Explanation:
ultraviolet radiation
Bombarded with energy
They stored the seeds in a single layer on the outside of the ISS behind a special kind of glass that let in ultraviolet radiation only at wavelengths between 110 and 400 nanometers
The kinetic energy of an object depends upon its speed and its... what?
Answer:
Mass
Explanation:
kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity (speed) squared.
Metal reactivity
a. increases
b. decreases
from left to right in the periodic table.
C. stays the same
d. can increase or decrease depending on the
element
Answer:
i think its A
Explanation:
When a ball is thrown straight up with no air Resistance, the acceleration is in what direction ?
Answer:
The ball has a velocity upwards but the acceleration is downwards. Gravity is giving the ball a downwards acceleration from the moment the ball leaves the hands.
The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3
.find the dimension of the constants a, b, c and d
Answer:
[tex]a[/tex] has units of distance
[tex]b[/tex] has units of distance over time
[tex]c[/tex] has units of distance over [tex]time^2[/tex]
[tex]d[/tex] has units of distance over [tex]time^3[/tex]
Explanation:
Since the expression for the distance is:
[tex]x = a+b\,t+c\,t^2+d\,t^3[/tex]
then:
[tex]a[/tex] has units of distance
[tex]b[/tex] has units of distance over time
[tex]c[/tex] has units of distance over [tex]time^2[/tex]
[tex]d[/tex] has units of distance over [tex]time^3[/tex]
because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.
What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?
m3/(kg⋅s)
m3/(kg2⋅s2)
m3/(kg2⋅s)
m3/(kg⋅s2)
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:
[tex]F=G*\frac{m_1m_2}{r^2}[/tex]
By writing a dimensional analysis with the proper algebra handling, we obtain:
[tex]N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}[/tex]
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
The items in a mixture can be returned to their original form.
True
False
Answer:
I believe it is true
If not, pls let me know. :)
Calculate the distance moved by a runner who runs with a speed of 5 km/h for a period of 1.5 hours.
7.5 km
10 km
2.5 km
5 km
Answer:
7.5 km
Explanation:
h5 per hour means that he traveled 5 km in 1 our. And then half of the hour, which means half an hour 5 km which is 2.5.
5 + 2.5 = 7.5
or just 1.5 x 5 = 7.5
The electric field everywhere on the surface of a thin, spherical shell of radius 0.710 m is of magnitude 936 N/C and points radially toward the center of the sphere. (a) What is the net charge within the sphere's surface? nC (b) What is the distribution of the charge inside the spherical shell? The positive charge has an asymmetric charge distribution. The negative charge has an asymmetric charge distribution. The positive charge has a spherically symmetric charge distribution. The negative charge has a spherically symmetric charge distribution.
Explanation:
Given that,
Radius of a spherical shell, r = 0.71 m
Electric field that points radially toward the center of the sphere is 936 N/C
(a) Let q is the net charge within the sphere's surface. Using Gauss's law, we can find it :
[tex]\dfrac{1}{q}\times \epsilon_0=EA\\\\q=\dfrac{\epsilon_0}{EA}\\\\q=\dfrac{8.85\times 10^{-12}}{936\times \pi \times (0.71)^2}\\\\q=5.97\times 10^{-15}\ C[/tex]
(b) Gauss's law is used to find the amount of charge enclosed within a surface. It doesn't say anything about the distribution of the charge inside the spherical shell.
2. A fish swimming at a constant speed of 18 m/s spots a sta-
tionary barracuda. Just as the fish passes the barracuda, the
predator begins swimming with a constant acceleration of
magnitude 2.2 m/s². The fish and barracuda are moving in the
same direction.
(a) How far does the barracuda swim before catching the fish?
(b) At what time will this occur? (Hint: Graphing may help you
visualize this problem.)
пип
Answer:
a) 294.55 meters
b) 16.4 seconds
Explanation:
The suggested graph shows ...
(a) The barracuda swims 294.55 meters before the it catches the fish.
__
(b) This occurs after about 16.4 seconds.
A cloud mass moving across the ocean at an altitude of 2000 m encounters a coastal mountain range. As it rises to a height of 3500 m to pass over the mountains, it undergoes an adiabatic expansion. The pressure at 2000 m is 0.802 atm and at 3500 is 0.602 atm. If the initial temperature of the cloud mass is 288 K, what is the cloud temperature as it passes over the mountains? Assume that Cp,m for air is 28.86 J K-1 mol-1 and that the air obeys the ideal gas law. If you are on the mountain, should you expect rain or snow?
