Answer:
8.46E+1
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 39 C
Charge 2 (q₂) = –53 C
Force (F) of attraction = 26×10⁸ N
Electrical constant K) = 9×10⁹ Nm²/C²
Distance apart (r) =?
The distance between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
26×10⁸ = 9×10⁹ × 39 × 53 / r²
26×10⁸ = 1.8603×10¹³ / r²
Cross multiply
26×10⁸ × r² = 1.8603×10¹³
Divide both side by 26×10⁸
r² = 1.8603×10¹³ / 26×10⁸
r² = 7155
Take the square root of both side
r = √7155
r = 84.6 m
r = 8.46E+1 m
<span>a = (v2 - v1)/t= acceleration formula
a = (70 - 0)/7
a = 10 km/hr/sec
-----
It's better to use as few units as possible.
10 km/hr = 10 km*1000 m/km/(1 hr*3600 sec/hr) = 25/9 m/sec
a= 25/9 m/sec/se</span>
The answer is A. Force is equal to mass x acceleration. The acceleration due to gravity is a constant so multiplying the mass by 2 would double the force.
Answer:
Velocity (magnitude) is 98.37 m/s
Explanation:
We use the vertical component of the initial velocity, which is:

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

Now we need to find
as a function of
. We use the horizontal velocity, which is always the same as follow:

We know the angle at 3 seconds:

Substitute
in
and then solve for 

With this expression we go back to the kinematic equation and solve it for initial speed

The expression of the electric flux is

Here,
Q = Total charge enclosed in the closed surface
= Permittivity due to free space
Rearranging to find the charge,

Replacing with our values we have finally



The charge enclosed by the box is 0.1684nC
The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.