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3 years ago

The most likely region of the radio spectrum for communication with other civilizations is in the "water hole."

Physics

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

Radio emissions of hydrogen and OH molecules

Explanation:

Water hole is a quiet region of the electromagnetic spectrum

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How are the three lines of defense the same

xz_007 [3.2K]

Here is an example of how the three lines of defense are the same:

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3 years ago

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Mammals must eliminate waste products that their bodies produce. Which organ helps mammals eliminate bodily waste?

Arada [10]

Answer:

Digestion helps mammals eliminate bodily waste.

Explanation:

Humans are considered mammals and we use digestion to eliminate bodily waste. Digestion has three main functions. First digestion of food, second absorption of nutrients, and third elimination of solid waste. There are two processes that take place Mechanical digestion and chemical digestion. Mechanical digestion is the chewing and breaking down of food into smaller molecules, whereas chemical digestion is the breakdown of the smaller molecules into simpler nutrients that the body can absorb and use.

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3 years ago

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

4 years ago

Different forces were applied to five balls, and each force was applied for the same amount of time. The data is in the table. A

Annette [7]

The magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Impulse:

The effect of applied force acting for a very small interval of time is known as Impulse.

Given data:

The initial magnitude of the force on Ball A is, F = 50 N.

The time interval for the applied force on Ball A is, t = 1.25 s.

And the impulse at initial on ball A is, I = 62.5 N-s.

Now, if force on ball A doubles, which means new magnitude of force is,

F' = 2F

F' = 2 × 50

F' = 100 N

For same time interval, the new impulse on Ball A is given as,

I' = F' × t

Solving as,

I' = 100 × 1.25

I' = 125 N-s

Taking the ratio of both the impulses as,

I' / I = 125/62.5

I' = 2 I

Thus, we can conclude that the magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Learn more about the impulse here:

brainly.com/question/904448

7 0

3 years ago

An incline plane has Ф = 40.0° and μ k = 0.15. Starting from rest, how long will it take a 4.0 kg block to reach a speed of 12 m

lesya [120]

The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:

$P_{t}-Fat=ma \\ mgsen\O-mgucos\O=ma \\ a=g(sen\O-ucos\O)$

Using the Velocity Hourly Equation, we get:

$V=V_{o}+a\Delta t \\ \Delta t= \frac{V}{a}$

Uniting the equations:

$\Delta t= \frac{V}{g(sen\O-ucos\O)}$

Entering the unknowns:

$\Delta t= \frac{V}{g(sen\O-ucos\O)} \\ \Delta t= \frac{12}{10(sen40^o-0.15cos^o)} \\ \Delta t= \frac{12}{10(0,64-0.15x0.77)} \\ \Delta t= \frac{12}{5.27} \\ \boxed {\Delta t=2.28s}$

Obs: Approximate results

If you notice any mistake in my english, please let me know, because i am not native.

5 0

3 years ago