Answer:
The acceleration of the cheetahs is 10.1 m/s²
Explanation:
Hi there!
The equation of velocity of an object moving along a straight line with constant acceleration is the following:
v = v0 + a · t
Where:
v = velocity of the object at time t.
v0 = initial velocity.
a = acceleration.
t = time
We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.
Let's convert mi/h into m/s:
50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s
Then, using the equation:
v = v0 + a · t
22.4 m/s = 0 m/s + a · 2.22 s
Solving for a:
22.4 m/s / 2.22 s = a
a = 10.1 m/s²
The acceleration of the cheetahs is 10.1 m/s²
The critical angle is the angle of incident in the optically dense media that produces the angle of refraction as 90°.
sin c =1/n, where n is the refractive index and c is the critical angle.
Also, n1sinФ1=n2sinФ2
In this case we have,
1.33×sinc = 1.309×sin90° Where c is the critical angle.
sin c =1.309/1.33
sin c = 0.981995498
c = 79.11°
The solution for this problem is:
Work it backwards: spring U = ½kx² = ½ * 4000N/m * (0.10m) ²
= 20 J
so the pre-impact of KE = 20 J = ½ * m * v² = ½ * 0.40kg * v²
v = 10 m/s → for the bullet/block.
Now conserve the momentum:
50g * u = 400g * 10m/s
u = 80 m/s is the arrow
velocity
Answer:
What is the magnitude of the player 's displacement? A baseball "diamond" is a square, each side of length 27.4 m, with home plate and if each length of the base is 27.4 m, and the baseball player ran to third base,Then, the baseball player is 27.4 feet away from home plate (where he started)below the horizontal.
Explanation: