The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).
<h3>
Net electric force on the charges</h3>
The net electric force on the charges is calculated as follows;
F = kq₁q₂/r²
where;
- k is coulomb's constant
- q₁ and q₂ are the charges
- r is the distance between the charges
<h3>Distance between the charges</h3>
Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).
Learn more about electric force here: brainly.com/question/17692887
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The resultant force on the object is
∑ <em>F</em> = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N
which has a magnitude of
<em>F</em> = √((6 N)² + (8 N)²) = √(100 N²) = 10 N
By Newton's second law, the acceleration has magnitude <em>a</em> such that
<em>F</em> = <em>m a</em>
10 N = (2 kg) <em>a</em>
<em>a</em> = (10 N) / (2 kg)
<em>a</em> = 5 m/s²
so the answer is B.
Answer:
0.4
Explanation:
F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.
Since the mass is moving with a constant velocity, there's no acceleration hence
where N is the weight of object and \mu is coefficient of kinetic friction.
![F=\mu N and making [tex]\mu](https://tex.z-dn.net/?f=F%3D%5Cmu%20N%20and%20making%20%5Btex%5D%5Cmu)
the subject
Substituting F for 8 N and N for 20 N
Therefore, coefficient of kinetic friction is 0.4
Answer:
90 meters
Explanation:
Given:
x₀ = 0 m
v₀ = 0 m/s
v = 30 m/s
t = 6 s
Find:
x
x = x₀ + ½ (v + v₀)t
x = 0 + ½ (30 + 0)(6)
x = 90
The car travels 90 meters.
Answer:
4 m/s
Explanation:
KE=1/2mv^2
720=1/2(90)
720=45v^2
divide by 45
16= v^2
over the square root
4=v
v= 4m/s
