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2 years ago

12

The temperature at one of the Viking sites on Mars was found to vary daily from -90.OF to -5.0 C. Convert these temperatures to

kelvin.
Plz guys help me write answer

1 answer:

kykrilka [37]2 years ago

6 0

Answer:

I don't know about this, this is which class question

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A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri

san4es73 [151]

Answer:

T = 8.55 N

Explanation:

When string makes an angle 40 degree with the vertical then it will have two forces on it

1) gravitational force (mg)

2) Tension force in string (T)

now we know that net force towards the center of the path is known as centripetal force and it is given as

$T - mg cos40 = F_c$

$T - (0.40\times 9.8)cos40 = \frac{mv^2}{L}$

$T = 3 + \frac{0.40\times 5^2}{1.8}$

$T = 3 + 5.55$

$T = 8.55 N$

6 0

3 years ago

If you use a 40x objective to magnify a slide, the image you see through the eyepiece will be magnified how much times?

sammy [17]

4x time 10x (I think)

Here, this will help-
http://www.cas.miamioh.edu/mbiws/microscopes/Magnification.html#totalmagnification

8 0

3 years ago

A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw

mezya [45]

Answer:

The mass is $m = 3.45408 kg$

Explanation:

From the question we are told that

The extension of the rod is $from , \ x_1 = 0 \to x_2 = 13.0$

The area is $A = 8.0 cm^2$

The density increase as follows $from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3$

The equation $\rho = B + Cx$

at $x_1= 0$ $\rho_1 =2.5 g/cm^2$

So

$2.5 = B + 0$

=> $B =2.5$

So at $x_2 = 13.0$ , $\rho_2= 19.0 g/cm^3$

So

$19.0 = 2.5 + C(13)$

=> $C = 1.27$

Now

$m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx$

$m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.$

$m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]$

$m = 3454.08 g$

$m = 3.45408 kg$

8 0

3 years ago

If a 2 x 10^-4C test charge is given 6.5J of energy, determine the electric potential difference.

Gwar [14]

Answer:

The electric potential difference is 32500 volt.

Explanation:

Given that,

Charge $q=2\times10^{-4}C$

Energy = 6.5 J

We need to calculate the electric potential difference

Potential difference :

Potential difference is equal to the energy divide by charge.

Using formula of potential difference

$V=\dfrac{E}{Q}$

$V=\dfrac{6.5}{2\times10^{-4}}$

$V=32500\ volt$

Hence, The electric potential difference is 32500 volt.

8 0

3 years ago

Explain an experiment of the phenomenon of rainfall

maks197457 [2]

Unclear/incomplete question. However, I inferred you need an explanation of the phenomenon of rainfall.

<u>Explanation</u>:

Basically, the phenomenon of rainfall follows a natural cycle called the water cycle. What we call <em>'rainfall'</em> occurs when water condensed (in liquid form) in the atmosphere is made to fall down on the ground as tiny droplets as a result of the forces of gravity.

<u>The water cycle makes rainfall possible:</u>

First, water on the earth's surface is evaporated (or is absorbed into) the atmosphere.

Next, it then condensed into liquid form; which later falls to the surface to the ground again. And the process continues.

7 0

3 years ago