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mr Goodwill [35]

3 years ago

10

What is all the mediums that light can travel through

1 answer:

LenKa [72]3 years ago

6 0

Answer:

Light travels as a wave. But unlike sound waves or water waves, it does not need any matter or material to carry its energy along. This means that light can travel through a vacuum—a completely airless space. (Sound, on the other hand, must travel through a solid, a liquid, or a gas.)

Explanation:

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An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4x106m/s2. What is the ma

Jet001 [13]

Answer:

Explanation:

Force on electron in an electric field E = eE where E is electric field .

acceleration = eE / m where m is mass of electron .

Putting the values

4 x 10⁶ = 1.6 x 10⁻¹⁹ x E / 9.1 x 10⁻³¹

E = 22.75 x 10⁻⁶ N/C

The direction of electric field will be towards west ( opposite to east )

because of negative charge on electron .

7 0

3 years ago

Tim’s cow is anemic. The cow is lacking which type of nutrient?

mixer [17]

Answer:

Iron deficiency

Explanation:

or more scientifically explained as decreased hemoglobin levels in your blood but still caused by lack of iron.

6 0

3 years ago

Read 2 more answers

A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect fricti

topjm [15]

Answer:

1.66 m/s

Explanation:

Work or kinetic energy = $\frac{1}{2} mv^{2}$

$4864=\frac{1}{2} (3540)v^{2}$

v = 1.66 m/s

6 0

3 years ago

Calculate how much work is required to launch a spacecraft of mass mm from the surface of the earth (mass mEmE, radius RERE) and

SIZIF [17.4K]

Answer:

Work done = (1/2)[(Gmm_e)/(R_e)]

Explanation:

I've attached the explanations below.

5 0

4 years ago

Read 2 more answers

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago