A particle is constrained to move round a circle radius 382400km and makes a single revolution in 27.3 days. (i). Find the veloc
1 answer:
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Answer:
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Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
Therefore, the mutual force between the two point charges is 319.64 N
Answer:
South = 1.5m
West =4.2m
Explanation:
Kindly see attached a rough draft of the situation
Step one
Given data
From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse
Step two:
Hence for an opponent to tackle him towards the south, he must be at
sin θ= opp/hyp
sin 20=x/4.5
x=sin 20*4.5
x=0.342*4.5
x= 1.5m
Also, for an opponent to tackle him towards the south, he must be at
cos θ= adj/hyp
cos 20=y/4.5
y=cos 20*4.5
y=0.93*4.5
y= 4.2m
