If the ratio of S.E to S.L is 2:5 what does the simple machine multiply

Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is

victus00 [196]

Answer:

3.15 N towards the positive x-axis

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = $\frac{-kQq}{r^2}$

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = $\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}$ = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = $\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}$ = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = 3.15 N towards the positive x-axis

3 0

3 years ago

PLEASE HELP ASAP!!!!!!!!!!!!

lianna [129]

The first choice on the list is the correct one.

7 0

3 years ago

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu

Aloiza [94]

In other words a infinitesimal segment dV caries the charge
dQ = ρ dV 

Let dV be a spherical shell between between r and (r + dr): 
dV = (4π/3)·( (r + dr)² - r³ ) 
= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) 
= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) 
drop higher order terms 
= 4·π·r²·dr 

To get total charge integrate over the whole volume of your object, i.e. 
from ri to ra: 
Q = ∫ dQ = ∫ ρ dV 
= ∫ri→ra { (b/r)·4·π·r² } dr 
= ∫ri→ra { 4·π·b·r } dr 
= 2·π·b·( ra² - ri² ) 

With given parameters: 
Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) 
= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) 
= 3.77×10⁻⁸C 
= 37.7nC

6 0

3 years ago

A bowling ball rolled with a force of 20 N accelerates at a rate of 5 m/sec2, a second ball rolled with the same force accelerat

emmasim [6.3K]

Answer:

sorry I need points

Explanation:

IM SO SORRY

5 0

2 years ago

I NEED HELP ASAP

34kurt

I think it’s gonna be b.

4 0

2 years ago