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2 years ago

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1. Give 3 examples of waves that carry small amounts of energy?

1 answer:

masya89 [10]2 years ago

4 0

ANSWER:

Radio waves
Micro waves
Infrared waves

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WHICH ONE!!! ASAP FOR A RETAKE FOR SCIENCE PLS

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I’m 95% sure it’s covalent bonds.

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2 years ago

To get maximum current in a circuit, the resistance should be in _____

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Answer:

Series is the correct answer

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2 years ago

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The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it

soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

$U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})$

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, $v_a$ as 0, 9.81 for g then

$58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s$

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2 years ago

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

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2 years ago

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A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient

lorasvet [3.4K]

Answer:

Change in mechanical energy = work done by friction

so it is equal to

$W = -8.16 J$

Explanation:

As we know that change in mechanical energy must be equal to the work done by non conservative forces only

So here when block moves down the inclined plane then the work done by friction force is given as

$W = F.d$

here we have

$F = \mu F_n$

here we know that

$F_n = mg cos\theta$

so we have

$F_n = 2.55(9.81)(cos60)$

$F_n = 12.5 N$

Now the friction force on the block is given as

$F_f = \mu F_n$

$F_f = 0.25 \times 12.5$

$F_f = 3.13 N$

now work done by the friction is given as

$W = -(3.13)(2.61)$

$W = -8.16 J$

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3 years ago