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Thepotemich [5.8K]

3 years ago

5

When placed in a lake, an object either floats on the surface or sinks. It does not float at some intermediate location between

the surface and the bottom of the lake. However, a weather balloon floats at some intermediate distance between Earth's surface and the top of its atmosphere. Explain.

2 answers:

STALIN [3.7K]3 years ago

3 0

The particle in the air match the mass with the particles around the balloon stoping it from floating any higher

Gnoma [55]3 years ago

3 0

Explanation:

i get your your qiestion but i am not trying to be annoying so please what is a wheather ballon

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A container is filled to a depth of 22.0 cm with water. On top of the water floats a 28.0-cm-thick layer of oil with specific gr

Ilya [14]

Answer:104.853 kPa

Explanation:

Given

Depth of water $h_2=22 cm$

on the top of water a layer of $h_1=28 cm$ oil layer is present

specific gravity of oil $\rho _o=0.5\rho _w$

Assuming Atmospheric Pressure to be $101.325 KPa$

We know Pressure due to Pressure Difference is given by

$\Delta P=\rho \times g\times h$

$\Delta P=\rho _0\times g\times h_1+\rho _w\times g\times h_2$

$\Delta P=0.5\rho _w\times g\times h_1+\rho _w\times g\times h_2$

$\Delta P=0.5\times 1000 \times 9.8\times 0.28+1000\times 9.8\times 0.22$

$\Delta P=1.372\times 10^3+2.156\times 10^3$

$\Delta P=3.528 kPa$

$Absolute\ Pressure=Atmospheric\ Pressure+\Delta P=101.325+3.528$

$Absolute\ Pressure =101.325+3.528=104.853 kPa$

5 0

3 years ago

Particle A of charge 3.30 10-4 C is at the origin, particle B of charge -6.24 10-4 C is at (4.00 m, 0), and particle C of charge

gulaghasi [49]

Answer:

a) $E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) zero

Explanation:

a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:

$E_T=E_1+E_2$

E1: electric field of charge 1

E2: electric field of charge 2

It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:

$E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]\\\\$

r13: distance between charges 1 and 3

r12: charge between charges 2 and 3

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

Thus, you first calculate the distance r13 and r23, and also the angles:

$r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°$

Next, you replace the values of all parameters in order to calculate E1 and E2:

$E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}$

finally, you obtain for ET:

$E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.

4 0

3 years ago

At 10:30 AM, detectives discover a dead body in a room and measure its temperature at 31∘C. One hour later, the body’s temperatu

STALIN [3.7K]

The reference time is 10:30 AM,

The body temperature is 31 ° C,

The room temperature ( $T_A$ ) is 20 ° c,

Newton and the law of cooling are useful in this situation, the equation says,

$\frac{dT}{dt} = -k(T-T_A)$

Where k is proportionality constant

t= time

T= Temperature of Body,

Thus,

$\frac{dT}{dt} = - k(T-20)$

$\frac{1}{T-20}dT = -kdt$

We can integrate here,

$\int \frac{1}{T-20}dT = -k\int dt$

$ln(T-20) = -kt+c$

$T-20=e^{-kt+c}$

We have our first equation,

$T=20+e^{-kt}e^c$

When the measurement t = 0 was made, the temperature T = 31 ° c

$31=20+e^{-k(0)}e^c$

11=e^c

We have know that $e^c = 11$

Our equation now is,

$T=20+11e^{-kt}$

When t=1hour then T=24.8, then

$24.8=20+11e^{-k(1)}$

$24.8-20=11e^{-k}$

$\frac{4.8}{11}=e^{-k}$

$ln(\frac{4.8}{11})=-k$

$-k=-0.83$

Then our equation is now,

$T=20+11e^{0.83t}$

Finally, we know can find the time when T=37, so

$37=20+11e^{-0.83t}$

$e^{-0.83t}=\frac{17}{11}$

-0.83t=ln\frac({17}{11})

$t=-0.52$

$t=31min$

<em>Therefore the time is 10:30-31min, we have that at 9:59AM the body was still alive</em>

6 0

3 years ago

What happens when the number of protons in an atom changes

MatroZZZ [7]

When you change the number of protons in an atom, you will change the atom from one element to a different element. Sometimes, when you add a proton to an element, the element will become radioactive. If you change the number of neutrons in an atom, you get an isotope of the same element.

7 0

3 years ago

The volleyball team practiced their serves, rotations, and ________. Which choice completes the sentence with correct parallel s

Phoenix [80]

Answer

Correct Answer is a.(spikes)

Explanation:

option a is correct because here serves, rotation and spikes all the three words are using as object.

The volleyball team practiced their serves, rotations, and spikes.

option b cannot be correct because here spiking is used as a verb and sentence(The volleyball team practiced their serves, ) is in past indefinite form and we cant use word spiking here because this form of verb is used in continues form of sentences only.

Option c cannot be correct as because in this option ""spikes the ball" spikes is used as a verb which can be used in present indefinite form of sentence only but here sentence is in past indefinite form.

Again option d cannot be correct because in this option ""spikes over the net" spikes is used as a verb which can be used in present indefinite form of sentence only but here sentence is in past indefinite form.

5 0

3 years ago

Read 2 more answers