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3 years ago

In what changes of state do atoms lose energy? Check all that apply

1 answer:

5 0

In this case, Sublimation, evaporation and melting are endothermic reactions (gain energy from the environment) while, freezing, deposition and condensation are exothermic (loss of energy to the surroundings) reactions.

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To make a ___________________ , position your car to the right side of the right lane

pashok25 [27]

A right turn would be the answer probably.

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3 years ago

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Mayan kings and many school sports teams are named for the puma, cougar, or mountain lion felis concolor, the best jumper among

denis23 [38]

To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
vi^2 - vf^2 = 2*g*h
Given that it has vf = 0 (it is not moving vertically at its maximum height), g = 32, and h = 13.8, we can solve for vi:
vi^2 = 29.72 ft/s
This is only its vertical speed, so this is equivalent to its original speed multiplied by the sine of the angle:
29.72 ft/s = (v_original)*(sin 42.2<span>°</span>)
v_original = 44.24 ft/s
Converting to m/s, this can be divided by 3.28 to get 13.49 m/s.

4 0

3 years ago

43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a

olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

$P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J$

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J$

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

$\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N$

The frictional force that acts on the sliding bear is 378.47166 N

5 0

3 years ago

Read 2 more answers

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e =

Svetllana [295]

Answer:

Explanation:

(a) Rigel, 2.7x10^32W, T = 11,000K

But L = 4pR²sT⁴

L = 2.7x10^32W, T = 11,000K, s= 5.67 x 10^-8, R= radius in meters

Rigel parallax, p = 0.00378 arc sec

Substituting the various values and making R the subject of the formula

R² = L/(4psT⁴)

R² = 2.7x10^32/(4 x 0.003878 x 5.67x10^-8 x (11,000)⁴)

R² = 2.7x10^32/1.2877x10^7

R² = 2.096761668 x 10^25

R = 4.579041021 x 10^12meters

(b)

Procyon B, 2.1x10^23W, T = 10,000K

But L = 4pR²sT⁴

L = 2.1x10^23W, T = 10,000K, s= 5.67 x 10^-8, R= radius in meters

Procyon B parallax, p = 0.00284 arc sec

R² = 2.1x10^23/(4 x 0.00284 x 5.67x10^-8 x (10,000)⁴)

R² = 2.1x10^23/(6.441 x 10^6)

R² = 3.26036 x 10^16

R = 1.80565 x 10^8 meters

(c) The radius of Rigel is given as 4.579041021 x 10^12meters and the radius of Procyon B is given by 1.80565 x 10^8 meters shows the remarkable difference between a super-giant star(Rigel) and a white dwarf star (Procyon B)

The radius of the sun a red star is 6.96 x 10^8meters which shows a certain level of resemblance with the size of a dwarf white star Procyon B.

The sun is larger than Procyon B as estimated above.

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3 years ago

Which of the following scientists studied physics in Egypt?

Margaret [11]

The answer for apex is ibn al-haytham.

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4 0

3 years ago