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4 years ago

When a sound wave encounters a barrier, what happens?

1 answer:

8 0

Sound waves will either be reflected (echo effect) or absorbed (dissipated) depending upon the material make-up of the barrier

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The magnification of a microscope is increased when_________.

azamat

Answer:

Option B

Explanation:

Magnification of Microscope is

$M = M_o \times M_e$

Mo= Magnification of objective lens and

Me= magnification of the eyepiece.

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.

Magnification,

$M \propto \dfrac{1}{f}$

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B

6 0

4 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

kap26 [50]

Answer:

a) $\omega \approx 219.911\,\frac{rad}{s}$ , b) $\alpha = 16.916\,\frac{rad}{s^{2}}$ , c) $a_{t} = 1.776\,\frac{m}{s^{2}}$ , d) $a_{n} = 5077.889\,\frac{m}{s^{2}}$ , e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation, f) Zero, g) $v = 23.091\,\frac{m}{s}$ .

Explanation:

a) The maximum angular velocity of the fan is:

$\omega = (35\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$\omega \approx 219.911\,\frac{rad}{s}$

b) The angular acceleration of the fan is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{219.911\,\frac{rad}{s}-0\,\frac{rad}{s}}{13\,s}$

$\alpha = 16.916\,\frac{rad}{s^{2}}$

c) The magnitude of the tangential aceleration is:

$a_{t} = (16.916\,\frac{rad}{s^{2}} )\cdot (0.105\,m)$

$a_{t} = 1.776\,\frac{m}{s^{2}}$

d) The magnitude of the centripetal acceleration is:

$a_{n} = (219.911\,\frac{rad}{s} )^{2}\cdot (0.105\,m)$

$a_{n} = 5077.889\,\frac{m}{s^{2}}$

e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation.

f) When fan is at full speed, it rotates at constant rate and, hence, there is no angular acceleration. Besides, the tangential acceleration experimented by the gum is zero.

g) The linear speed of the gum is:

$v = (219.911\,\frac{rad}{s} )\cdot (0.105\,m)$

$v = 23.091\,\frac{m}{s}$

5 0

3 years ago

In which range is the peak wavelength of a star that is much hotter than the sun most likely to be?

Mazyrski [523]

<span>ultraviolet light for sure is the right answer </span>

3 0

4 years ago

How can a driver best be prepare to enter sharp curves

Eduardwww [97]

Slow down before entering the curve. Stay close to the outside. Use the gas pedal as you enter the middle of the curve.

7 0

3 years ago

A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57

ololo11 [35]

Answer:

47.76°

Explanation:

Magnitude of dipole moment = 0.0243J/T

Magnetic Field = 57.5mT

kinetic energy = 0.458mJ

∇U = -∇K

Uf - Ui = -0.458mJ

Ui - Uf = 0.458mJ

(-μBcosθi) - (-μBcosθf) = 0.458mJ

rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

μB * (cosθf - cosθi) = 0.458mJ

θf is at 0° because the dipole moment is aligned with the magnetic field.

μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

θ = cos⁻ 0.6722

θ = 47.76°

7 0

3 years ago

Read 2 more answers