Answer:
5. -24 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s².
mathematically,
a = dv/dt ............................ Equation 1
Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.
But
v = dx(t)/dt
Where,
x(t) = 27t-4.0t³...................... Equation 2
Therefore, differentiating equation 2 with respect to time.
v = dx(t)/dt = 27-12t²............. Equation 3.
Also differentiating equation 3 with respect to time,
a = dv/dt = -24t
a = -24t .................... Equation 4
from the question,
At the end of 1.0 s,
a = -24(1)
a = -24 m/s².
Thus the acceleration = -24 m/s²
The right option is 5. -24 m/s²
Answer:
31.404 seconds
Explanation:
To answer this equation, SUVAT is your best option utilizing and rearranging the known values to solve for the unknown.
here we have the values for
s=895
u=22
v=35
t= the unknown value
in this instant the equation s=0.5 x (u+v)t is the best equation to use
so we sub in the known values
895=0.5 x (22+35)t
rearrange to solve for t
895=28.5t
895/28.5=t
t=31.404 seconds (rounded to 3 decimal places)
<h2>Answer:</h2>
<u>By wrapping the wire along a solenoid and connecting it to electricity</u>
<h2>Explanation:</h2>
If you wrap a copper wire into coils and run an electrical current through it, you will create a magnetic field. If you rotate a permanent magnet as opposed to an item that has been magnetized inside a coil of copper wire, you can create an electrical current. The strength of magnetic field generated is proportional to the amount of current through the winding.
Answer:
The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.
Explanation:
The application of coulonb's law is used to approach the question as shown in the attached file.
