All categories

3 years ago

A submarine is at an ocean depth of 3.00 km.

Physics

1 answer:

Alexxx [7]3 years ago

7 0

Answer:

1.#(#('#;#+#('#;#+jwhsjjshajahkqhwmskek

Explanation:

jsjsnanakamakwjKbakqwigjgodrgoshoihrdohrf

tjfjrhgcorfhptgkc

tgj

crlfkhc

rjv

tkgjv

kyjv

tgjvhrlkrfjhirghogrtiguotightjptrjpghpf

rofrirhfohrihithfitgprirl

ceohr

fbojofjigoitigotjog

irojfofjkrhrjoejoqjodirnkfnkrjoritpruixhor

firhofeiidiehdijeihr
jdhxjdhjchfihvidjfdoigorjgiifugiruogu
ifgijrojiru

ifhfojrorior

bifjibjtgitkriwgi eror iurh oryi ehiwyrfoerhfoefoeuis in het begin van een tui ueen uitgebreide beschrijving ygt grup u een uitgebreide keuze u guugu

sjksjsosjsnshf ruor uw eigen website aanmaken r y fav u can u to uf the owner u will tu tutururu ba sa inyo ay walang naging y rin h results ry from y yifuirfjofjkrjirhieifr Fu t I

You might be interested in

A capacitor consists of two concentric cylinders. The inner cylinder has a radius of 0.001 m and the outer cylinder a radius of

AlexFokin [52]

Answer:

The capacitance is 1.75 nF

Explanation:

From the question we are given that

The inner radius is $r_{in} = 0.001$

The outer radius is $r_{out} = 0.0011 \ m$

Length of the capacitor is $L = 1m$

The dielectric constant is $Di = 2 \ for \ 0 < \phi < \pi$

The dielectric constant is $Di_2 = 4 \ for \ \pi < \phi < 2\pi$

Generally the capacitance of a capacitor can be mathematically represented as

$C = \frac{\pi \epsilon_0 Di_1 L}{ln\frac{r_{out}}{r_{in}} } + \frac{\pi \epsilon_0 Di_2L}{ln\frac{r_{out}}{r_{in}} }$

$= \frac{\pi \epsilon_0 L (Di_1 + Di_2)}{ln\frac{r_{out}}{r_{in}} }$

$= \frac{(3.142)(8.85*10^{-12})(1)(2+4)}{ln\frac{0.0011}{0.001} }$

$=1.75*10^{-9} F$

$1.75nF$

5 0

3 years ago

A positive charge is fixed at point (−4,−3) and a negative charge − is fixed at point (−4,0). Determine the net electric force ⃗

Masteriza [31]

The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

<h3>Net electric force on the charges</h3>

The net electric force on the charges is calculated as follows;

F = kq₁q₂/r²

where;

k is coulomb's constant
q₁ and q₂ are the charges
r is the distance between the charges

<h3>Distance between the charges</h3>

$|r| = \sqrt{(x_2-x_1)^2+ (y_2-y_1)^2} \\\\|r| = \sqrt{(-4--4)^2+ (0--3)^2} \\\\|r| = \sqrt{(0)^2 + (3)^2} \\\\|r| = 3 \ units$

$F_{net} = \frac{kq_1q_2}{3^2} \\\\F_{net} = \frac{1}{9} (kq_1q_2) , \ N$

Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

Learn more about electric force here: brainly.com/question/17692887

#SPJ1

3 0

2 years ago

A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com

blagie [28]

Answer:

4500.5 nutritional calories per gram

Explanation:

Heat lost by the new candy = heat gained by the bomb calorimeter.

Heat gained by the bomb calorimeter = c×ΔT

where c = heat capacity of the calorimeter = 32.20 KJ/K = 32200 J/K

ΔT = change in temperature = 2.69°C = 2.69 K.

Heat gained by the bomb calorimeter = 32200 × 2.69 = 86618 J

Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories

4.60 g of the new candy lost this amount of calories by undergoing combustion,

The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram

8 0

3 years ago

What is an example of a wave that is not mechanical and how is it different?

vovikov84 [41]

Answer:

light is an example of a wave that is not mechanical .

it is different as it does not need material medium for its propagation

5 0

3 years ago

A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed

Marina86 [1]

Answer:

a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

Magnetic filed ,B= 0.003 T

Speed ,V= 500 m/s

θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

$a=\dfrac{qVB\ sin\theta}{m}$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2$

a=0.212 m/s²

4 0

3 years ago