99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
= (0²-27.5²)/(2x50.0)
=-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!
Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :
v is the maximum speed
Hence, the maximum speed of the ball is 5.86 m/s.
Constant = straight line
“Travels at constant negative acc.”
Which is negative slope
Solution: B. Straight line w/ neg. slope
Answer:
The incident light ray which lands upon the surface is said to be reflected off the surface. The ray that bounces back is called the reflected ray. If a perpendicular were to be drawn on reflecting surface, it would be called normal. The figure below shows the reflection of an incident beam on a plane mirror.
Explanation: