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2 years ago

What rule should be used to transform a table of data to represent the

Physics

1 answer:

adoni [48]2 years ago

4 0

Answer:

Multiply the x values with -1.

Explanation:

By multiplying the numbers by one, you are changing them to be the opposite of their original state.

You multiply the numbers that are in the x-value column because you are reflecting the image over the y-axis.

Hope this helped and good luck!

Read more about Speed

brainly.com/question/4931057

7 0

1 year ago

_____ In order to form an electric circuit, you need to have a power source. a light bulb or some resistance. wires or conductor

My name is Ann [436]

Answer:

True

Explanation:

The satement is valid.

8 0

2 years ago

state the principle of the Conservation of Linear Momentum and show how it follows from Newton's second law of motion

photoshop1234 [79]

Explanation:

The linear momentum of a particle is defined as the product of the mass of the particle times the velocity of that particle. Conservation of momentum of a particle is a property exhibited by any particle where the total amount of momentum never changes. Linear momentum of a particle is a vector quantity and is denoted by →p

6 0

2 years ago

A 26.5-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same dire

Dahasolnce [82]

Answer:

v₁ =0.19 m/s and v₂ = 0.18 m/s

Explanation:

By conservation of energy and conservation of momentum we can find the velocity of each object after the collision:

Momentum:

Before (b) = After (a)

$m_{1}v_{1b} + m_{2}v_{2b} = m_{1}v_{1a} + m_{2}v_{2a}$

$26.5 kg*0.205 m/s + 12.5 kg*0.150 m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ $7.31 kg*m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ (1)

Energy:

Before (b) = After (a)

$\frac{1}{2}m_{1}v_{1b}^{2} + \frac{1}{2}m_{2}v_{2b}^{2} = \frac{1}{2}m_{1}v_{1a}^{2} + \frac{1}{2}m_{2}v_{2a}^{2}$

$26.5 kg*(0.205 m/s)^{2} + 12.5 kg*(0.150 m/s)^{2} = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$ (2)

From equation (1) we have:

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg}$ (3)

Now, by entering equation (3) into (2) we have:

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*(\frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg})^{2}$ (4)

By solving equation (4) for $v_{1a}$ , we will have two values for

$v_{1a} = 0.16$

$v_{1a} = 0.21$

We will take the average of both values:

$v_{1a} = 0.19 m/s$

Now, by introducing this value into equation (3) we can find $v_{2a}$ :

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*0.19 m/s}{12.5 kg}$

$v_{2a} = 0.18 m/s$

Therefore, the velocity of object 1 and object 2 after the collision is 0.19 m/s and 0.18 m/s, respectively.

I hope it helps you!

6 0

2 years ago

A laboratory dish, 20 cm in diameter, is half filled with water. One at a time, 0.49 μL drops of oil from a micropipette are dro

irakobra [83]

Explanation:

Formula for path difference is as follows.

x = 2tn

and, refractive index (n) = $\frac{\lambda}{2t}$

Thickness is calculated as follows.

Thickness (t) = $\frac{volume}{area}$

Area = $\pi r^{2}$

= $\pi \times (\frac{d}{2})^{2}$

= $\frac{0.49 \times 10^{-6}}{3.14 \times 0.01 m}$

= $1.56 \times 10^{-8}$ m

Now, the refractive index will be calculated as follows.

For drop, n = $\frac{\lambda}{2t}$

For B drop, n = $\frac{\lambda}{26t}$

So, n = $\frac{640 \times 10^{-9}}{26 \times 1.56 \times 10^{-8}}$

= $\frac{640 \times 10^{-9}}{40.56 \times 10^{-8}}$

= 1.5

Thus, we can conclude that index of refraction of the oil is 1.5.

6 0

2 years ago