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finlep [7]
3 years ago
6

Which of the following has greater kinetic energy?

Physics
1 answer:
uysha [10]3 years ago
5 0
I aint got no clue ask someone that knows about this.
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Humpback whales are known to produce a collection of elaborate and repeating sounds with frequencies starting at 20 Hz. The soun
jasenka [17]

Answer:

70 m.

Explanation:

Given,

Frequency, f = 20 HZ

speed of sound, v = 1400 m/s

wavelength of the waves = ?

we know,

v = f λ

\lambda= \dfrac{v}{f}

\lambda= \dfrac{1400}{20}

\lambda=70\ m

Hence, the wavelength of the wave is equal to 70 m.

8 0
4 years ago
Compare the weight of a mountain climber when she is at the bottom of a mountain with her weight when she is at the top of the m
kakasveta [241]

Answer:

The correct answer is  a. Both are the same

Explanation:

For this calculation we must use the gravitational attraction equation

    F = G m M / r²

Where M will use the mass of the Earth, m the mass of the girl and r is the distance of the girl to the center of the earth that we consider spherical

To better visualize things, let's repair the equation a little

     F = m (G M / r²)

The amount in parentheses called acceleration of gravity, entered the force called peos

     g = G M / r²

     F = W

    W = m g

When analyzing this equation we see that the variation in the weight of the girl depends on the distance, which is the radius of the earth plus the height where the girl is

    r = Re + h

    Re = 6.37 10⁶ m

    r² = (Re + h)²

    r² = Re² (1 + h / Re)²

Let's replace

    W = m (GM / Re²)   (1+ h / Re)⁻²

    W = m g   (1+ h / Re)⁻²

This is the exact expression for weight change with height, but let's look at its values ​​for some reasonable heights h = 6300 m (very high mountain)

     h / Re = 10 ⁻³

     (1+ h / Re)⁻² = 0.999⁻²

Therefore, the negligible weight reduction, therefore, for practical purposes the weight does not change with the height of the mountain on Earth

The correct answer is a

4 0
3 years ago
A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centr
GREYUIT [131]
The centripetal acceleration of the car is 48,800 mi/h2
4 0
4 years ago
The table shows the specific heat capacities of various substances. How much energy is required to raise the temperature of 5g o
satela [25.4K]

Answer:

Q = 50.25 [J]

Explanation:

To solve this problem we must use the following equation that relates the temperature change with the mass and with the specific heat.

Q = m*Cp*(DT)

where:

Q = energy in form of heat [J]

m = mass = 5 [g] = 0.005 [kg]

Cp = specific heat = 1005 [J/kg*°C]

DT = temperature change = 10 [°C]

Now replacing:

Q = 0.005*1005*10

Q = 50.25 [J]

4 0
3 years ago
Use the worked example above to help you solve this problem. A car traveling at a constant speed of 27.9 m/s passes a trooper hi
Sidana [21]
H h h huh h jvjvhvuvuvuvyv
7 0
3 years ago
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