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shepuryov [24]
3 years ago
15

Which of the following energy sources is expected to have the highest rate of consumption in 2030?

Physics
2 answers:
strojnjashka [21]3 years ago
8 0
<u>Answer</u>
<span>d. renewable energy
</span>
<u>Explanation</u>
Energy can be said to be one of the most basic commodity in the 21st century. Every household uses energy in almost all its doing eg. cooking lighting, cleaning, entertainment such televisions and radios, communication etc. 
For this reason, the scientist have to find the most sustainable source of energy. Some are not sustainable as they can be used up and other pollute the environment. 
By the year 2030, the most expected source of energy to have the highest rate of consumption is renewable energy. 
viva [34]3 years ago
8 0

just took the test answer is a.coal on e2020 and edge Explanation:

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Mr.Z holds 200 kg above his head for 5 seconds. What is the work done on the weights?
Alenkasestr [34]

m = mass held by mr. Z above his head = 200 kg

g = acceleration due to gravity = 9.8 m/s²

F = force applied by mr. Z to hold the mass

Using equilibrium of force , force equation is given as

F = mg

F = (200) (9.8)

F = 1960 N

Since the mass is not moved,

d = displacement of the mass = 0 m

we know that , work done is given as

W = F d

inserting the values

W = (1960) (0)

W = 0 J

8 0
3 years ago
Three uniform spheres of radius 2R, R, and 3R are placed in a line, in the order given, so their centers are lined up and the sp
kolezko [41]

Answer:

x = 2.33 R from the center of mass of the smallest sphere.

Explanation:

Due to the symmetry of the spheres, the center of mass of any of them, is located just in the center of the sphere.

If we align the centers of the spheres with the x-axis, the center of mass of any of them will have only coordinates on the x-axis, so the center of  mass of the system will have coordinates on the x-axis only also.

By definition, the x-coordinate of the center of mass of a set of discrete masses m₁, m₂, m₃, can be calculated as follows:

Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}

In this case, we need to get the coordinates of the center of mass of each sphere:

If we place the spheres in such a way that the center of the first sphere has the x-coordinate equal to its radius (so it is just touching the origin), we will have:

x₁ = 2*R

For the second sphere, the center will be located at a distance equal to the diameter of  the first sphere plus its own radius, as follows:

x₂ = 4*R + R = 5*R

Finally, for the third sphere, the center will be located at a distance equal to the diameter of  the first sphere, plus the diameter of the second sphere,  plus its own radius, as follows:

x₃ = 4*R + 2*R + 3*R = 9*R

We can calculate the mass of each sphere (assuming that all are from the same material, with a constant density), as the product of the density and the volume:

m = ρ*V

For a sphere, the volume can be calculated as follows:

\frac{4}{3} *\pi *(r)^{3}

So, we can calculate the masses of the spheres, as follows:

m₁ = ρ*\frac{4}{3} *\pi *(2r)^{3}

m₂ = ρ*\frac{4}{3} *\pi *(r)^{3}

m₃ = ρ*\frac{4}{3} *\pi *(3r)^{3}

The total mass can be calculated as follows:

M= ρ*\frac{4}{3} *\pi * (8*r³ + r³ + 27*r³) =ρ*\frac{4}{3} *\pi * 36*r³

Replacing by the values, and simplifying common terms, we can calculate the x-coordinate of the center of mass of the system as follows:

Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}

Xcm = \frac{(8*R^{3} *2*R)+(R^{3}*(5*R))+27*R^{3}*(9*R))}{36*R^{3} }=\frac{264*R^{4} x}{36*R^{3}} = 7.33 R

As the x-coordinate of the center fof mass of the entire system is located at 7.33*R from the origin, and the center of mass of the smallest sphere is located at 5*R from the origin, the center of mass of the system is located at a distance d:

d = 7.33*R - 5*R = 2.33 R

4 0
3 years ago
Briefly outline the caloric theory about the nature of heat
Stells [14]
Briefly outline the caloric theory about the nature of heat
4 0
3 years ago
Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples.
Angelina_Jolie [31]
In scientific terms, ultrasound is a sound pressure, cyclic in nature, that has a greater frequency than the limit at the top of human hearing capabilities. What this means is that an ultrasonic sound can’t be heard by the human ear because their frequency is too high for our ears to pick up. In healthy young adults, this upper hearing capability is an average of 20 kilohertz. Ultrasound has many applications in several fields. Perhaps the best known application for ultrasound is sonography. This is where medical staff use the high pitched noise to produce a picture of a fetus while in the mother’s womb. Another use however, doesn’t directly concern humans at all. Bats use the high pitched noises to see in the dark and get an accurate reading on their preys internal structure. A popular belief is that an ultrasonic sound has the ability to turn the locking mechanism in a door lock, as demonstrated on some spy movies. On the opposite side of this are infrasonic sounds. These are noises with a frequency less than the lowest level of human hearing capabilities is 20 hertz. It is possible for humans to perceive infrasonic sounds, but only if the air pressure is sufficient. Although the war is the main tool for hearing these low sounds, it is possible for other parts of the body to “feel them”. Infrasound can be used to send signals in the army to special machines that can pick them up. These can be used to transmit vital data. Animals are able to pick up some low infrasonic noises which warn them of natural disasters before they happen, generally earthquakes and tsunamis.


I hope some of this information I gave you can help you. I came up with everything myself to help you.
4 0
3 years ago
A 1.3 kg ball drops vertically onto a floor, hitting with a speed of 21 m/s. It rebounds with an initial speed of 9.8 m/s. (a) W
Anarel [89]

Answer:

a) Impulse(J)=40.04 kg.m/s

b)F=2566.66 N

Explanation:

Given that

V_1=21 m/s,V_2=9.8m/s

We know that impulse(J) impulse is a vector quantity

J=P_2-P_1

We know that P=mV

SoJ=m(V_2-V_1)

Now by putting the values

J=1.3(9.8-(-21))

So impulse(J)=40.04 kg.m/s

Force is given as

F=\dfrac{dP}{dt}

or we can say that

F=m\left(\dfrac{v_2-V_1}{dt}\right)

So F=\dfrac{40.04}{0.0156}

F=2566.66 N

3 0
3 years ago
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