<u>Answer:</u>
The velocity is 30.279 m/s
<u>Explanation</u>:
Consider the initial speed of the semi-trailer be v
Then, initial kinetic energy = 
According to question, the semi-trailer coast along a ramp, which is inclined at an angle of 170, and to a distance of 160m to stop
Change in vertical position =
= 46.779m
Final potential energy of semitrailer = mgh
Applying principle of conservation of energy,
= mgh
Solving for v, we get 
= 2gh = 2*9.8*46.779 = 916.8684
= 916.8684
v = 30.279 m/s
Therefore, the velocity is 30.279 m/s
Answer:
26.82m/s
Explanation:
Given
Mass = m= 0.4kg
Initial Velocity = u = 0
Charge = 4.0E-5C
Distance= d = 0.5m
Object Charge = 2E-4C
First, we'll calculate the initial energy (E)
E = Potential Energy
PE = kQq / d
Where k = coulomb constant = 8.99E9Nm²/C²
Energy is then calculated by;
PE = 8.99E9 * 4E-5 * 2E-4 / 0.5
PE = 143.84J
Energy = Potential Energy = Kinetic Energy
K.E = ½mv² = 143.84J
½mv² = ½ * 0.40 * v² = 143.85
0.2v² = 143.85
v² = 143.85/0.2
v² = 719.25
v = √719.25
v = 26.81883666380777
v = 26.82m/s
Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge
from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"
here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Answer:
(A) 26 m/s
(B) 32.4 m
(C) v = 15.4 m/s
Explanation:
initial speed (u) = 6.4 m/s
acceleration due to gravity (a) = 9.9 m/s^[2}
time (t) = 2 s
(A) What is its speed after falling for 2.00s?
from the equation of motion v = u + at we can get the speed
v = 6.4 + (9.8 x 2) = 26 m/s
(B) How far does it fall in 2.00s?
from the equation of motion 
we can get the distance covered
s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)
s = 12.8 + 19.6 = 32.4 m
c) What is the magnitude of its velocity after falling 10.0m?
from the equation of motion below we can get the velocity
v = 15.4 m/s
Answer: 3.142656 × 10^16 feet
Explanation: Given that the
Speed = 982,080,000 ft/s, and
Time = 32,000,000 seconds
The formula for speed is:
Speed = distance/ time
Make distance the subject of formula
Since the time is second in one year and speed is ft/s, substitute both into the formula
Distance = speed × time
distance = 982,080,000 × 32,000,000
Distance = 3.142656 × 10^16 feet.
The distance of one light year in feet is 3.142656 × 10^16
<u>Correct Question:</u>
Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour
<u>Answer:</u>
The total distance covered by Charlie is 8 km in 1 hour.
<u>Explanation:</u>
The average velocity as given in the question is,
v = 8 km/hr
Total time taken,
As we know the formula to evaluate the total distance d when the average velocity and time is given;
Hence, the total distance covered by Charlie in 1 hour will be 8 km.
