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2 years ago

Anyone know this lmk ASAP !!!

Physics

1 answer:

Vesna [10]2 years ago

8 0

Answer:1. Transparent

2. Transparent

3. Opaque

Explanation: In a transparent medium light can pass through. For opaque objects it does not allow to pass through instead it is reflected back.

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How did the iron filing patterns show attractive forces between magnetic poles?

nasty-shy [4]

When the magnet is placed on the glass, it is attracted to the iron filings. The pattern of the iron filings shows that the lines of force that make up the magnetic field of the magnet. Also, The lines of force of north and south poles attract each other whereas those of two north poles fend off each other.

4 0

3 years ago

Read 2 more answers

A car speeds up from 12.0 m/s to 16.0 m/s in 8.0 s. the acceleration of the car is?

maria [59]

Vi=12m/s Vf=16m/s t=8s a=? a=Vf-Vi/t=16-12/8=4/8=1/2 a=0.5m/s^2

6 0

3 years ago

A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

nalin [4]

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

$v_{s}=rw\\v_{s}=r(2\pi f)$

Substitute the given values

$v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s$

For approaching generator the frequency is calculated as:

$f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz$

On the other hand,for the receding generator,Doppler's effect is expressed as:

$f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz$

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

8 0

3 years ago

The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

Verizon [17]

Answer:

The value is $E_i = 1.5596 *10^{-18} \ J$

Explanation:

From the question we are told that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$

The velocity is $v = 2.371*10^6 \ m/s$

The mass of electron is $m_e = 9.109*10^{-31} \ kg$

Generally the energy of the incident light is mathematically represented as

$E = \frac{h * c}{\lambda}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

h is the Planck constant with value $h = 6.62607015 * 10^{-34 } J\cdot s$

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$

Generally the kinetic energy is mathematically represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$

=> $E_k = 2.56 *0^{-18} \ J$

Generally the ionization energy is mathematically represented as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$

7 0

3 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago