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a_sh-v [17]
3 years ago
12

Under what atmospheric condition is fog most often formed in the san joaquin valley?

Physics
1 answer:
GalinKa [24]3 years ago
4 0
It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway,here is the answer. The atmospheric condition in which <span>fog is most often formed in the san Joaquin valley is stable stability. Hope this helps.</span>
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Eva makes a graphic organizer to compare electromagnets with solenoids. A venn diagram with 2 intersecting circles. The circle o
Eva8 [605]

The metalcore is the central component of any metal that is covered by wires. Metalcore is found in the X denoted area.

<h3>What is a solenoid?</h3>

A solenoid is a coil of wire that conducts an electric current. A solenoid is an electromagnet made out of a wire or helical coil.

When an electric current is sent through the coil, it produces a magnetic field. A solenoid is a coil that generates a magnetic field when an electric current passes through it.

When a conductive wire is used to build a loop, a solenoid is formed.

Hence the metalcore is found in the given X area.

To learn more about the solenoid refer to the link ;

brainly.com/question/16015159

5 0
2 years ago
A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an
anyanavicka [17]

Answer:

The value of tangential acceleration \alpha_{t} =  40 \frac{m}{s^{2} }

The value of radial acceleration \alpha_{r} = 80 \frac{m}{s^{2} }

Explanation:

Angular acceleration = 50 \frac{rad}{s^{2} }

Radius of the disk = 0.8 m

Angular velocity = 10 \frac{rad}{s}

We know that tangential acceleration is given by the formula \alpha_{t} = r \alpha

Where r =  radius of the disk

\alpha = angular acceleration

⇒ \alpha_{t} = 0.8 × 50

⇒ \alpha_{t} = 40 \frac{m}{s^{2} }

This is the value of tangential acceleration.

Radial acceleration is given by

\alpha_{r} = \frac{V^{2} }{r}

Where V = velocity of the disk = r \omega

⇒ V = 0.8 × 10

⇒ V = 8 \frac{m}{s}

Radial acceleration

\alpha_{r} = \frac{8^{2} }{0.8}

\alpha_{r} = 80 \frac{m}{s^{2} }

This is the value of radial acceleration.

7 0
3 years ago
The velocity time graph of a car shown below a) Calculate the magnitude of displacement of the car in 40 seconds. b) During whic
Gekata [30.6K]

Answer:

a) 0 metres

b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s

c) 20 m/s

Explanation:

a) <em>Formula of displacement= velocity x time</em>

time=40 s

velocity =0 m/s

∴ displacement= 0 x 40 = 0 m

Magnitude of displacement is 0 m

b) The increase in velocity shows that there has been acceleration.

c) The average velocity of the car is =\frac{0+40}{2\\}   {initial velocity + final velocity}

                                                            =\frac{40}{2}

                                                             =20

Therefore, the magnitude of the average velocity  of the car is 20 m/s

3 0
3 years ago
A doppler effect occurs when a source of sound moves. True or False
Shtirlitz [24]
<h2>Answer: True </h2>

The <u>Doppler effect</u> refers to the change in a wave perceived frequency when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other.

In other words, it is the variation of the frequency of a wave due to the relative movement of the source of the wave with respect to its receiver.

It should be noted that this effect  bears its name in honor of the Austrian physicist <u>Christian Andreas Doppler</u>, who in 1842 proposed the existence of this effect for the case of light in the stars. Another important aspect is that the effect occurs in all waves (including light and sound). However, it is more noticeable to humans with sound waves.

4 0
2 years ago
A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th
Usimov [2.4K]

Answer:

\omega_f = 3.584\ rad/s

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed  = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

\theta = \omega_i t + \dfrac{1}{2}\alpha t^2

5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2

\alpha= \dfrac{10}{2.8^2}

       α = 1.28 rad/s²

now, calculation of angular velocity

\omega_f = \omega_i + \alpha t

\omega_f =0 +1.28\times 2.8

\omega_f = 3.584\ rad/s

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0
3 years ago
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