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ki77a [65]
3 years ago
9

Two charged objects have a repulsive force of .080 N. if the charge of both of the objects is doubled, then what is the new forc

e?
Physics
1 answer:
Viefleur [7K]3 years ago
4 0
This was kinda hard thinking it might be that the new force is going to 4 times bigger than the original one.
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A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum
Kryger [21]

Given data

*The given mass of the pendulum is m = 3 kg

*The given height is h = 0.3 m

The formula for the maximum speed of the pendulum is given as

v_{\max }=\sqrt[]{2gh}

*Here g is the acceleration due to the gravity

Substitute the values in the above expression as

\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}

Hence, the maximum speed of the pendulum is 2.42 m/s

7 0
1 year 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
A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta
-BARSIC- [3]

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

          = 107 cos 13.5°= 104 N

Power = \dfrac{work\ done}{time}

69.4 = \dfrac{W}{33}

W = 2290.2 J

Work done= Force x displacement

displacement= \dfrac{2290.2}{104}

                      = 22.02 m

6 0
3 years ago
The number of moons a planet has orbiting around it is determined by...
Pavlova-9 [17]

We have no idea.  There are probably many things involved ... the planet's mass,
the availability of small bodies in the neighborhood that can be captured, etc.


7 0
3 years ago
A running student has half the kinetic energy that his brother has. The student speeds up by 1 m/s, at which point he has the sa
hoa [83]

Answer:

V = (√2) + 1) m/s

Explanation:

Let the mass and speed of the running student be M and V respectively.

We are told that when he speeds up by 1 m/s, he has the same kinetic energy as his brother.

Thus, his speed at which he mow has the same kinetic energy as his brother is (V + 1) m/s

Now, we are told that the mass of the student is twice as large as that of his brother. Thus, his brother's mass is; M/2

Since kinetic energy is given by the formula K.E = ½mv²

Therefore, since we want to find the original speed of both students and that the initial condition says that the running student had half the kinetic energy of the brother, we now initial condition as;

½MV²= ½(½(M/2)V²) - - - - (eq 1)

Since he has sped up by 1 m/s, and has a kinetic energy now equal to that of his brother, we have;

(½M(V + 1)²) = (½(M/2)V²) - - - - (Eq2)

Dividing eq 1 by eq 2 gives;

V²/(V + 1)²= 1/2

Taking square root of both sides gives;

V/(V + 1) = 1/√2

Cross multiply to give;

(√2)V = V + 1

(√2)V - V = 1

V((√2) - 1) = 1

V = 1/((√2) - 1)

Simplifying this using surfs gives;

V = [1/((√2) - 1)] × ((√2) + 1))/((√2) + 1))

V = ((√2) + 1))/1

V = (√2) + 1) m/s

8 0
3 years ago
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