Answer:
In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.
Explanation:
The magnitude of the air drag is 784 N
Explanation:
An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:
where
is the magnitude of the air drag
m is the mass of the object
g is the acceleration of gravity
In this problem, we have
m = 80 kg is the mass of the airman
is the acceleration of gravity
Substituting into the formula, we find:
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Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
Answer:
A
Explanation:
In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.
Hope this helps!
Answer:
The answer to your question is given below
Explanation:
Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.
From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.
Remember, both objects must get to the ground at the same time..!
Let the time taken for both objects to get to the ground be t.
Time A = Time B = t
But B falls through time 2t
Therefore,
Time A = Time B = 2t
Height = 1/2gt^2
For A:
Time = 2t
dA = 1/2 x g x (2t)^2
dA = 1/2g x 4t^2
For B
Time = t
dB = 1/2 x g x t^2
Equating dA and dB
dA = dB
1/2g x 4t^2 = 1/2 x g x t^2
Cancel out 1/2, g and t^2
4 = 1
4dA = dB
Divide both side by 4
dA = 1/4 dB
