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3 years ago

Please heeelllllppppppppp

Physics

1 answer:

vagabundo [1.1K]3 years ago

8 0

Answer:They repel each other

Step by Step Explanation :

If you bring a positive object near the electroscope, it will attract the electrons toward the terminal. The leaves will move away from each other. You can also use this process with a positive object to charge the electroscope.

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A cruise ship has a mass of 100,000,000 kg and a kinetic energy of 5,000,000 kJ. Calculate the speed of the cruise ship.

KiRa [710]

Explanation:

K.E = 5,000,000 KJ

MASS = 100,000,000 Kg

K.E = 1/2 MV^2

$5000000 = \frac{1}{2} \times 100000000 \times v {}^{2}$

$5000000 = 50000000 \times v {}^{2}$

$v {}^{2} = \frac{5000000}{50000000}$

$v {}^{2} = \frac{5}{50}$

therefore

$v {}^{2} = \frac{1}{10}$

$v = \sqrt{0.1}$

v = 0.316 m/s^2

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3 years ago

If the temperature of the solar surface is 5800 K and Wien's law for the peak wavelength of the spectrum of the Sun, assumed to

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Answer:

The dominant wavelength of the sun is $499.65nm$

Explanation:

Wien's law is defined as:

$\lambda_{max} T = c$ (1)

Where $\lambda_{max}$ is the maximum wavelength, c is the Wien's constant and T is the temperature.

Therefore, $\lambda_{max}$ can be isolated from equation 1.

$\lambda_{max} = \frac{c}{T}$ (2)

$\lambda_{max} = \frac{2.898x10^{-3}m\cdot K}{T}$

Notice that it is necessary to express the Wien's constant in units of meters

$c = 2.898x10^{-3}m\cdot K . \frac{1x10^{9}nm}{m}$ ⇒ $2.898x10^{6} nm \cdot K$

Finally, equation 2 can be used:

$\lambda_{max} = \frac{2.898x10^{6} nm \cdot K}{5800 K}$

$\lambda_{max} = 499.65nm$

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4 years ago

When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

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Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

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