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3 years ago

5) help plz I need help I don't know if I am right so if I am not can you plz help me

Physics

2 answers:

seraphim [82]3 years ago

7 0

the answer is waning Gibbous

The Waning Gibbous is an intermediary Moon phase. It starts right after the Full Moon, and it lasts until the Third Quarter.

inessss [21]3 years ago

7 0

Answer:

the answer is waning gibbous

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A man pulls on his dog's leash to keep him from running after a bicycle. Which term best describes this example? Select one: A.

madreJ [45]

C. Negative force. The dog isn't going to learn that way.

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3 years ago

In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os

GalinKa [24]

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

$\omega=\dfrac{1}{\sqrt{CL}}$

By putting the values

$\omega=\dfrac{1}{\sqrt{CL}}$

$\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}$

ω = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33 A

I= 6.49 A

I = 6.5 A

E) 6.5 A

3 0

3 years ago

A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?

Olenka [21]

To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.

7 0

3 years ago

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Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

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3 years ago

Position vs Time

viktelen [127]

Answer: 1. the object is moving away from the origin

4. the object started at 2 meters

5. the object is traveling at a constant velocity

Explanation:

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3 years ago

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