Ask question

Ask question

All categories

3 years ago

5

HERES THE ANSWER AND QUESTION

2 answers:

tatyana61 [14]3 years ago

5 0

Answer:

agree

Explanation:

kobusy [5.1K]3 years ago

5 0

Answer:

THE ANSWER IS A,B,F

Explanation:

You might be interested in

The National Aeronautics and Space Administration (NASA) studies the physiological effects of large accelerations on astronauts.

m_a_m_a [10]

Answer:

$v = 23.95 m/s$

Explanation:

As we know that when astronaut is revolving in circular path then the acceleration of the astronaut is due to centripetal acceleration

so it is given as

$a_c = \frac{v^2}{R}$

here we know that

$a_c = 4.50 g$

also we know that

$R = 13 m$

now we have

$4.50 \times 9.81 = \frac{v^2}{13}$

$v = 23.95 m/s$

3 0

4 years ago

What is the definition of "gravitational potential energy"?

AVprozaik [17]

According to Georgia State University, gravitational potential energy<span> is the energy an object possesses because of its position in a gravitational field. This is most commonly in reference to an object near the surface of the Earth, where the gravitational acceleration is assumed to be constant at about 9.8 m/s2.</span>

5 0

3 years ago

Read 2 more answers

The higher the temperature of an object the

Allushta [10]

higher temp = higher energy = higher frequency = shorter wavelength

4 0

4 years ago

A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

<u>Now the no. of moles in this many molecules:</u>

$n_m=\frac{n}{N_A}$

where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

7 0

3 years ago

An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of th

Marizza181 [45]

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. $d_{i}=-0.0566cm$

b. $M=441.69$

Explanation:

For this type of numerical we will use the following formulas

$\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1} }{R}$ ......... Eq1

where,

$n_{1}$ $=$ refractive index of the medium surrounding refracting surface/object

i.e. $n_{1}$ = $n_{air}$

$n_{2}$ $=$ refractive index of the refracting surface/object

i.e. $n_{2}=n_{glass}$

$d_{0}=$ distance of object from the vertex of the refracting surface

$d_{i}=$ distance of image from the vertex of the refracting surface

$R=$ radius of curvature of the refracting surface

$M=\frac{d_{0} }{d_{i} }$ ........... Eq2

where,

$M=$ magnification

Convention:

$R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R$

Given:

$n_{1}$ = $n_{air}$ $=1.0$

$n_{2}=n_{glass}$ $=1.5$

$d_{0}=$ $25cm$

$R=-11$ $cm$ because refraction surface is concave

Required:

a. $d_{i}=$ $?$

b. $M=?$

Solution:

a. putting values in eq1, we get

$\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}$

$\frac{1.5}{d_{i} }=-0.045-0.040$

$d_{i}=(-0.085)(\frac{1}{1.5} )$

$d_{i}=-0.0566$ cm

b. $M=\frac{25}{-0.05665}$

$M=441.69$

5 0

4 years ago