / (time for the change)
Change in speed = (speed at the end) minus (speed at the beginning.
The cart's acceleration is
(0 - 2 m/s) / (0.3 sec)
= ( -2 / 0.3 ) (m/s²) = -(6 and 2/3) m/s² .
Newton's second law of motion says
Force = (mass) x (acceleration) .
For this cart: Force = (1.5 kg) x ( - 6-2/3 m/s²)
= ( - 1.5 x 20/3 ) (kg-m/s²)
<span> = </span>- 10 newtons .
<span>The force is negative because it acts opposite to the direction </span>
<span>in which the cart is moving, it causes a negative acceleration, </span>
<span>and it eventually stops the cart.</span>
Answer:
It will be more than deta t
Explanation:
Because
deta t' = န deta t
But န= 1/√ (1 - v²/c²
So the observers in all the initial frames will be more than deta t
The answer is D because i'm in the same studies right now.
Answer: 0.29 kN
Explanation:
We have the following data:
is the weight of the astronaut on Earth
is the free fall acceleration due gravity on Earth (directed downwards)
is the free fall acceleration due gravity on Zuton (directed downwards)
is the acceleration of the spaceship at litoff (directed upwards)
We have to find the <u>magnitude of the force</u>
the space ship exerts on the astronaut.
Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:
(1)
Where
is the mass of the atronaut.
Isolating
:
(2)
(3)
(4)
Now that we know the mass of the astronaut, we can find its weight on Zuton:
(5)
(6)
(7)
Then, we can calculate the force the space ship exerts on the astronaut by the following equation:
(8)
Isolating
:
(9)
(10)
(11)
Finally:
The slope represents the acceleration