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3 years ago

What kind of weather would an occluded front likely bring?

Physics

2 answers:

IRISSAK [1]3 years ago

5 0

The answer would be b

LuckyWell [14K]3 years ago

3 0

B. much precipitation

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What is the motion and arrangement of molecules in a liquid

Viktor [21]

Answer:

particles

Explanation:

in liquids, particle are close together

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3 years ago

How many atoms are in SO 3

kkurt [141]

Answer:

There are 6.87 x 1023 atoms in 1.14 mol SO3, or sulfur trioxide :

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3 years ago

A spinning coil of wire is what moves the cone in a speaker, producing the sound. True/False

Georgia [21]

False, the spinning coil of wire that moves the cone in a speaker does not produces sound.

Explanation:

The wire coil is an electromagnet that is fixed to speaker cone. A normal magnet attached to the back of the speaker cone.When audio is sent in the form of short bursts of electric current to the speaker cone through the wire.

A magnetic field is induced when electric current allowed to pass through the coil. This magnetic field is repelled by the other magnet. This repulsion cause the cone to move forward. In the absence of electric current in the coil, the cone moves backward.

Thus sound waves are produced due to the resulting rarefaction and compression. So it is not the spinning coil of wire but he permanent magnet that produces the sound.

5 0

4 years ago

Read 2 more answers

Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2

Svetllana [295]

Answer:

I had to search the Figure on Google to solve this question.

a) The magnitude of the force F₃ is:

$F_{3} = 87.47 N$

And the direction of F₃:

$\alpha = 79.04 ^{\circ}$ (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N

Explanation:

I had to search the Figure on Google to solve this question.

a) We can find the force of the third person as follows:

$\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0$

$\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0$

So, in x-direction we have:

$\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0$

$F_{3x} = -85.87 N$

In y-direction we have:

$\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0$

$F_{3y} = -16.63 N$

The magnitude of the force F₃ is:

$F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N$

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

$tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}$

$\alpha = 79.04 ^{\circ}$

b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a.

To find the weight of the cart when it accelerates at 200 m/s², we need to consider: F₃ = 0.

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

$\Sigma F_{x} = F_{1x} + F_{2x} = ma$