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artcher [175]
3 years ago
15

In a solution,a Is the substance being dissolved

Physics
1 answer:
Elena-2011 [213]3 years ago
4 0
In a solution, a SOLUTE is the substance being dissolved.

In a solution, a SOLVENT is the substance in which something is dissolved.

For example in salt water, the salt is the solute. And the water is the solvent. Usually, which substance has a larger amount in the solution can be called a solvent. It doesn't matter if it's a gas, solid, or liquid. They can all be solute or solvents.
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What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi
Murrr4er [49]

Complete Question:

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from the spacecraft s 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth.

What s the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

3 0
3 years ago
An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,

where h_1 and h_2 are the height of the column of oil and the unkown liquid, respectively. Writing for \gamma_{unk}, we have

\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0
3 years ago
Why do you think the combined wave is more powerful than either the transverse or longitudinal wave with the same amplitude
liq [111]

Answer:

Explanation:

The combined wave only end up been more powerful than the Longitudinal wave. This means, the transverse wave is more powerful than the combined wave. In transverse wave, the oscillation is perpendicular to the direction of the wave, while in longitudinal wave, the motion of the movement of the object is parallel to the movement of the wave. And in combined wave, the movement of the medium is in a circular manner,

6 0
3 years ago
Thin Layer Chromatography consists of three parts: The analyte, the stationary phase, and mobile phase. Match each of these term
Margaret [11]

Answer:

Analyte⇒ one of analgesics

stationery phase⇒ silica

mobile phase⇒ solvent

Explanation:

during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.

8 0
3 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
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