Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:
and the time taken to cover the horizontal distance d is
So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by
where
is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:
The height of the crossbar is h = 3.05 m, so the ball passes
above the crossbar.
Hi! The answer is ‘B’! Because the nucleus is found at the center and contains protons (positive charge) and neutrons (no charge)
Answer:
The gravitational acceleration of the planet is, g = 8 m/s²
Explanation:
Given data,
The distance the object falls, s = 144 m
The time taken by the object is, t = 6 s
Using the III equations of motion
S = ut + ½ gt²
∴ g = 2S/t²
Substituting the given values,
g = 2 x 144 /6²
= 8 m/s²
Hence, the gravitational acceleration of the planet is, g = 8 m/s²
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V
Answer:
<h2>The current required winding is
</h2>
Explanation:
We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil and
n= N/L (number of turns per unit length)
Given data
The number of turns n= 1200 turns
length L= 0.42 m
magnetic field B= 1*10^-4 T
μ₀= 
Applying the equation B=μ₀*n*I
I= B/μ₀*n
I= B*L/μ₀*n
