Answer:
s=4.44 m
Explanation:
Given that
Coefficient of the kinetic friction ,μ = 0.13
Initial velocity ,u= 3.4 m/s
Final velocity of the box ,v= 0 m/s
The acceleration due to friction force
a= - μ g
Now by putting the values in the above equation
a= - 0.13 x 10 ( take g= 10 m/s²)
a= - 1.3 m/s²
We know that
v²= u ² + 2 a s
s=distance
a=acceleration
v=final speed
u=initial speed
Now by putting the values in the above equation
0²= 3.4² - 2 x 1.3 x s
s=4.44 m
The distance cover by box will be 4.44 m.
a) The motion along the vertical direction and the motion along the horizontal direction.
b) The object remains in the air for a time period of 2usin(θ)/g.
Any object that is thrown in the air when gravity is acting on it is called a projectile. The motion of this projectile is called projectile motion.
When the projectile is thrown in the air at some angle θ, then there are two independent motions taking place at the same time. First is the component of motion along the vertical direction along which gravity acts. Second is the component of motion along the horizontal direction along which the object moves with a constant velocity. No force acts along the horizontal direction. The horizontal motion does not affect the vertical motion and the converse is also true. So these are independent of each other.
The time of flight is the time during which a projectile remains in the air. This time of flight is calculated using the formula,
T=2usin(θ)/g
where T is the time of flight, u is the initial velocity and g is the acceleration due to gravity.
Hence, the object remains in the air for a time period of 2usin(θ)/g.
Learn more about projectile.
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D. 5.098 x 106 is the correct answer when reduced to the proper notation.
Answer:
the answer for the question is C
