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2 years ago

7

The Earth’s internal __________ source provides the energy for our dynamic planet, providing it with the driving force for on-go

ing disastrous events such as earthquakes and volcanic eruptions and for plate-tectonic motion.

1 answer:

Lesechka [4]2 years ago

4 0

The internal (heat) source

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A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the

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2 years ago

Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)

Gala2k [10]

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

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2 years ago

A particular field is 111 yards long. Express this length in meters.

romanna [79]

Answer:

101.498 m

Explanation:

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2 years ago

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The distance from earth to teh sun usb approxximately 93 million miles. a scientist would write that number as

erastova [34]

A scientist would write that number as 1.49 x 10⁸ kilometers .

(Or, if the scientist is in France or the UK, he might write it as 1.49 x 10⁸ kilometres .)

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2 years ago

A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?

rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec$

Maximum acceleration is given by $a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2$

So maximum acceleration in the simple harmonic motion will be $0.854rad/sec^2$

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2 years ago

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