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Grace [21]
2 years ago
9

1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer

depend on whether the charges were of the same magnitude or different? How does this relate to Newton’s 3rd law?
Physics
1 answer:
Aloiza [94]2 years ago
7 0

Answer:

Following are the solution to the given question:

Explanation:

Its strength from both charges is equivalent or identical. The power is equal. And it is passed down

F=\frac{kq_1q_2}{r^2}

Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.

|F_{12}| = |F_{21}|

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A force of 10 N causes a spring to extend by 20 mm. Find: a) the spring constant of the spring in N/m​
Mars2501 [29]

Answer:

formula used K=F/∆l

∆l is the elongation of the spring

  1. F=10N
  2. ∆l=20mm===> 0.02m
  3. K=10N divided 0.02m= 500N/m
6 0
2 years ago
3. Take sugar, oil, corn syrup, a glass and water. Pour the water in the glass and then add each of the above the substances one
hoa [83]

Here are the observations

<u>S</u><u>u</u><u>g</u><u>a</u><u>r</u><u>:</u><u>-</u>

  • Sugar is soluble in water
  • so It will dissolve in water .

<u>C</u><u>o</u><u>r</u><u>n</u><u> </u><u>s</u><u>y</u><u>r</u><u>u</u><u>p</u><u>:</u><u>-</u>

  • Corn syrup is also basically a sugar.
  • It will dissolve in water too .
  • If we shake the mixture in glass then corn syrup will be dissolved.

<u>O</u><u>i</u><u>l</u><u>:</u><u>-</u>

  • Oil is not soluble in water
  • Hence it won't dissolve in water.
  • It will float over water and make two layers
7 0
2 years ago
Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el
Rina8888 [55]
<span>this  may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
8 0
3 years ago
Read 2 more answers
A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'
qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s


Check the calculations again !


7 0
3 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
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