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3 years ago

Why are ocean currents important?

1 answer:

4 0

For the considerably longer periods– decades to millennia – which are relevant for climate change, the slightly larger heat capacity of the deep ocean is important. Ocean currents and mixing by winds and waves can transport and redistribute heat to deeper ocean layers.

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To fill the medication prescription, what information must the pharmacy technician need to obtain? A. The name of the medication

shepuryov [24]

C. Patient info, name of med, dosage & route, special instructions, prescriber’s DEA#, and number of refills

3 0

3 years ago

Which of the following statements is/are true? Check all that apply. A nonconservative force permits a two-way conversion betwee

saul85 [17]

Answer:

A conservative force permits a two-way conversion between kinetic and potential energies.

The work done by a nonconservative force depends on the path taken.

A potential energy function can be specified for a conservative force.

Explanation:

A conservative force is defined as a force whose work done does not depend on the path taken, but only on the initial and final position of motion.

This means that for a conservative force, it is possible to defined a potential energy function U which depends only on the position of the object. An example of conservative force is gravity: the gravitational potential energy of an object, in fact, depends only on its position in the field, not on the path taken.

This behaviour also implies that when an object moves from A to B and then back from B to A, the potential energy gained (or lost) moving from A to B is lost (or re-gained) when moving from B to A. This means that the total mechanical energy (sum of kinetic energy and potential energy) of the object is conserved, and therefore there is a constant conversion between potential and kinetic energy during the motion.

A non-conservative force instead does not show this properties, as the work done by it depends on the path taken, and therefore it is not possible to define a potential energy function. An example of non-conservative force is friction.

According to what we wrote above, therefore, the only correct statements are:

A conservative force permits a two-way conversion between kinetic and potential energies.

The work done by a nonconservative force depends on the path taken.

A potential energy function can be specified for a conservative force.

3 0

3 years ago

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

Aleksandr-060686 [28]

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

$\rho$ = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N$

The drag force on the athlete is 33 N

3 0

3 years ago

How does a helicopter fly ? ( I need it explained in the view of physics)

Zielflug [23.3K]

Answer:

because blades of helicopter create a high pressure difference between lower and higher portion of blade so due to this pressure difference helicopter became able to fly.

Explanation:

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3 years ago

Read 2 more answers

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$