If a force acts constantly on a stationary object, the object will

If object a has more mass than object b does object a contain more matter?

aev [14]

The answer should be a yes

8 0

3 years ago

A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. <em>The correct form is: A </em> $+10\,\mu C$ <em> charge and a </em> $-10\,\mu C$ <em> at a distance of 0.3 meters. </em>

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$F$ - Electrostatic force, in newtons.

$\kappa$ - Electrostatic constant, in newton-square meters per square coulomb.

$|q_{A}|,|q_{B}|$ - Magnitudes of electric charges, in coulombs.

$r$ - Distance between charges, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C$ and $r = 0.3\,m$ , then the magnitude of the electrostatic force is:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$

$F = 9.987\,N$

In consequence, correct answer is B.

4 0

2 years ago

a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o

Fofino [41]

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

$x_1 = 50 cm$ represents an actual distance of

$d_1 = 1.0\cdot 10^5 km$

The actual distance between Mars and the Sun is 228 million km, therefore

$d_2=228\cdot 10^6 km$

On the scale model, this would corresponds to a distance of $x_2$ .

Therefore, we can write the following proportion:

$\frac{x_1}{d_1}=\frac{x_2}{d_2}$

And solving for $x_2$ , we find:

$x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m$

Learn more about distance:

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4 0

2 years ago

What happens if :<br> . The test charge is not tiny.

docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

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3 0

2 years ago

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat

RSB [31]

Answer:

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

$C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\$

where;

V is the potential difference between the plates

The charge on the plates is given as;

$Q = \frac{V\epsilon _0 A}{d}$

The energy stored in the capacitor is given as;

$E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2$

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

3 0

2 years ago