Answer:
The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Explanation:
Given that,
Weight Fg = mg
Acceleration = a
Tension = T
Drag force = Fa
Vertical force = L
We need to find the correct relationships
Using balance equation
In horizontally,
The acceleration is a
...(I)
In vertically,
No acceleration
Put the value of mg
....(II)
Hence, The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Answer:
Explanation:
Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.
Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:
where
is the Coulomb's constant
is the charge on the sphere
is the radius of the sphere
is the distance from the surface of the sphere
Substituting, we find
Answer:
Explanation:
40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.
<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
Answer:
Explanation:
= 250 moles.
N = n×6.02×
= 1.505×
Total charge = (1.505×) × (1.6×
)
= 2.4×
C.
