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3 years ago

Ions can have a positive or negative charge. Please select the best answer from the choices provided T F

1 answer:

5 0

From among the choices provided, the more appropriate
answer is ' T ', the initial letter often used to represent
words that include 'true', 'truth', 'trust', etc., (as well as
'tree', 'train', 'transmit', 'Transylvania', 'trachea', 'travesty',
and 'trick', which are irrelevant to the present discussion).

This response is the most fitting and appropriate, because
the statement that precedes the list of allowable choices is
exemplary in its accuracy and veracity. An ion can, in fact,
have a positive or negative charge, although the same ion
cannot have both.

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An earthquake causes a 3 kg book to fall from a shelf. If the book lands with

Anna [14]

The correct answer is C

7 0

3 years ago

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago

Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely

m_a_m_a [10]

The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

Space long= 84 ft
Space wide= 42 ft
Space deep= 9 ft
Cost by yard3 = $39/yd3
Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about volume at: brainly.com/question/12628341

#SPJ4

4 0

1 year ago

Which statements describe what is occurring at t=5 seconds? Check all that apply

Alex73 [517]

Answer: The object changed directions

The object sped up

Explanation:

7 0

3 years ago

"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much

san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, $x=\dfrac{\lambda D}{d}$

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and $\lambda$ is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :

$x=\dfrac{n\lambda D}{d}$

$\lambda$ is the wavelength

n is the order

If n = 1,

$x=\dfrac{\lambda D}{d}$

So, the the number of bright fringes per unit width on the screen is $\dfrac{\lambda D}{d}$ . Hence, the correct option is (B).

6 0

3 years ago