Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25

= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Answer:
(a) 
(b) 5220 j
(c) 1740 watt
(d) 3446.66 watt
Explanation:
We have given mass m = 290 kg
Initial velocity u = 0 m/sec
Final velocity v = 6 m/sec
Time t = 3 sec
From first equation of motion
v = u+at
So 
(a) We know that force is given by
F = ma
So force will be 
(b) From second equation of motion we know that

We know that work done is given by
W = F s = 580×9 =5220 j
(c) Time is given as t = 3 sec
We know that power is given as

(d) Time t = 1.5 sec
So 
Answer:
Explanation:
Work
Other units Foot-pound, Erg
In SI base units 1 kg⋅m2⋅s−2
Derivations from other quantities W = F ⋅ s W = τ θ
Dimension M L2 T−2
Idk if this is what u are looking for but i hope this help.:)
No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.