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Which characteristic does inertia describe

nadezda [96]

Answer:

Inertia is the resistance of any physical object to any change in its velocity. This includes changes to the object's speed, or direction of motion. An aspect of this property is the tendency of objects to keep moving in a straight line at a constant speed, when no forces act upon them.

Explanation:

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3 0

3 years ago

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Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied

Karolina [17]

Ans:

12500 N/C

Explanation:

Side of square, a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

$E = \frac{Kq}{r^2}$

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD = $\sqrt{2}\times a$

where, a be the side of the square

So, Electric field at B due to charge at A

$E_{A}=\frac{Kq}{a^2}$

$E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}$

EA = 6531.32 N/C

Electric field at B due to charge at C

$E_{C}=\frac{Kq}{a^2}$

$E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}$

Ec = 6531.32 N/C

Electric field at B due to charge at D

$E_{D}=\frac{Kq}{2a^2}$

$E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}$

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

$E=\sqrt{E_{x}^{2}+E_{y}^{2}}$

$E=\sqrt{8840.5^{2}+8840.5^{2}}$

E = 12500 N/C

Explanation:

8 0

4 years ago

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The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Natalka [10]

Answer:

$EMF = 1684.67 Volts$

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

$EMF = \frac{d\phi}{dt}$

now we have

$\phi = BA$

here since magnetic field is constant so we have

$EMF = A\frac{dB}{dt}$

now we have

$A = (190 \times 10^3)(190 \times 10^3)$

$A = 3.61 \times 10^{10} m^2$

now we have

$EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})$

$EMF = 1684.67 Volts$

6 0

3 years ago

A car drives to the east in a time of 6 hours. Then, immediately (not realistic, but just assume this is the case for this probl

garri49 [273]

Answer:

Average speed of car in the first trip is 10 km/hr

Explanation:

It is given that first the car drives 6 hours to the east

Then travels 12 km to west in 3 hours

Average speed for the entire trip = 8 km/hr

Total time = 3+6 = 9 hour

So distance traveled in 9 hour = 9×8 = 72 km

As the car travel 12 km in west so distance traveled in east = 72-12 = 60 km

Time by which car traveled in east = 6 hour

So speed $=\frac{distance}{time}=\frac{60}{6}=10km/hr$

So average speed of car in the first trip is 10 km/hr

7 0

4 years ago

What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t

vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0

3 years ago

Someone HELPPPPPPPPP MARKING BRAINLIST JUST CHECK WORK (15 points)