When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force. hence the statement is true.
Answer:
The rock's speed after 5 seconds is 98 m/s.
Explanation:
A rock is dropped off a cliff.
It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².
Speed after 5 seconds = V
We know that acceleration = average speed/time
In our case,
g = ((0+V)/2)/5
9.8*5 = V/2
=> V = 2*9.8*5
V = 98 m/s
Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.
Answer:
HELLO! I WILL EXPLAIN WHY GOLF BALLS HAVE DIMPLES.
Explanation:
DIMPLES ON A GOLF BALL CREATE A THIN BOUNDARY LAYER OF AIR THAT CLINGS TO THE BALL'S SURFACE. THIS ALLOWS THE AIR TO FOLLOW THE BALL'S SURFACE A LITTLE FARTHER AROUND THE BACKSIDE OF THE BALL, THEREBY DECREASING THE SIZE OF THE WAKE.
I AM SO SORRY ABOUT THE CAP LOCKS I THINK MY CAP LOCKS BUTTON IS PUSHED, AND IT IS NOT POPPING UP FOR SOME REASON.
ANYWAYS, HAVE A GREAT DAY!!!!!!
Missing figure and missing details can be found here:
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Solution:
(a) The work done by the spring is given by

where k is the elastic constant of the spring and

is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:

where the negative sign is given by the fact that

points in the opposite direction of the displacement of the cart, and where

therefore, the work done by the weight is