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34kurt
1 year ago
8

Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau

se of the electrostatic charge on the hairspray droplets. If two drops of hairspray repel each other with a force of 9 x 10^-9 N at a distance of 0.07 cm ( 7 x 10^ -4 m), what is the charge on each of the equally charged drops of hairspray?
Please show the steps as well. Written out, if possible.
Physics
1 answer:
____ [38]1 year ago
7 0

The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

Chrge,q₁ = q₂ =q C

From Columb's law;

\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9}  = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\  q = 7 \times 10^{-13} \ C

Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

To learn more about Columb's law refer to the link;

brainly.com/question/1616890

#SPJ1

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Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
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Answer with Explanation:

Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.

For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as  pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference

While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.  

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3 years ago
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(I even worked on a microwave system in South America where huge grid dishes were used on a 90-mile link.)

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