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neonofarm [45]
3 years ago
3

Object A, which has been charged to +13.5 nC, is at the origin. Object B, which has been charged to -22.05 nC, is at x=0 and y=1

.84 cm. What is the magnitude of the electric force on object A?
Physics
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

7.9 x 10^-3 N attractive in nature

Explanation:

QA = 13.5 nC = 13.5 x 10^-9 C

QB = - 22.05 nC = - 22.05 x 10^-9 C

d = 1.84 cm = 0.0184 m

Let the electric force on object A is F.

By use of Coulomb's law in electrostatics

F = K x QA x QB / d^2

F = (9 x 10^9 x 13.5 x 10^-9 x 22.05 x 10^-9) / (0.0184 x 0.0184)

F = 7.9 x 10^-3 N attractive in nature

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gladu [14]
 i thinkits an instrument called seismograph. not sure
3 0
3 years ago
You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame
Novay_Z [31]

Answer:

The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

\phi=NAB

Put the value into the formula

\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5

\phi=0.003180

We need to calculate the emf

Using formula of emf

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{0.003180}{0.14}

\epsilon =-0.000227\ V

\epsilon=-2.27\times10^{-4}\ V

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

8 0
4 years ago
Como foi a despedida do bilhete?​
Dmitrij [34]
Can you translate to english ?
8 0
3 years ago
A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the po
Alex

Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

\vec{F_B}=q\vec{v}\ X\ \vec{B}       (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

8 0
3 years ago
A parallel-plate capacitor is formed from two 9.1 cm-diameter electrodes spaced 1.3 mm apart. The electric field strength inside
vitfil [10]

Answer:

2.87nC

Explanation:

See attached file

6 0
3 years ago
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