The acceleration is 
Explanation:
This is a uniformly accelerated motion (constant acceleration), therefore we can apply the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the scooter in this problem, we have
u = 0 (it starts from rest)
v = 10.0 m/s is the final velocity
t = 4.0 s is the time interval
Solving for a, we find the acceleration:

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Answer:

Explanation:
Given that

As both charges are negative so there exist force of repulsion in direction as shown in figure.

Angle at which force F12 is acting is



Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction


Answer:
A
Explanation:
Because Jessica pushes a 25 kilogram cart a distance of 10 meters.
Answer:
d = 4.9 m
Explanation:
It is mentioned that a stone is dropped from a certain height. It is required to find the distance covered by it in one second.
The initial speed of the stone is equal to 0 as it was at rest. Let d is the distance covered by the stone.
Using second equation of motion :

Put u = 0 and a = g

So, the distance covered by it in one second is 4.9 m.
Answer:
1.144 A
Explanation:
given that;
the length of the wire = 2.0 mm
the diameter of the wire = 1.0 mm
the variable resistivity R = ![\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]](https://tex.z-dn.net/?f=%5Crho%20%28x%29%20%3D%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%20%5C%20m%7D%29%5E2%5D)
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as = 
![dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}](https://tex.z-dn.net/?f=dR%20%3D%20%5Cfrac%7B%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%7D%29%5E2%5D%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%2810%5E%7B-3%7D%29%5E2%7D)
Taking the integral of both sides;we have:
![\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5ER_0%20%20dR%20%3D%20%5Cint%5Climits%5E2_0%203.185%20%5C%20%5B1%2Bx%5E2%5D%20%5C%20dx)
![R = 3.185 [x + \frac {x^3}{3}}]^2__0](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5Bx%20%2B%20%5Cfrac%20%7Bx%5E3%7D%7B3%7D%7D%5D%5E2__0)
![R = 3.185 [2 + \frac {2^3}{3}}]](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5B2%20%2B%20%5Cfrac%20%7B2%5E3%7D%7B3%7D%7D%5D)
R = 14.863 Ω
Since V = IR


I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A