Answer:
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Answer:
<u>We are given:</u>
u = 2.5 m/s
a = 0.2 m/s/s
t = 25 seconds
v = v m/s
<u>Solving for 'v':</u>
From the first equation of motion:
v = u + at
Replacing the values
v = 2.5 + (0.2)(25)
v = 2.5 + 5
v = 7.5 m/s
Answer:
An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.
Explanation:
Answer: hello the complete question is attached below
answer :
r2 = 4r1
Explanation:
Electric field strength = F / q
we will assume the rod has an infinite length
For an infinitely charged rod
E ∝ 1/ r
considering two electric fields E1 and E2 at two different locations as described in the question
E1/E2 = r1/r2 ----- ( 2 )
<u>Calculate for r2 when E2 = E1/4 </u>
back to equation 2
E1 / (E1/4) = r1 / r2
∴ r2 = 4r1
Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
