Answer:
Explanation:
The first part of this question is simply asking us to convert the speed from miles per hour to meters per second:
choice C.
The next part wants us to use the equation for acceleration and find the acceleration:
where v is final velocity, v0 is initial velocity, and t is time in seconds (which was one of the reasons we had to convert the initial velocity from 60.0 mph to m/s):
and
a = 10.7 m/s/s, choice B.
Explanation:
Acceleration is defined as the change in velocity over time.
When there is an increment or increase in the magnitude of velocity of a moving body then it is known as positive acceleration.
Whereas when there is a decrease in magnitude of velocity of a moving body then it is known as negative acceleration.
Thus, we can conclude that positive acceleration occurs when an object speeds up.
k = spring constant of the spring = 100 N/m
m = mass hanging from the spring = 0.71 kg
T = Time period of the spring's motion = ?
Time period of the oscillations of the mass hanging is given as
T = (2π) √(m/k)
inserting the values in the above equation
T = (2 x 3.14) √(0.71 kg/100 N/m)
T = (6.28) √(0.0071 sec²)
T = (6.28) (0.084) sec
T = 0.53 sec
hence the correct choice is D) 0.53
Answer:
For these reasons at 98 mph the path is straighter
Explanation:
To solve this problem we are going to use the kinematic equations, specifically those of projectile launches, let's calculate the distances that the ball travels
X = Vox t
Y = Yo + Voy t - ½ g t²
They tell us that the only parameter that changes is the speed, so the distance to the plate is known
t = Vox / x
We replace
Y- Yo = Voy (Vox / X) - ½ g (Vox / x)²2
Y -Yo = Vo² sinθ cos θ / x - ½ g Vo² sin²θ / x²
Y -Yo = Vo² (sinθ cosθ / x - ½ g sin²θ / x²)
The trajectory will be flatter when Y is as close as possible to Yo, when examining the right side of the equation, the amount in Parentheses is constant and to what they tell us that the angles and the distance the plate does not change.
Consequently, of the above, the only amount changes is the initial speed if it increases the square of the same increases, so that the height Y approaches the height of the shoulder, that is, DY decreases. For these reasons at 98 mph the path is straighter
Its b i belive
because it the only thing i saw on the list that conduts