Answer:
snow
Explanation:
Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.
We can sole this problem using the following derivation:
[tex]ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\[/tex]
[tex]ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\[/tex]
Since T2 = 265 K, we should expect a snow
An object experiences a net acceleration to the left. Which of the following statements about this object are true? There may be more than one true statement below. Group of answer choices If there is one force to the left on the object, there must at least be one force on it to the right as well. There must be two or more forces on the object. The object cannot be slowing down. The net force on the object is to the right. If there is one force down on the object, there must at least be one force up on it as well. If the mass of the object was doubled, it would experience an acceleration of half the magnitude. Assume the forces on the object are unchanged. The object might be in equilibrium. If the mass of the object were halved, it would experience an acceleration to the right. Assume the forces on the object are unchanged. The net force on the object is to the left.
Answer:
When an object experiences acceleration to the left, the net force acting on this object will also be to the left.If the mass of the object was doubled, it would experience an acceleration of half the magnitudeExplanation:
When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
From Newton's second law of motion, the acceleration of the object is given as;
a = ∑F / m
a = -F / m
The negative value of "a" indicates acceleration to the left
where;
∑F is the net force on the object
m is the mass of the object
At a constant force, F = ma ⇒ m₁a₁ = m₂a₂
If the mass of the object was doubled, m₂ = 2m₁
a₂ = (m₁a₁) / (m₂)
a₂ = (m₁a₁) / (2m₁)
a₂ = ¹/₂(a₁)
Therefore, the following can be deduced from the acceleration of this object;
When an object experiences acceleration to the left, the net force acting on this object will also be to the left.If the mass of the object was doubled, it would experience an acceleration of half the magnitudeSandra shot a rocket so that it moved with an initial velocity of 9.81 m/s straight upward. The rocket leaves and returns to ground level. What is the total time the rocket was in the air before it strikes the ground? (Ignore air resistance.
Answer:
2.00 seconds.
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 9.81 m/s
Total time in air (T) =..?
Next, we shall determine the time taken for the rocket to reach it's maximum height. This can be obtained as follow:
Note: At maximum height, the final velocity is zero.
Initial velocity (u) = 9.81 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time to reach the maximum height (t) =.?
v = u – gt (since the rocket is going against gravity)
0 = 9.81 – 9.8t
Rearrange
9.8t = 9.81
Divide both side by 9.8
t = 9.81/9.8
t = 1.00 s
Therefore the time taken to reach the maximum height is 1.00 second.
Finally, we shall determine the total time spent by the rocket in the air as follow:
Time to reach the maximum height (t) = 1.00 s
Total time in air (T) =..?
T = 2t
T = 2 × 1.00
T = 2.00 s
Therefore, the total time spent by the rocket in the air is 2.00 seconds.
What the density of pure water?
a)1.0kg/l
b) 0.1 kg/l
c)100kg/l
d)10kg/l
Answer:
a) 1.0(kg/l) = 1 [kg/Lt]
Explanation:
The density of pure water is taken as 1000 [kg/m3], now we must convert the units to (kg/liters)
[tex]1000[\frac{kg}{m^{3}}]*\frac{1m^{3} }{1000Lt} = 1000[\frac{kg}{lt} ][/tex]
A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4 m/s^2. *
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
Imagine you are standing in the hallway. If the gravitational field strength were to suddenly double, what would change?
Answer:
If the gravitational field strength were to suddenly double while I was standing in the hallway, I'd suddenly feel like heavier.
Explanation:
The weight of a matter is given as the production between it's mass and the acceleration of gravity.
Mathematically, this is expressed as:
m x g that is m.g or mg.
Where m = mass and
g = gravity.
If the value of gravity was to double then weight (w) will become
w = m×2g
Assume for a moment that before the increase m was 5 and g was 5, it means that w would be 25.
But if g doubled, then we would have
w = 5 x 10
which is equals 50.
Thus, with the increase in gravitational field strenght, I'd find myself exerting more effort to stand upright, jump or run.
Cheers!
A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the
x axis be parallel to the ground and +y axis point upward. In the time interval from t=0 to t=4 s the trajectory of the caralong a certain section of the track is given by
→r=A(1 m/s)t^i+A((1 m/s3)t3−6(1 m/s2)t2)^j, where A is positive dimensionless constant.a. At t=2.0s is the roller coaster car ascending or descending?
b. Derive a general expression for the speed v of the car. Make sure that your expression would give the correct value for the speed (in m/s), but you don't need to put in anything explicitly about units (unlike what you see in the original expressionfor →r. Express your answer in meters per second in terms of A and t.
