Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
A planet orbiting a star in a eclipse and sometimes it is closer to the star but sometimes it is farther. When it is closer the gravity on the planet from the star is stronger and it speeds up. The area the planet sweeps over is equal because when it speeds up the length covered along the orbital path is greater, but it is also closer to the star, and that dimension is decreased.
Answer:
-40,000 N
Explanation:
First, use the kinematics equation v(f) = v(i) + at. Final velocity is 0, initial is 8, and time is 0.2 seconds. Solving for a, you get -40 m/s^2. Then, use Newton’s second law, F=ma, to find the force. F = (1000)(-40) = -40,000 N.
1) find speed (8.8 m/s)
2) find acceleration (38.7 m/s^2)
answer is about 38.7 m/s^2
The answer is c ok
I wrote this for 20 word rule