A box sliding on a horizontal floor surface starts out moving at 4.9 m/s and stops after 2 seconds. The surface and the box have a kinetic friction has friction coefficient of 0.25 (g=9.8 ms-2).
The speed of an object at the start of a measurement, or an initial state, is known as the initial speed. The ratio of distance travelled to travel time, also known as the average speed, is the sum of the initial and final speeds. The difference between initial and ending speeds is the speed change. A force that acts between moving surfaces is referred to as kinetic friction. A force acting in opposition to the direction of a moving body on the surface is felt.
Interval (t) equals 2 seconds
Kinetic friction coefficient is 0.25.
gravity-induced acceleration (g) = 9.8 m/s2.
Coming to a stop at a final velocity of 0 m/s.
Regular force equals mg
Kinetic friction force is given by = k.
N = μkmg
I = (delta P) (delta P)
Vo = (0.25)(9.8)(2) (2)
Vo = 4.9m/s
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Answer:
9m/s^2
Explanation:
solution,
Given. Initial velocity (u)=0m/s
Final velocity (v)=45m/s
Time taken(t)=5sec
Acceleration (a)=?
we know,
a=v-u/t
=45-0/5
=45/5
=9m/s^2
Therefore acceleration (a)= 9m/s^2
Answer:
<h3>Ion beams and Electrochemical are two examples of direct current electricity.</h3>
Answer:
Explanation:
When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components
now we know that
also we know that
now plug in all data in above equations
Answer:
Approximately 
, assuming that the rocket had no propulsion onboard, and that air resistance on the rocket is negligible.
Explanation:
Initial velocity of this rocket: 
.
When the rocket is at its maximum height, the velocity of the rocket would be equal to 
. That is: 
.
The acceleration of the rocket (because of gravity) is constantly downwards, with a value of 
.
Let 
denote the distance that the rocket travelled from the launch site to the place where it attained maximum height. The following equation would relate 
to 
, 
, and 
:
.
Apply this equation to find the value of :
.
In other words, the maximum height that this rocket attained would be 
.
Again, assume that the air resistance on this rocket is negligible. The rocket would return to the ground along the same path, and would cover a total distance of 
.