Answer:
Explanation:
r=A(1 m/s)t^i+A((1 m/s3)t3−6(1 m/s2)t2)^j,
dr / dt = A i + (3 A t² - 6 x 2 t ) j
At t = 2
dr / dt = A i + (12A - 24 A ) j = A i - 12 A j .
Since slope is negative so velocity is downwards , hence it is descending
b ) velocity = dr / dr = . A i + (3 A t² - 6 x 2 t ) j
Vx = A
Vy = 3A t² - 12 t
give five characteristics of a metal
Answer:
are good conductors of heat,are you used to make doors utensils e.t.c.
Using a set of observations to test a hypothesis.
PLEASE HELP!!! THANKS I GIVE BRAINLIEST !!A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experiment, four circuits were built with 1, 2, 3, and 4 batteries respectively. For each trial, the brightness of the light bulb was measured using a light meter. Which variable should the student keep constant
Answer:
The batteries
Explanation:
Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all the available space. If only 74.0 percent of the space inside the cube is taken up by Pt atoms, calculate the radius in picometers of a Pt atom. The mass of a single Pt atom is 3.240 × 10−22 g. [The volume of a sphere of radius r is (4/3) πr3. The volume of a cube is l3, where l is the length of a side. Avogadro's number is 6.022 × 1023.]
Answer:
A)6.6×10^22atoms of Pt in the cube
B)1.4×10^-8m
Explanation:
(a) Calculate the number of Pt atoms in the cube.
an edge length of platinum (Pt) = 1.0 cm.
Then Volume= 1.0 cm×1.0 cm×1.0 cm=1cm^3
Then we have volume of the cube as 1cm^3
Given:
The density Pt = 21.45 g/cm3
the mass of a single Pt atom =3.240 x 10^-22 g
Then with 1atom of the platinum element, we can calculate the number of Pt atoms in the cube as
Density of pt/mass of a single Pt atom
=(21.45 /=3.240 x 10^-22)
=6.6×10^22atoms of Pt in the cube
B)Volume of cube V=4/3πr^3
V= 4/3 ×π×r^3
V= 4.19067r^3
r^3= V/4.19067
But volume is not total volume but just 74% of it, then With 74% of the space inside the cube is taken up by Pt atoms, then we need to find 74% of volume of the cube which is 1cm^3
74/100 ×1= 0.74cm^3
Then our new volume V is 0.74cm^3
r^3=0.74/4.19067×6.620 x 10^22
r^3=2.6674×10^-24
r= 3√2.6674×10^-24
r=1.4×10^-8m
3. How does the resistance of the light bulbs differ when the bulbs are cold and when the bulbs are hot? Why do you think this happens?
Answer:
the hot bulb will have high resistance to the flow of current. While the cold bulb will have a low resistance to the flow of current.
Explanation:
A conductor that does not obey Ohm's law is described as non - ohmic. An example is a filament lamp. It glows as the current passes through it.
How does the resistance of the light bulbs differ when the bulbs are cold and when the bulbs are hot ?
The resistance of the light bulbs increase gradually as its temperature is increased.
So, the hot bulb will have high resistance to the flow of current. While the cold bulb will have a low resistance to the flow of current.
Because the resistance of an impure metal wire is greater than the resistance of a pure metal wire of the same dimension.
a piano dropped from a plane in the air. As it falls, upward force of air resistance gets greater as the piano picks up speed eventually, the resistance forces equal to the downward force of gravity. the piano now
Answer:
When the piano is dropped, the first force acting on it will be the gravitational force, that accelerates the piano at 9.8m/s^2 downwards.
As the piano accelerates, the velocity increases, now appears other force, the air resistance, that opposes to the motion of the piano.
As the velocity of the piano increases, also does the force that the air applies on the piano.
There is a point where the velocity of the piano is such that the air resistance is equal in magnitude, but in the opposite direction, to the force of gravity.
Then the net force on the piano is zero, which means that there is no acceleration, so the piano will keep falling down at constant velocity after this point.
2. The visible region of the hydrogen spectrum results from relaxation of electrons from excited states to energy level 2 (n1). Use the Rydberg equation and your measured wavelengths to determine the energy transitions associated with each of your observed wavelengths for hydrogen. In other words, calculate the excited state energy level (n2) for each of your observed wavelengths for hydrogen. n has integer values; so, calculate it first with appropriate significant digits, then round it to an integer. Use the key to show your work for at least one calculation. Must show energy levels for each hydrogen wavelength.
