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3 years ago

Which energy resources are found above the Earth's surface?

Physics

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

Hey

Your answer would be B) B) solar, wind, biomass

Hydroelectric, uranium Geothermal=bellow earth's surface.

Coal, Oil, Natural gas=bellow earth's surface.

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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

magnetic field intensity, $B=0.07\ T$

kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

<em>Now, form the mathematical expression of Kinetic Energy:</em>

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

6 0

3 years ago

How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =

zlopas [31]

Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

___________________________________________

coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10 Joule= 100 Joule

Thus, 100 joule heat must be added to a 5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

8 0

3 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago

An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish

VikaD [51]

The answer is 7.1 I just took the test

8 0

3 years ago

Read 2 more answers

Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

mash [69]

Answer:

The value is $} N = 66 \ blocks$

Explanation:

From the question we are told that

The weight of the block is $W_b = 100 \ N$

The dimension of the block is $d = 0.400 m \ X \ 0.250 \ m \ X \ 0.130 \ m$

Generally two atmosphere is equivalent to

$P_{2atm} = 2 * 1.013 *10^{5} = 202600 \ N/m^2$

Generally 1 atm = $1.013 *10^{5} N/m^2$

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

$A = 0.250 * 0.130$

=> $A = 0.0325 \ m^2$

Generally the force due to this blocks is mathematically represented as

$F = N * W_b$

Here N is the number of blocks

$} 202600 = \frac{N * 100 }{ 0.0325}$

=> $} N = 66 \ blocks$

3 0

3 years ago