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3 years ago

15

Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this

polar air?

1 answer:

LenaWriter [7]3 years ago

5 0

Answer:

That the polar air has has more pressure than the air at the equator.

Explanation:

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Why do we feel the force of gravity from the Earth but we don’t feel the force of gravity from the moon?

Zigmanuir [339]

Answer:

The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.

Explanation:

Hope this helped!!

8 0

2 years ago

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A certain 60.0 Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 11.6 kV/m.

yuradex [85]

Explanation:

Given that,

Frequency of the power line, f = 6 Hz

Value of maximum electric field strength of 11.6 kV/m

(a) The wavelength of this very low frequency electromagnetic wave is given by using relation as :

$c=f\lambda$

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8\ m/s}{60\ Hz}$

$\lambda=5\times 10^6\ m$

(b) As its can be seen that the wavelength of this wave is very high. It shows that it is a radio wave.

(c) The relation between the maximum magnetic field strength and maximum electric field strength is given by :

$B_0=\dfrac{E_0}{c}\\\\B_0=\dfrac{11.6\times 10^3}{3\times 10^8}\\\\B_0=3.86\times 10^{-5}\ T$

So, the maximum magnetic field strength is $3.86\times 10^{-5}\ T$ .

7 0

4 years ago

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A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago

Differenciate between fundamentals quantities and derived quantities

REY [17]

Answer:

Fundamental quantities are the base quantities of a unit system, and they are defined independent of the other...

• Derived quantities are based on fundamental quantities, and they can be given in terms of fundamental quantities.

• In SI units, derived units are often given names of people such as Newton and Joule.

Explanation:

3 0

3 years ago

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A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of

grigory [225]

Answer:

$a=2.9\ m/sec^2$

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force $F_r$ acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

$\vec F_r=m\vec a$

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force $F_r$ is computed as the hypotenuse of a right triangle

$|F_r|=\sqrt{256^2+104^2}$

$|F_r|=276.32\ Nw$

The acceleration can be obtained from the formula

$F_r=ma$

Note we are using only magnitudes here

$\displaystyle a=\frac{F_r}{m}$

$\displaystyle a=\frac{276.32Nw}{95.3Kg}$

$\boxed{a=2.9\ m/sec^2}$

7 0

3 years ago