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3 years ago

Which of the following best describes wind?

Physics

1 answer:

Afina-wow [57]3 years ago

5 0

Answer:

The answer is D the rising of warm air pushing down cool air.

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The planet Jupiter has an acceleration due to gravity that is approximately 2.4 times as much as the earth (23.2 m s2 ). Which o

kumpel [21]

On Jupiter, C. your weight would increase by a factor of 2.4 . Weight is a product of mass and gravity. Mass does not change dependent upon location.

6 0

3 years ago

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Below is the data from a gas law experiment comparing the pressure and the volume of a gas at a given temperature.

Wewaii [24]

Answer:

The combined gas equation relates three variables pressure, temperature and volume when the number of moles is constant.

The equation is PV / T = constant. Which is valid for a fixed number of moles of the gas.

You can derive the combined gas equation from the combination of Bolye's law, Charles' law and Gay-Lussac's law, which needs some algebra.

Explanation:

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3 years ago

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Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point

Neko [114]

Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

$B = \frac{\mu_oI}{2\pi R}$

At the center, the magnetic field strength is twice

$B_c = 2(\frac{\mu_oI}{2\pi R}) =\frac{ \mu_oI}{\pi R}$

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

$B_c = \frac{ \mu_oI}{\pi R}\\\\I = \frac{B_c\pi R}{\mu_o} = \frac{370 *10^{-6}* \pi *0.043}{4\pi *10^{-7}}\\\\I = 39.8 \ A$

Therefore, current needed is 39.8 A

6 0

2 years ago

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0

3 years ago

alexgriva [62]

Answer:

Get your answer for $1 from www.gotit-pro.com

Explanation:

6 0

3 years ago