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3 years ago

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A pathogen has entered your body and you become sick. You develop a fever and vomit. This is your body's way of trying to mainta

in

1 answer:

larisa [96]3 years ago

4 0

Homeostasis is the answer

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If the force that propels the cannonball forward is 500N, how much force will move the cannon backward?

Licemer1 [7]

Answer:

Well it would be equal to 500N because pushing forward the ball (or whatever maybe a body) would push the canon back an even 500N backwards...

Explanation:

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3 years ago

Five locations are marked on the world map. which location is most prone to hurricanes? five locations are marked on the world m

Alex17521 [72]

Five locations are marked on the world map. The spot that is prone to hurricane mostly A.

<h3>What is hurricane?</h3>

A hurricane is a cyclone with winds of 74 miles (119 kilometers) each hour or more prominent that is typically joined by downpour, thunder, and lightning, and that occasionally moves into calm scopes.

The locations on the world map spot A, B, C, D, E are regions with country prone to hurricane.

The SPOT A, is GULF of Mexico, a sea bowl nearer to the Atlantic sea, encompassed with the North American landmass, the Hawaiian island is likewise inside that locale.

The SPOT B, is the southern Pacific sea with nations like BRAZIL, PERU, ECUADOR, CHILE, in that equivalent locale. This is likewise visited by the tropical storm.

The SPOT C, is inside the locales of AFRICA, nations at the edge are likewise impacted by typhoon.

The SPOT D, is Asian landmass with Russia at the edge. They are likewise inclined to typhoon.

The SPOT E, is the North Pacific sea locale situated at the upper left hand side of the world guide, with nations like Canada, Alaska and co. Near the artic circle.

Thus, location A is the most prone to hurricanes.

Learn more about hurricane.

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2 years ago

A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt

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Kinwtic energy = (40kg*2*2)/2=40*2=80J

4 0

4 years ago

Read 2 more answers

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

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<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

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<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

<h3>Step B:</h3>

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre

IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0

3 years ago