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3 years ago

The weather is warm and dry. Which change would a cold front bring?

Physics

2 answers:

astra-53 [7]3 years ago

8 0

Answer:

im prob too late lol but the answer would be rain or thunderstorms

Explanation:

i took the test :p

andreev551 [17]3 years ago

3 0

Answer:

rain or thunder

Explanation:

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three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

5 0

3 years ago

Air at 27oC and 1 atm flows over a flat plate 40 cm in length and 1 cm in width at a speed of 2 m/s. The plate is heated over it

irga5000 [103]

Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

$T_1$ = 27°C = (273 + 27) K = 300 K

$T_2$ = 600°C = (600 +273) K = 873 K

speed v = 2 m/s

length x = 40 cm = 0.4 cm

width = 1 cm = 0.001 m

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

$k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg$

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = $\dfrac{\rho \times v \times x }{\mu}$

reynolds number = $\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}$

reynolds number = $\dfrac{0.98}{18.6 \times 10^{-6}}$

reynolds number = 52688.11204

Prandtl number = $\dfrac{c_p \mu}{k}$

Prandtl number = $\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}$

Prandtl number = $\dfrac{0.018693}{0.02476}$

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate can be computed as follows:

Nusselt no = $\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} =161.4252008}$

$h =\dfrac{161.4252008 \times 0.02476}{ 0.4}$

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = $h\times A \times (T_2-T_1)$

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

6 0

3 years ago

An object weighs 15N in air and 13N when completely

anastassius [24]

Answer:

The volume of water displaced = 0.204 m³

Explanation:

From Archimedes' principle,

Upthrust = lost in weight = weight of water displaced.

U = W₁ - W₂..................... Equation 1

Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.

Given: W₁ = 15 N, W₂ = 13 N

Substituting these values into equation 1

U = 15 - 13

U = 2 N

But,

Weight of water displaced = mass of water displaced. × acceleration due to gravity.

Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

8 0

3 years ago

Substances which naturally attract each other called what

grandymaker [24]

Answer:

Ferromagnetism, physical phenomenon in which certain electrically uncharged materials strongly attract others. Two materials found in nature, lodestone (or magnetite, an oxide of iron, Fe3O4) and iron, have the ability to acquire such attractive powers, and they are often called natural ferromagnets.

Hope this help :)

5 0

3 years ago

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$