All categories

3 years ago

The windowpanes are___________ a. opaque b. transparent c. absorbent .

Physics

2 answers:

kogti [31]3 years ago

4 0

Answer:

The windowpanes are- transparent.

The color of the panes are due to the wavelengths of light that the glass- allows to pass through

Explanation:

Just answered the question.

Naddik [55]3 years ago

4 0

Answer:

The two answers are:

Explanation:

1) transparent

2) allows to pass through

I also just did this question. No lie. I hope this helps!

- sincerelynini

You might be interested in

Which statement is true about acceleration?<br> who is an underated physicist?

wlad13 [49]

Answer: A vehicle's capacity to gain speed within a short time...

4 0

3 years ago

You hold a metal block of mass 40 kg above your head at a height of 2 m.

Kitty [74]

Answer:

The work done by gravity is 784 J.

Explanation:

Given:

Mass of the block is, $m=40\ kg$

Height to which it is raised is, $h=2\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,

$\textrm{Work by gravity}=F_g\times h$

Force of gravity is given as the product of mass and acceleration due to gravity.

$\therefore F_g=mg=40\times 9.8=392\ N$ . Now,

$\textrm{Work by gravity}=F_g\times h=392\times 2=784\ J$

Therefore, the work done by gravity is 784 J.

5 0

3 years ago

In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0

2 years ago

According to the Theory of Plate Tectonics, which of the following statements is correct?

kramer

Answer:

It's D. The geothermal energy from Earth's core loses most of its heat before reaching the outer layer. The plates on the outer layer have been proven to be moving due to evidence such as earthquakes, volcanic eruptions, and tsunamis.

Explanation:

8 0

3 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

2 years ago