Answer:
The magnitude of the electric force between these two objects
will be: 181.274 N.
i.e. F 
N
Step-by-step explanation:
As
Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.
so
q₁ = 4.5 μC = 4.5 × 10⁻⁶ C
q₂ = 2.8 μC = 2.8 × 10⁻⁶ C
separated distance = d = 2.5 cm
Calculating the magnitude of the force between two charged objects using the formula:
∵ 
∵ 
∵ 
∵ 
F N
Therefore, the magnitude of the electric force between these two objects will be: 181.274 N.
i.e. F N
Answer:
x=3
y=5
Step-by-step explanation:
x+5y=28 (i)
-x-2y=-13. (ii)
add equation 2 from equation 1
x+5y=28
-x-2y=-13
3y=15
y=5
put the value of y in equation 1
x+5y=28
x+5*5=28
x+25=28
x=28-25
x=3
The formula for distance is equal to:
d = v * t
where d is distance, v is velocity or speed, and t is
time
Since the distance travelled by the two airplane is
similar, therefore we can create the initial equation:
v1 * t1 = v2 * t2
We know that v1 = 496, and v2 = 558 so:
496 t1 = 558 t2
or
x = 558 t2 / 496
We also know that airplane 1 travelled 30 minutes (0.5
hours) earlier than airplane 2, therefore:
x = t2 + 0.5
Hence,
496 (t2 + 0.5) = 558 t2
496 t2 + 248 = 558 t2
t2 = 4 hours
x = t2 + 0.5 = 4 + 0.5
x = 4.5 hours
So the equation is:
x = 558 t2 / 496
And the first plane travelled:
x = 4.5 hours
