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Arte-miy333 [17]

3 years ago

10

A block of aluminum with a mass of 1 kg is placed in a beaker of water filled to the brim. Water overflows. The same is done in

another beaker with a 1-kg block of lead. Does the lead displace more, less, or the same amount of water?

1 answer:

RSB [31]3 years ago

3 0

Answer:Aluminium displaces more water

Explanation:

Given

mass of aluminium block=1 kg

mass of lead block= 1 kg

Also we know density of lead $=11.34 gm/cm^3$

Density of Aluminium $=2.70 gm/cm^3$

As the density of lead is more than that of Aluminium therefore its volume is lesser than Aluminium for same mass.

Thus aluminium displaces more amount of water as compared to lead.

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State the term using to describe the turning force force exerted by the man

defon

Answer:

The distance between the line of action of force and the axis of rotation (or pivoted point)

Explanation:

The distance between the line of action of force and the axis of rotation (or pivoted point) .

4 0

3 years ago

What is the practical applications of radio waves

Keith_Richards [23]

Radio waves have many uses—the category is divided into many subcategories, including microwaves and electromagnetic waves used for AM and FM radio, cellular telephones and TV.

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Extremely low frequency (ELF) radio waves of about 1 kHz are used to communicate with submerged submarines. The ability of radio waves to penetrate salt water is related to their wavelength (much like ultrasound penetrating tissue)—the longer the wavelength, the farther they penetrate. Since salt water is a good conductor, radio waves are strongly absorbed by it; very long wavelengths are needed to reach a submarine under the surface.

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5 0

3 years ago

A boat sails along the shore. To an observer, the boat appears to move at a speed of 11 m/s, and a man on the boat walking forwa

amid [387]

You first subtract the speed at which the man is moving (11 m/s) from the rate the boat is moving (12.4 m/s). Which equals 1.4, then divide it by 6 meters, as the man is moving relative to the boat.

It therefore equals 4.29 s

4 0

3 years ago

Read 2 more answers

7. A spring balance reads force in Newton. The scale is 20 cm long and reads from 0 to

GarryVolchara [31]

Answer:

The potential energy when it reads 40 N is $PE = 5.33 \ J$

Explanation:

From the question we are told that

The lowest reading of the spring balance is 0 N and this is at 0 cm = 0 m

The height reading of the spring balance is 60 N and this is at 20 cm = 0.20 m

Generally the length corresponding to the reading of 40 N is mathematically represented as

$d = \frac{40 * 0.20 }{60 }$

=> $d = 0.133 \ m$

Generally the potential energy is mathematically represented as

$PE = m * g * d$

Here

$F = mg = 40 N$

So

$PE = 40 * 0.133$

=> $PE = 5.33 \ J$

7 0

3 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

<u>Now the force due to second charge:</u>

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

<u>Now the force due to third charge:</u>

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

<u>Now the its horizontal component:</u>

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

3 0

3 years ago