All categories

3 years ago

Development is best described as:

Physics

1 answer:

Ede4ka [16]3 years ago

7 0

Answer:

A. Change that normally occurs in people throughout their lifetimes.

Explanation:

APEX verified

You might be interested in

A combination of two identical resistors connected in series has an equivalent resistance of 12. ohms. What is the equivalent re

SIZIF [17.4K]

Answer:

R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical

2 R1 = 12 and R1 = R2 = 6 ohms

1 / R = 1 / R1 + 1 / R2 for resistors in parallel

R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3

The equivalent resistance would be 3 ohms if connected in parallel

7 0

2 years ago

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

elena55 [62]

Answer:

$3.2\times 10^{-7}\ m$ or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = $9.38\times 10^{14}\ Hz$

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = $3\times 10^8\ m/s$

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

$v=f\lambda$

Now, expressing the above equation in terms of wavelength 'λ', we have:

$\lambda=\frac{v}{f}$

Now, plug in the given values and solve for 'λ'. This gives,

$\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m$

Therefore, the wavelength of the radiations absorbed by the ozone is nearly $3.2\times 10^{-7}\ m$ or 0.32 μm.

7 0

3 years ago

Which of the following is an example of kinetic energy?

tresset_1 [31]

B. Sound, because everything else sits still and sound waves move

5 0

3 years ago

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

3 years ago

If the moon is rising at midnight, the phase of the moon must be

Nadya [2.5K]

The phase is called 3rd quarter.

Hope this helps:)

7 0

3 years ago