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2 years ago

Oil tends to float on water because the density of oil is _____ the density of water.

Physics

1 answer:

Rama09 [41]2 years ago

7 0

Oil is less dense than water so it tends to float on the top of the water. Hope this Helps!

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I need help thanks :)))))))))

dsp73

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end. It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one. It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

6 0

2 years ago

A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e

ddd [48]

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J$

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

$E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J$

The energy extracted by the turbine every second is 2649600 Joules

8 0

2 years ago

What is the mass of an atom that has 2 protons, 3 neutrons, and 2 electrons

marusya05 [52]

You can find the mass of an atom by adding the number of protons and neutrons. In this case the atom has 2 protons and 3 neutrons so the mass is 5.

3 0

3 years ago

What is the net force on the purple ring in the picture below. _________

Setler79 [48]

The dogs i think haha

5 0

2 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

2 years ago