The answer is d that might help you
The solution to the problem is as follows:
<span>Average = 80
So Sum = 80 * 5 = 400
Mode = 88, so two results are 88 (if three results were 88, then the median would be 88).
Three results are 81, 88, and 88.
That leaves 143. We could still have one 81 score, so that leaves the lowest score as 62.
Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 - 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground?
d = 144 - 16t2
0 = 144 - 16t2
16t^2=144
t^2=9
t=3</span>
The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, 
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:
where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s
Substituting,
Learn more about projectile motion:
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Answer:
-5.8868501529 m/s² or -5.8868501529g
0.118909090909 s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Dividing by g
The acceleration is -5.8868501529 m/s² or -5.8868501529g
The time taken is 0.118909090909 s
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
