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3 years ago

1. why is experimentation so important to science

Physics

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer: to provide evidence to a theory

Explanation: Experimentation allows for multiple trials to provide evidence to a scientific theory. Without experimentation there would be no data to back up your hypothesis.

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A paper filled capacitor is charged to a potential difference of 2.1 V and then disconnected from the charging circuit. The diel

Goryan [66]

Answer:

Part a)

$V' = 7.77 Volts$

Part b)

$k' = 6.27$

Explanation:

As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery

So here we can say that charge on the plates will remain conserved

So we will have

$Q = kC(2.1)$

now dielectric is removed between the plates of capacitor

so new potential difference between the plates

$V' = \frac{Q}{C'}$

$V' = \frac{kC(2.1)}{C}$

$V' = 3.7 \times 2.1$

$V' = 7.77 Volts$

Part b)

Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates

So again charge is same so potential difference is given as

$V" = \frac{Q}{k'C}$

$0.59 V = \frac{kCV}{k'C}$

$0.59 = \frac{3.7}{k'}$

$k' = 6.27$

5 0

2 years ago

What is the objects average velocity?

lyudmila [28]

Answer:

answer should be 10 because the line goes from (0,0) then to (1,10) and so on

5 0

2 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$