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3 years ago

Help please help me out with this

Physics

1 answer:

Soloha48 [4]3 years ago

3 0

The first picture would be 1/15

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As temperature increase what energy increases

garik1379 [7]

Kinetic energy would increase sir.

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4 years ago

4. The Mariana trench is in the Pacific Ocean and has a depth of approximately 11,000 m. The density of seawater is approximatel

Alex73 [517]

Explanation:

It is known that relation between pressure and density is as follows.

P = $\rho gh$

where, P = pressure

$\rho$ = density

g = acceleration due to gravity

h = height

Putting the given values into the above formula as follows.

P = $\rho gh$

= $1025 \times 9.8 \times 11000$

= 110495000 Pa

Now, relation between pressure and force is as follows.

P = $\frac{F}{A}$

or, F = PA

F = $110495000 \times \pi \times (0.1)^{2}$

= $3.47 \times 10^{6} N$

Thus, we can conclude that a force of $3.47 \times 10^{6} N$ can be experienced at such depth.

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3 years ago

All matter in the Universe consists of many substances called elements. <br><br> true or faluse

Mrrafil [7]

The answer to this statement is true!

8 0

3 years ago

Read 2 more answers

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

Read 2 more answers

If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force betw

LiRa [457]

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

4 0

3 years ago