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2 years ago

Why don’t trade winds blow straight toward the equator Apex

1 answer:

6 0

Trade winds near equator blows in curve path instead of straight path. This is because of earth rotation. This effect of earth rotation that cause wind to move in curve motion is called Coriolis effect. These kind of wind blows at the northeast of the North hemisphere and southeast of the South hemisphere. The trade wind are warm and it blows due to rising of hot air from equator.

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If a girl carries groceries up a flight of stairs, is she doing work on the groceries? Explain.

nikdorinn [45]

Answer:

Explanation:

Yes she is doing work. With or without the groceries, she is still doing work. She does more work with the groceries than without because Work is defined by F which is defined by mass. The mass increases with the groceries.

The work done is against the force of gravity.

4 0

2 years ago

Please help me with this physics problem

Vesna [10]

Answer:

idk

Explanation:

4 0

2 years ago

What is a true statement about the rotation period of the moon?

mrs_skeptik [129]

The time it takes for the Moon to rotate once around its axis is equal to the time it takes for the Moon to orbit once around Earth

5 0

2 years ago

A 17.0 resistor and a 6.0 resistor are connected in series with a battery. The potential difference across the 6.0 resistor is m

tia_tia [17]

Answer:

V= 57.5 V

Explanation:

If the resistors are in the linear zone of operation, the potential difference across them, must obey Ohm's law:

$V = I*R$

For the 6.0 Ω resistor, if the potential difference across it is 15 V, we can find the current flowing through it as follows:

$I = \frac{V}{R} = \frac{15 V}{6.0 \Omega} = 2.5 A$

In a series circuit, the current is the same at any point of it, so the current through the battery is I = 2.5 A
The equivalent resistance of a series circuit is just the sum of the resistances, so, in this case, we can write the following equation:

$R_{eq} = R_{1} +R_{2} = 17.0 \Omega + 6.0 \Omega = 23.0 \Omega$

Applying Ohm's Law to the equivalent resistance, we can find the potential difference through it, that must be equal to the potential difference across the battery, as follows:

$V = I* R_{eq} = 2.5 A * 23.0 \Omega = 57.5 V$

8 0

2 years ago

A 150 g copper bowl contains 210 g of water, both at 24.0°C. A very hot 430 g copper cylinder is dropped into the water, causing

Dahasolnce [82]

Answer:

A. 15969.22 cal

B. 1052,22 cal

C. 528,87 °C

Explanation:

To solve this kind of question, a proper method is to work from the data that you have towards the data that you need. Also, it is recommended to analyze related equations as they could give us clues on how to find the missing information or the information that the problem is asking us.

Let us start with Question A. It is important to remember that energy transfers with the environment are being neglected; this means that all the energy that the cylinder lose is picked up by the water and the copper bowl. To find the amount of energy transferred to the water, we first find the amount of energy necessary to raise the water’s temperature to 100°C and then we find the amount of energy necessary to evaporate the 17.1 g of water indicated by the question. This would be:

Q = m_water * CP_water *∆T =210g *1 cal/(g K) * (100°C-24°C) = 15960 cal

Q_evap = m_wat * L = 17,1 g * 539 cal/kg* (1 kg)/(1000 g) =9.2169 cal

Therefore, the total energy that was transferred to the water is the sum of these components, that would be Q_tot = 15960 cal + 9.2159 cal = 15969.22 cal. Let´s also remember that a temperature difference in K is equal to a temperature difference in ° C

To solve Question B, we use the same method. We must find the amount of energy necessary to raise the temperature from its initial temperature to the one stated by the problem to be the equilibrium temperature of the system (100°C):

Q= m_copper *CP_copper *∆T = 150g * 0.0923 cal/(g K) * (100°C-24°C) = 1052,22 cal

If we add the components we just found in questions A and B, we can find the amount of energy than the Copper cylinder lost, this would be: Q_tot = 15969.22 cal + 1052.22 cal = 17021.44 cal.

The question C asks us to find the initial temperature of the cylinder and Q_tot will help us to find it.

We know that Q_tot is the energy lost by the cylinder and we also know that Q_tot = m_cylinder * CP_copper * ∆T. Therefore, what we need to do is clear the last term of the equation and find the initial temperature.

Q_tot = m_cylinder *CP_copper *∆T → T_fin-T_initial = Q_tot/(m_cylinder*CP_copper ) = (-17021.44 cal)/(430g*0.0923 cal/(g K))

→ T_initial = 100°C + (-17021.44 cal)/(430g * 0.0923 cal/(g K)) = 528,87 °C

If we convert the 100°C to K before we do the calculation, the result would be the same one, You would only need to add 273,15 to the final result to check it out.

Hope everything was clear. If you have any further question, I'll be happy to help :D

5 0

3 years ago