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3 years ago

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The islands of Java and Sumatra in Indonesia are an example of A. a rift zone B.a hot spot C. an ocean ridge D. a volcanic islan

d arc

1 answer:

marissa [1.9K]3 years ago

6 0

The correct answer is D.

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The effort is always less than the load for a second-class lever. <br><br> true or false

ddd [48]

False

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3 years ago

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A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below

AlexFokin [52]

Answer:

The magnetic flux through the desk surface is $6.6\times10^{-4}\ T-m^2$ .

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

$\phi=BA\costheta$

Where, B = magnetic field

A = area

Put the value into the formula

$\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}$

$\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927$

$\phi=6.6\times10^{-4}\ T-m^2$

Hence, The magnetic flux through the desk surface is $6.6\times10^{-4}\ T-m^2$ .

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3 years ago

Why does the frequency of a wave increase as the wavelength decreases

olchik [2.2K]

Explanation:

We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that

Velocity = wavelength * frequency

or, frequency = velocity / wavelength

Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.

So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.

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3 years ago

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Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

$r'=\frac{1}{4}r$

So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

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3 years ago

Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:

miss Akunina [59]

Answer:

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3 years ago