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Vadim26 [7]
3 years ago
5

The picture above shows 3 sets of balloons, all with a particular charge. Which of the picture(s) is true? Explain. Then explain

why the other 2 pictures are incorrect. Make sure to explain all 3 sets.

Physics
2 answers:
IrinaVladis [17]3 years ago
8 0
Among all of the others Picture 3 is correct.

We know, Like charges repel each other whereas unlike charges attract each other, In picture 3 they are like charges (both are positive) so, we can repulsion hence, picture is correct!

In picture 1, they are unlike charges, & showing repulsion which is wrong, same In picture 2, they are like charges & showing attraction so they are wrong as well

Hope this helps!
MariettaO [177]3 years ago
7 0
C is correct because they would repel each other A is wrong be they wouldn't repel And B is wrong because they shouldn't be repelling each other
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Given a volume of 1000. cm^3 of an ideal gas at 300. K, what volume would it occupy at a temperature of 600. K?
Inessa [10]

Answer:2000cm3

Explanation: this is Charles' law which has the expression V1/T1 = V2/T2

Making V2 the subject of the formula, we have that

V2 = V1 x T2/ T1

= 1000 x 600 / 300

= 2000cm3

Where:

V1= initial velocity

V2= Final velocity

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6 0
3 years ago
The distance between two adjacent peaks on a wave is called the wavelength. The wavelength of a beam of ultraviolet light is 113
Vesna [10]

Answer:

0.000000113 or 1.13*10^{-7} meters

Explanation:

One nanometer is 10^{-9} meters. So 113 nanometers would be 113*10^{-9}, or 1.13*10^{-7} meters. That's expessed on "cientific notation." On the "usual" notation, it will be 0.000000113 meters.

3 0
3 years ago
A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H
Wewaii [24]

Answer:

(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s

Explanation:

Given data

v_{o}=7.14m/s\\\alpha =36.5^o

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2

Using equation of motion

v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s

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