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Deffense [45]
3 years ago
12

What is work?

Physics
1 answer:
WITCHER [35]3 years ago
5 0
It’s applying force to move an object
You might be interested in
I need help with this answer
7nadin3 [17]

Answer:

double replacement

Explanation:

sorry if im wrong

8 0
3 years ago
A book is launched up along the rough incline. Kinetic energy given to a book at initial point is 100 J. Book comes to stop at s
den301095 [7]

Answer:

(A) 60 J

Explanation:

At state 1

KE₁=100 J

At state 2

KE₂ = 0

U₂=80 J

Given that surface is rough so friction force will act in opposite to the direction of motion

Lets take work done by friction = Wfr

From work power energy

Work done by all forces = Change in kinetic energy

Wfr + U₂=ΔKE

Wfr+80 = 100

Wfr= 20 J

Now when book slides from top position then

Wfr+ U = KEf - KEi

-20 + 80 = KEf-0

KEf= 60 J

(A) 60 J

7 0
3 years ago
Why does the large number of hydrogen atoms in the universe suggest that other elements?
lidiya [134]

Answer:

Explanation:

The abundance of the chemical elements is a measure of the occurrence of the chemical elements relative to all other elements in a given environment. Abundance is measured in one of three ways: by the mass-fraction (the same as weight fraction); by the mole-fraction (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the volume-fraction. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and ideal gas mixtures. Most abundance values in this article are given as mass-fractions.

For example, the abundance of oxygen in pure water can be measured in two ways: the mass fraction is about 89%, because that is the fraction of water's mass which is oxygen. However, the mole-fraction is about 33% because only 1 atom of 3 in water, H2O, is oxygen. As another example, looking at the mass-fraction abundance of hydrogen and helium in both the Universe as a whole and in the atmospheres of gas-giant planets such as Jupiter, it is 74% for hydrogen and 23–25% for helium; while the (atomic) mole-fraction for hydrogen is 92%, and for helium is 8%, in these environments. Changing the given environment to Jupiter's outer atmosphere, where hydrogen is diatomic while helium is not, changes the molecular mole-fraction (fraction of total gas molecules), as well as the fraction of atmosphere by volume, of hydrogen to about 86%, and of helium to 13%.[Note 1]

The abundance of chemical elements in the universe is dominated by the large amounts of hydrogen and helium which were produced in the Big Bang. Remaining elements, making up only about 2% of the universe, were largely produced by supernovae and certain red giant stars. Lithium, beryllium and boron are rare because although they are produced by nuclear fusion, they are then destroyed by other reactions in the stars.[1][2] The elements from carbon to iron are relatively more abundant in the universe because of the ease of making them in supernova nucleosynthesis. Elements of higher atomic number than iron (element 26) become progressively rarer in the universe, because they increasingly absorb stellar energy in their production. Also, elements with even atomic numbers are generally more common than their neighbors in the periodic table, due to favorable energetics of formation.

The abundance of elements in the Sun and outer planets is similar to that in the universe. Due to solar heating, the elements of Earth and the inner rocky planets of the Solar System have undergone an additional depletion of volatile hydrogen, helium, neon, nitrogen, and carbon (which volatilizes as methane). The crust, mantle, and core of the Earth show evidence of chemical segregation plus some sequestration by density. Lighter silicates of aluminum are found in the crust, with more magnesium silicate in the mantle, while metallic iron and nickel compose the core. The abundance of elements in specialized environments, such as atmospheres, or oceans, or the human body, are primarily a product of chemical interactions with the medium in which they reside.

4 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

4 0
3 years ago
Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction
Pavel [41]

Answer:

3.2\cdot 10^{-8} N

Explanation:

The inital electrostatic force between the two spheres is given by:

F=k\frac{q_1 q_2}{r^2}

where

F=8\cdot 10^{-9} N is the initial force

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have

r'=\frac{r}{2}

Therefore, the new electrostatic force will be

F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F

So the force has increased by a factor 4. By using F=8\cdot 10^{-9} N, we find

F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N

6 0
3 years ago
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