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3 years ago

12

Which best describes what occurs when an object takes in a wave as the wave hits it? O transmission O absorption O reflection O

refraction

1 answer:

MatroZZZ [7]3 years ago

4 0

Answer:

Absorbtion

Explanation:

Just did the test on Edg.

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Name three situations where rolling friction would be more advantageous than sliding friction

german

Well one situation i can think of is a tire rolling on a car
(ill try to find more)

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4 years ago

Early in 1981 the Francis Bitter National Magnet Laboratory at M.I.T. commenced operation of a 3.3 cm diameter cylindrical magne

Igoryamba

Answer:

The maximum electric field $E_{max}= 0.132V/m$

Explanation:

From the question we are told that

The diameter is $d = 3.3 cm = \frac{3.3}{100} = 0.033m$

The magnetic field of the cylinder is $B = 30 T$

The frequency is $f = 15Hz$

The radial distance is $d_r = 1.4cm = \frac{1.4}{100} = 0.014m$

This magnetic field can be represented mathematically as

$B(t) = B_i + B_1sin (wt + \o_i)$

The initial magnetic field is the average between the variation of the magnetic field which is represented as

$B_i = \frac{30 + 29.6}{2}$

$= 29.8T$

Then $B_1$ is the amplitude of the resultant field is mathematically evaluated as

$B_1 = \frac{30.0 - 29.6}{2}$

$= 0.200T$

The electric field induced can be represented mathematically as

$E = \frac{1}{2} [\frac{dB }{dt} ]d_r$

$= \frac{d_r}{2} \frac{d}{dt} (B_i + B_1 sin (wt + \o_o))$

$= \frac{1}{2} (B wr cos (wt + \o_o))$

At maximum electric field $cos (wt + \o_o) = 1$

$E_{max} = \frac{1}{2} B_1 wd_r$

$E_{max} = \frac{1}{2} B_1 2 \pi f d_r$

$= \frac{1}{2} (0.200) (2 \pi (15 ))(0.014)$

$E_{max}= 0.132V/m$

6 0

3 years ago

In preparation for building a space station, an astronaut removes a self-telescoping uniform rod from the cargo bay and releases

emmainna [20.7K]

To solve this problem it is necessary to apply the concepts related to the conservation of angular momentum. This can be expressed mathematically as a function of inertia and angular velocity, that is:

$L = I\omega$

Where,

I = Moment of Inertia

$\omega$ = Angular Velocity

For the given object the moment of inertia is equivalent to

$I = \frac{mr^2}{12}$

Considering that the moment of inertia varies according to distance, and that there are two of these without altering the mass we will finally have to

$L_i = L_f$

$I_i \omega_1 = I_f \omega_2$

$(\frac{mr_{initial}^2}{12})(\omega_1)=(\frac{mr_{final}^2}{12})(\omega_2)$

$(r_{initial}^2})(\omega_1)=(r_{final}^2)(\omega_2)$

Our values are given as,

$r_{initial} = 3m\\\omega_1 = 0.05rad/s \\r_{final}=1.5m$

Replacing we have,

$(3^2})(0.05)=(1.5^2)(\omega_2)$

$\omega_2 = 0.2rad/s$

Therefore the angular speed after the catch slips is 0.2rad/s

7 0

3 years ago

Science assessment help pls

DENIUS [597]

Answer:

Work Done = W

force = F

Distance = d

W = Fd

or W = F*d

W (in joules) = 3.5*4 = 14 Nm (or J)

1Nm = 1J

so newton meters and joules are the same

Power = Work (in joules) /time (in seconds)

i don’t know the time so i can’t solve it

3 0

3 years ago

Read 2 more answers

suppose the pilot starting again from rest opens the throttle part.way at constant acceleration the airboat then covers a distan

mr_godi [17]

Answer:

Acceleration is 1.2 m/s^2.

Explanation:

initial velocity, u = 0

distance, d = 60 m

time, t = 10 s

Let the acceleration is a.

use second equation of motion

$s= u t +0.5 at^2\\\\60 = 0 + 0.5 \times a \times 10\times 10\\\\a = 1.2 m/s^2$

Now according to the Newton's second law

Force = mass x acceleration

Let the mass is m.

F = m x 1.2 = 1.2 m Newton

3 0

3 years ago