X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
a. We can calculate the amount of work by calculating the area under the graph.
first area (rectangular): 2.5 x 6 = 15
second area(trapezoid): 1/2 x (6+10) x 2.5 =20
total work done: 35 J
b. the force was first applied = 6 N
F = m.a
a = 6 : 3 = 2 m/s²
vf²=vi²+2as
vf²=6²+2.2.5
vf²=56
vf=7.5 m/s
Answer:
The answer is the option a.
Explanation:
We know that magnetic force (Fm) is defined as
Fm = q (v x B)
Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.
"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is
v x B = vBsin (thetha)
So,
Fm = qvBsin (thetha)
And, in case in which v and B are parallel vectors, thetha is zero, and,
sin (thetha)=sin (0) = 0
So, Fm=0
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
![\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bate%5E%7B-6t%7D%5D%3Da%5B%281%29e%5E%7B-6t%7D%2Bt%28e%5E%7B-6t%7D%28-6%29%29%5D)
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
