Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is
Explanation:
From the question we are told that
The radius of the inner circle is ![r_i = 0.80 \ m](https://tex.z-dn.net/?f=r_i%20%20%3D%20%200.80%20%20%5C%20m)
The radius of the outer circle is ![r_o = 1.20 \ m](https://tex.z-dn.net/?f=r_o%20%20%3D%20%201.20%20%5C%20m)
The charge on the spherical shell
The magnitude of the point charge at the center is ![q_c = + 300 nC = + 300 * 10^{-9} \ C](https://tex.z-dn.net/?f=q_c%20%3D%20%20%2B%20300%20nC%20%20%3D%20%20%2B%20300%20%2A%2010%5E%7B-9%7D%20%5C%20C)
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as
![E = \frac{k * q_c }{x^2}](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7Bk%20%2A%20%20q_c%20%20%20%7D%7Bx%5E2%7D)
substituting values
![E = \frac{k * q_c }{x^2}](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7Bk%20%2A%20%20q_c%20%20%20%7D%7Bx%5E2%7D)
where k is the coulomb constant with value ![k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.](https://tex.z-dn.net/?f=k%20%3D%209%2A10%5E%7B9%7D%20%20%5C%20kg%5Ccdot%20m%5E3%5Ccdot%20s%5E%7B-4%7D%20%5Ccdot%20A%5E%7B-2%7D.)
substituting values
![E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B9%2A10%5E9%20%20%2A%20%20300%20%2A10%5E%7B-9%7D%7D%7B0.6%5E2%7D)
![E = 7500 \ N/C](https://tex.z-dn.net/?f=E%20%3D%207500%20%5C%20N%2FC)
-- Electric field lines DO never cross. <em>(A)
</em>
-- Electric field lines that are close together DO indicate a stronger electric field. <em>(B)
</em>
-- Electric field lines DO not affect the charge that created them. <em>(C)</em>
-- Electric field lines DON'T begin on north poles and end on south poles. North and South "poles" are the way we talk about magnets, not electric charges.
Answer:
Supercells
Explanation:
supercells are rotating thunderstorms that has a well-defined radar circulation called a mesocyclone. They can sometimes produce destructive hail, severe winds, frequent lightning, and flash floods.
Answer:
t = 4.08 s
R = 40.8 m
Explanation:
The question is asking us to solve for the time of flight and the range of the rock.
Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:
We have these known variables:
- (v_0)_y = 0 m/s
- a_y = -9.8 m/s²
- Δx_y = -20 m
And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.
- -20 = 0t + 1/2(-9.8)t²
- -20 = 1/2(-9.8)t²
- -20 = -4.9t²
- t = 4.08 sec
The time it takes for the rock to reach the ground is 4.08 seconds.
Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.
List out known variables:
- v_0 = 10 m/s
- t = 4.08 s
- a_x = 0 m/s
We are trying to solve for:
By using the same equation, we can plug these known values into it and solve for Δx.
- Δx = 10 * 4.08 + 1/2(0)(4.08)²
- Δx = 10 * 4.08
- Δx = 40.8 m
The rock lands 40.8 m from the base of the cliff.
Answer:
3675 J
Explanation:
Gravitational Potential Energy =
× mass × g × height
( g is the gravitation field strength )
Mass = 50 kg
G = 9.8 N/kg ( this is always the same )
Height = 15 m
Gravitational Potential Energy =
× 50 ×9.8 × 15
= 3675 J