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dem82 [27]
3 years ago
8

Find the difference in temperature between the water at the top and the bottom of a waterfall 100 m high.

Physics
2 answers:
Alekssandra [29.7K]3 years ago
7 0
The answer would be 60 cuz a watetfall is a waterfall

posledela3 years ago
7 0
The temperature at the top of the waterfall would be warmer than at the bottom because while the water is falling its causing the temperature to drop to lower Degrees
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Find the frequency of a wave of wavelength 2.5m and speed 400 m/s
Sedaia [141]

Answer:

The frequency of wave is 160Hz.

Explanation:

Given that the formula of speed is V = f×λ where V represents speed, f is frequency and λ is wavelength.

So first thing, you have to make frequency the subject by dividing wavelength on both sides :

v = f \times λ \:

v \div  λ =  f \times λ \div λ

f =  \frac{v}{λ}

Next you have to substitute the value of v and f into the formula :

Let λ = 2.5m,

Let v = 400m/s,

f =  \frac{400}{2.5}

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7 0
3 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
nert xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH o f values for xenon difluorid
loris [4]

Answer:

For Xenon fluoride, the average bond energy is 132kj/mol

For tetraflouride,the average bond energy is 150.5kj/mol.

For hexaflouride, the average bond energy is 146.5 kj/mol

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For tetraflouride

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For hexaflouride

402/6 = 67

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C. Gamma rays have more energy than infrared rays
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3 years ago
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