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Tasya [4]
3 years ago
12

Charges of +2 µC and +3 µC are 4 mm from each other. Raina’s group draws four diagrams trying to represent the electrical force

between the charges.

Physics
2 answers:
BlackZzzverrR [31]3 years ago
6 0

Answer: I believe that on Edge it is X

Explanation:

ahrayia [7]3 years ago
4 0
The correct diagram is shown below:

The charges of +2 µC and +3 µC are 4 mm from each other. The diagram below represents the electrical force between the charges. i.e. repulsive force. However the force of repulsion exerted by charge +3 µC on +2<span> µC will be more. The same charges repel each other and opposite charges attract each other.</span>

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A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of th
zheka24 [161]

The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m  = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Learn more on momentum here: brainly.com/question/402617

5 0
2 years ago
At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta
Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

  • u = initial speed of the automobile = 0 m/s
  • a = constant acceleration of the automobile = 2.5\ m/s^2
  • v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

  • t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0
3 years ago
What’s the difference between distance and displacment
Brut [27]
Distance is how far you are and displacment is separate like for someone oh From somewere
3 0
3 years ago
An experiment is conducted using 2 plants to determine if the amount of sunlight they receive affects how fast they grow. Which
Phantasy [73]

Greetings!

The correct answer choice is Choice 4.

<em>Why?</em>

In a scientific experiment the only thing being changed is the independent variable. Everything else should stay the same.

In this experiment, the independent variable is the amount of sunlight each plant should receive. <em>Here's a tip</em>- when looking for and independent variable look for whats being changed on purpose.

Hope this helps!

~Fluerie

4 0
3 years ago
Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor
rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

e_{1}=\frac{E}{4\pi r^{2}}

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}

Hence we sense it as 4 times fainter than the nearer star.

5 0
3 years ago
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