a train starts from rest and accelerates uniformly until it has traveled 2.1 km and acquired a forward velocity of 24 m/s.
1 answer:
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Answer:
Explanation:
Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as
W = F×d
or
W = .................................. Equation 1
Where b = upper limit, a = lower limit, Fx = expression of force.
<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>
Substituting these values into equation 1
<em>W = </em>
W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ
Work = upper limit - lower limit
Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2
Substituting the values of a and b into equation 2
Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]
Work = [5.2 + 26.53 -3.29 + C] - C
Work = 28.44 J
Work done by the force = 28.44 J.
Explanation:
Formula for angle subtended at the center of the circular arc is as follows.
where, S = length of the rod
r = radius
Putting the given values into the above formula as follows.
=
=
=
Now, we will calculate the charge density as follows.
=
=
Now, at the center of arc we will calculate the electric field as follows.
E =
=
= 34.08 N/C
Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.
Answer:
a= - 6.667 m/s²
Explanation:
Given that
The initial speed of the box ,u= 20 m/s
The final speed of the box ,v= 0 m/s
The distance cover by box ,s= 30 m
Lets take the acceleration of the box = a
We know that
v²= u ² + 2 a s
Now by putting the values in the above equation we get
0²=20² + 2 a x 30
a= - 6.667 m/s²
Negative sign indicates that velocity and acceleration are in opposite direction.
Therefore the acceleration of the box will be - 6.667 m/s² .