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3 years ago

Is there life on other planets ?

Physics

2 answers:

RUDIKE [14]3 years ago

7 0

Supposedly there is life on Mars but it’s early to say for sure

ki77a [65]3 years ago

7 0

Answer:

Yes

Explanation:

maybe not in the planets that we know of but the universe is huge. Our galaxy is made up of so many solar systems. There could be life in our galaxy, and in many other galaxies.

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What do you think happens to the temperature when water changes to gas?

Olenka [21]

Answer:

it gets hot and more hot until it turns to gas

Explanation:

5 0

3 years ago

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b. $v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

c.Work don=Final K.E-Initial K.E

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

4 0

3 years ago

Read 2 more answers

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$

Generally, the equation for Rate of flow of Liquid is mathematically given as

$\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}$

Where dP is pressure difference r is the radius

$\mu$ is the viscosity of water

L is the length of the pipe

$Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}$

$Q=5.2 \mathrm{~mL} / \mathrm{s}$

In $30s the quantity that flows out of the tube

$&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}$

In conclusion, the quantity that flows out of the tube

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

Read more about the flows rate

brainly.com/question/27880305

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5 0

2 years ago

Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places

Olenka [21]

Answer:

The magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage $V_{out} = 17.33$

Resistance $R_{1} = 3$ kΩ

Voltage gain $A_{v} = 4.33$

For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

$A_{v} = 1+\frac{R_{f} }{R_{1} }$

$4.33 = 1+ \frac{R_{f} }{3 k }$

$R_{f}$ ≅ 10 kΩ

For finding input voltage we use,

$A_{v} = \frac{V_{out} }{V_{2} }$

$V_{2} = \frac{17.33}{4.33}$

$V_{2} = 4$ V

The Phase of $V_{2}$ is zero because output voltage phase is 360°

Therefore, the magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

7 0

3 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago