Answer:
1. 5 m/s
2. 8 m/s
3. 10 s
4. 3 m/s
Explanation:
1. Average speed = distance / time
s = 20 m / 4 s
s = 5 m/s
2. First, find the time to travel from A to B:
60 m = 10 m/s × t
t = 6 s
Now find the average speed from A to C:
s = 80 m / (6 s + 4 s)
s = 8 m/s
3. If t is the time traveling from C back to B, and the average speed of the whole trip is 5 m/s, then:
5 m/s = 100 m / (6 s + 4 s + t)
t = 10 s
4. Average velocity = displacement / time
v = 60 m / 20 s
v = 3 m/s
Answer:
Wilhelm Conrad Roentgen (1845-1923)
Antoine Henri Becquerel (1852-1908)
Pierre (1859-1906) and Marie (1867-1934) Curie
Explanation:
Wilhelm Conrad Roentgen (1845-1923)
Contribution: Discovery of x-rays in 1901.
Antoine Henri Becquerel (1852-1908)
Contribution: He discovered that radioactivity is the separation of x-rays and document and the difference between two.
Pierre (1859-1906) and Marie (1867-1934) Curie
Contribution: She discovered Polonium and Radium in1911
This problem is going to be pretty long to solve. So, prepare.
We’re interested in the change in our x position. So we have to break the velocity vector up into its components. Do cosine of 50 and then multiply by the magnitude of the velocity. I got 20.57m/s. That’s our initial velocity. And remember, horizontal acceleration is zero. The vertical acceleration, or any vertical component, has no effect on the horizontal components. In order to solve this problem, we want to utilize this equation:
Change in x-position = Vix*t
Let’s solve for time, which is dependent on the vertical components. The projectile will stop when it vertically hits the ground. Generally you want to use this equation for solving for time:
Yf = Yi + Viy*t + 1/2at^2
We didn’t solve for the vertical component yet, so let’s do that now. (Sine of 50)*(32) = 24.51m/s
Let’s now plug everything in:
0 = 0 + 24.5t - 4.9t^2
0 = 24.5t - 4.9t^2
0 = t(24.5 - 4.9t)
-24.5 = -4.9t
t = 5 seconds
The hard stuff is pretty much over. Put that 5 seconds into the other equation I said we wanted to use to solve the problem
Change in x-position (range) = (20.57)*(5)
= 102.85 meters
Answer B
Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
