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3 years ago

What type of climate would be near cold ocean currents? PLEASE HELP QUICK!!

Physics

1 answer:

olya-2409 [2.1K]3 years ago

5 0

Answer:

hot humid with lots of rain.

Explanation:

ocean currents act as conveyer belts of warm and cold water sending heat to the polar regions and helping the tropical areas cool off, thus influencing both weather and climate. the tropics are particularly rainy because heat absorption , and thus ocean evaporation, is highest.

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A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

IrinaK [193]

Answer : The specific heat capacity of the alloy $1.422J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$C_1$ = specific heat of alloy = ?

$C_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of alloy = 21.6 g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of system = $31.10^oC$

$T_1$ = initial temperature of alloy = $93.00^oC$

$T_2$ = initial temperature of water = $22.00^oC$

Now put all the given values in the above formula, we get

$21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC$

$c_1=1.422J/g^oC$

Therefore, the specific heat capacity of the alloy $1.422J/g^oC$

6 0

3 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the insta

rusak2 [61]

From an energy balance, we can use this formula to solve for the angular speed of the chimney

ω^2 = 3g / h sin θ

Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s

The formula for radial acceleration is:
a = rω^2

So,
a = 53.2 (0.987) = 52.494 /s^2

The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887

The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2

If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°

7 0

3 years ago

A car moved 60 meters west in 2 hours. What is its average velocity?

Fofino [41]

Explanation: Velocity is the displacement of an object during a specific unit of time. Two measurements are needed to determine velocity. Displacement and time. Displacement includes a direction, so velocity also includes a direction. Speed with direction. Velocity can be an average velocity or an instantaneous velocity. Units for velocity are the same as for speed: m/s, km/h, and mph. Delta x(Δx) is the symbol used for displacement. Delta (Δ) means to "change in." Δx means to "change in position." Δx is calculated by final position minus initial position. Velocity formula: → v=Δx/t as a fraction.

v=Δx/t

$v=\frac{xf-xi}{t}= \frac{60m}{2}=30m$

<em><u>Final answer is 30.</u></em>

Hope this helps!

Thanks!

Have a great day!

-Charlie

5 0

3 years ago

Our sun is unusual because it does not belong to a(n) ____

lesantik [10]

The answer is binary star system

5 0

3 years ago

Read 2 more answers