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2 years ago

b) With the aid of diagrams, briefly describe an experiment to determine the density of an irregular shaped object (eg. stone).

Physics

1 answer:

Ludmilka [50]2 years ago

4 0

Answer:

check the explanation

Explanation:

To find density you need mass and volume

to find mass use a digital balance and record it. Ex - 50g
to find the volume of the object
use a measuring cylinder and lets say you fill it up to 50cm³
then add the irregular object to the cylinder and record the new volume Ex - 54cm³
to find the volume of the object subract the final volume from the intial volume

54cm³-50cm³=4cm³

now that you have mass and volume use the equation D=M/V

Density = Mass = 50g = 12.5g/cm³
 Volume 4cm³

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A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

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2 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

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3 years ago

A 10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead, which is much denser than aluminum.

zalisa [80]

Answer:

Aluminium

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The amount of buoyant force depends on the volume of body immersed, density of liquid and the value of acceleration due to gravity.

Here, the density of liquid is same in both the cases and g be the same. So, here the amount of buoyant force depends on the volume of body immersed.

As the density of lead is more than the density of aluminium, so the volume of aluminium is more than lead, as volume is equal to mass divided by density. So, the buoyant force acting on the aluminium is more than lead.

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3 years ago

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Answer:

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3 years ago

A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

stiks02 [169]

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3 years ago

b) With the aid of diagrams, briefly describe an experiment to determine the density of an irregular shaped object (eg. stone).​

b) With the aid of diagrams, briefly describe an experiment to determine the density of an irregular shaped object (eg. stone).