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3 years ago

What is the magnitude and direction of an electric field that exerts a 2.00 * 10^-5 N upward force on a −1.5 μC charge?

Physics

1 answer:

Over [174]3 years ago

4 0

The relation between electric field intensity, electric force and charge is given as follow,

$E = \frac{F}{q}$

Where,

F = Force

q = Charge

The direction of force on a charge placed in an electric field is same as the direction of electric field if the charge has positive polarity and opposite if charge has negative polarity.

Replacing we have that

$E = \frac{2*10^{-5}N}{-1.5*10^{-6}C}$

$E = -13.3N/C$

As direction of force is upward and polarity of charge is negative, hence the direction of field will be opposite to that of force. Therefore direction of field is downward.

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Some believe that the positions of the planets at the time

laila [671]

Answer:

$1.4007\times 10^{-8}\ N$

$1.50075\times 10^{-6}\ N$

$0.000000667\ N$

Explanation:

$m_1$ = Mass of baby = 3 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Distance between objects

Gravitational force of attraction is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 70\times 3}{1^2}\\\Rightarrow F=1.4007\times 10^{-8}\ N$

The force between baby and obstetrician is $1.4007\times 10^{-8}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(6\times 10^{11})^2}\\\Rightarrow F=1.50075\times 10^{-6}\ N$

The force between the baby and Jupiter is $1.50075\times 10^{-6}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(9\times 10^{11})^2}\\\Rightarrow F=0.000000667\ N$

The force between the baby and Jupiter is $0.000000667\ N$

7 0

3 years ago

How does the earth orbit the sun?

scoray [572]

Answer:

The Sun's gravity pulls on the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

Explanation:

Hope that helps

6 0

2 years ago

A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the cen

Liono4ka [1.6K]

Answer:

Part a)

$\theta = 8.05 degree$

Part b)

$a = 1.37 m/s^2$

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

$mg sin\theta - F_f = ma$

also we have torque equation on it

$F_f R = I\alpha$

for pure rolling

$a = R \alpha$

$F_f = \frac{Ia}{R^2}$

now we have

$mg sin\theta = ma + \frac{Ia}{R^2}$

now we have

$mg sin\theta = (m + \frac{2}{5}m)a$

$a = \frac{5}{7}g sin\theta$

now given that

$a = 0.10 g$

so we have

$0.10 g = \frac{5}{7} g sin\theta$

$sin\theta = 0.14$

$\theta = 8.05 degree$

Part b)

If the inclined plane is frictionless then the acceleration is given as

$a = g sin\theta$

$a = 9.8(0.14)$

$a = 1.37 m/s^2$

5 0

3 years ago

Read 2 more answers

Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci

EleoNora [17]

We subtract the velocity of car 1 from the velocity of car 2:
$v=(30\ mph\ North)-(20\ mph\ South)$
$=(30\ mph\ North)+(20\ mph\ North)$
$=50\ mph\ North$

4 0

2 years ago

Read 2 more answers

The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel

aniked [119]

Answer:

= 2.52 x 10^ 6 m/s

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv) ÷ d

Here, d = distance between the two plates

q = charge of the charged particle

Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a = (q) (Δv) ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv) ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv) ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 = ut + 1/2 at^2 [proton}

put d/2 = ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron) ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

Note, e = 1.60 x 10^-19

m(electron) = 9.11 X 10^-31

m(proton) = 1.67 X 10^-27

Input these values into the formula above, initial speed, UI =

= 2.52 x 10^ 6 m/s

7 0

3 years ago