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notsponge [240]
4 years ago
15

When you place an egg in water, it sinks. If you add salt to the water, after some time the egg floats. Choose the correct expla

nation
A. Adding salt to the water increases its density When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats
B. Adding salt to the water decreases Its density When the densily of the water matches that of the egg, the egg becomes neutrally buoyant and floats
C. Adding salt to the water decreases its volume When the volume of the water matches that of the egg, the egg becomes neutrally buoyant and floats
D. Adding salt to the water increases ts volume When the volume of the water matches that of the egg, the egg becomes neutraly buoyant and floats
Physics
1 answer:
lora16 [44]4 years ago
5 0

Answer:

B

Explanation:

This happens because adding salt to the water decreases Its density. When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats.

The weight of the egg becomes equal to the upward buoyant force by the water on to the egg and hence, the egg floats.

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A wire carries a current of 10 amps in a direction of 90 degrees with respect to the direction of an external magnetic field of
Oksi-84 [34.3K]

Answer:

15 N

Explanation:

The magnetic force on a piece of current-carrying wire is given by:

F=ILB sin \theta

where

I is the current in the wire

L is the length of the piece of wire

B is the magnetic field strength

\theta is the angle between the direction of B and I

In this problem:

I = 10 A

B = 0.3 T

L = 5 m

\theta=90^{\circ}

Substituting into the equation, we find

F=(10 A)(0.3 T)(5 m) sin 90^{\circ}=15 N

6 0
3 years ago
b) A satellite is in a circular orbit around the Earth at an altitude of 1600 km above the Earth's surface. Determine the orbita
asambeis [7]

Explanation:

The orbiting period of a satellite at a height h from earth' surface is

T=2πr32gR2

where r=R+h.

Then, T=2π(R+h)R(R+hg)−−−−−−−−√

Here, R=6400km,h=1600km=R/4

T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√

Putting the given values,

T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h

Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.

Let it be nR.Then

T=2π(R+nR)R(R+nRg)−−−−−−−−−−√

=2π(Rg)−−−−−√(1+n)32

=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32

(5075s)(1+n)32=(1.41h)(1+n)32

For T=24h, we have (24h)=(1.41h)(1+n)32

or (1+n)32=241.41=17

or 1+n(17)23=6.61

or n=5.61

The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.

3 0
3 years ago
Which two aspects of sound go together?
Varvara68 [4.7K]

Which two aspects of sound go together?

  • <em>P</em><em>i</em><em>t</em><em>ch and frequency</em>

<u>Pitch</u><u> </u><u>and</u><u> </u><u>frequency</u><u> </u><u>are</u><u> </u><u>two</u><u> </u><u>aspects</u><u> </u><u>of</u><u> </u><u>sound</u><u> </u><u>that</u><u> </u><u>go</u><u> </u><u>together</u><u>.</u><u> </u><u>Pitch</u><u> </u><u>means</u><u> </u><u>Higness</u><u> </u><u>or</u><u> </u><u>Lowness</u><u> </u><u>of</u><u> </u><u>sound</u><u>.</u><u> </u><u>F</u><u>r</u><u>e</u><u>q</u><u>u</u><u>e</u><u>n</u><u>c</u><u>y</u><u> </u><u>is</u><u> </u><u>the measurement of the number of times that a repeated event occurs per unit of time.</u><u>.</u><u>.</u><u>~</u>

4 0
3 years ago
Read 2 more answers
A ball is thrown straight up into the air with a speed of 10 m/s . If the ball has a mass of 0.3 kg how high does the ball go ?
Damm [24]
<h3><u>Answer;</u></h3>

= 5.102 m

<h3><u>Explanation;</u></h3>

The total energy, i.e. sum of kinetic and potential energy, is constant.

That is; E = KE + PE

Initially, PE = 0 and KE = 1/2 mv^2

At maximum height, velocity=0, thus, KE = 0 and PE = mgh

Since, total energy is constant (KE converts to PE when the ball is rising).

Therefore, KE = PE

or, 1/2 mv^2 = mgh

or, h = v^2 /2g

        = 10^2 / (2x9.8)

        =<u> 5.102 m</u>

5 0
3 years ago
Read 2 more answers
A diatomic ideal gas initially at pressure 3 atm and volume 2 L expands adiabatically to a volume of 8 L. How much work is done
SSSSS [86.1K]

Answer:

option d is correct

work done is 647 J

Explanation:

Given data

pressure P = 3 atm

volume V1 = 2 L

volume V2  = 8 L

to find out

work is done by the gas

solution

we know that work done formula that is

work done = k ( V_{f}^{n}  - V_{i}^{n} ) / n

here n = 1 - γ

and k = PV^γ  

and we know for diatomic gas γ = 1.4

so k will be = ( 3× 101325 ) × (0.002)^{1.4}

k = 50.62

so that now

work done = k ( V_{f}^{n}  - V_{i}^{n} ) / n

work done = 50.62 ( (0.008)^{1-1.4}  - (0.002)^{1-1.4} ) / 1-1.4

so work done will be 647 J

so option d is correct

work done is 647 J

3 0
4 years ago
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