Answer:
E = 1.89 eV , E = 2.56 eV , E = 2.86 eV
Explanation:
The emission of light in the visible range, explained by the Balmer series with expression
1 /λ= [tex]R_{H}[/tex] (1/2² - 1 / n²)
n = 3, 4, 5 ...
the constant R_{H} called Rydberg's constant and is equal to 1,097 10⁷ m⁻¹
These transitions are clearly explained by Bohr's atomic model, where the empirical series of Balmer and Rydberg are deduced from a theoretical model of the hydrogen atom in natural form.
Let's calculate the wavelengths for each transition
State
initial final λ (10⁻⁷ m)
3 2 6.5634
4 2 4.8617
5 2 4.3408
Let's calculate the energy of each of these wavelengths using Planck's equation
E = h f = h c /λ
λ = 6.5634 10⁻⁷ m
E = 6.63 10⁻³⁴ 3 10⁸ / 6.5634 10⁻⁷
E = 3.03 10⁻¹⁹ J
we reduce to eV
E = 3.03 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E = 1.89 eV
λ = 4.8617 10⁻⁷m
E = 6.63 10⁻³⁴ 3 10⁸ / 4.8617 10⁻⁷
E = 4.09 10⁻¹⁹ J
E = 2.56 eV
λ= 4.3408 10⁻⁷ m
E = 6.63 10⁻³⁴ 3 10⁸ / 4.3408 10⁻⁷
E = 4.582 10⁻¹⁹J
E = 2.86 eV
What is the acceleration at the apex of a vertical up and down problem?
Answer:
The correct option is +9.81 m/s²
Explanation:
The acceleration of a vertical (up and down) plane depends on if the object is going up or down. Acceleration due to gravity is 9.81 m/s/s or 9.81 m/s². When an object falls/comes down (vertically) without interference, the acceleration of such an object is same as acceleration due to gravity (+9.81 m/s²). However, when an object is thrown/goes up, the acceleration of such objects goes against the gravity (of earth) hence the acceleration is -9.81 m/s².
At the top/apex of a vertical up and down problem, the object will be pulled back down (because of gravity) and hence, it's acceleration becomes +9.81 m/s² (changing from negative while coming up to positive).
A train running along a straight track at 32.90 m/s is slowed at a rate of 0.783 m/s/s to a stop how far did the train travel before coming to rest ?
Answer:
The distance traveled by the train is 691.2 m
Explanation:
Given;
initial velocity of the train, u = 32.9 m/s
deceleration of the train, a = - 0.783 m/s²
final velocity of the train when brought to a stop, v = 0
The distance traveled by the train is given by;
v² = u² + 2as
0 = (32.9)² + 2(-0.783)s
0 = 1082.41 - 1.566s
1.566s = 1082.41
s = 1082.41 / 1.566
s = 691.2 m
Therefore, the distance traveled by the train is 691.2 m
A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The maximum acceleration the truck can have so the bag does not slip is 2.40 m/s2 . Calculate the (a) angle of the hill relative to horizontal and (b) coefficient of static friction between the bag and the truck. (c) The truck is now travelling on level ground at constant speed. The sand bag is tossed forward sliding along the truck bed with a horizontal speed of 2.55 m/s. If the coefficient of kinetic friction is 0.350, how far does the bag slide before coming to rest
Answer:
a
[tex]\theta = 23.32^o [/tex]
b
[tex] \mu_s = 0.27 [/tex]
c
[tex] s = 0.948 \ m [/tex]
Explanation:
From the question we are told that
The mass of the bag is [tex]m_b = 25.0 \ kg[/tex]
The normal force experienced is [tex]F_n = 225 \ N[/tex]
The maximum acceleration of the bag is [tex]a = 2.40 \ m/s^2[/tex]
Generally this normal force experience by the bag is mathematically represented as
[tex]F_n = mg cos \theta[/tex]
=> [tex]225 = (25 * 9.8) cos \theta[/tex]
=> [tex] 0.9183 = cos \theta[/tex]
=> [tex]\theta = cos^{-1}[0.9183][/tex]
=> [tex]\theta = 23.32^o [/tex]
Generally for the bag not to slip , it means that the frictional force is equal to the sliding force
[tex]F_f = F_s[/tex]
Hence [tex]F_f [/tex] is mathematically represented as
[tex]F_f = \mu_s * F_n [/tex]
While [tex]F_s [/tex] is mathematically represented as
[tex]F_s = m * a [/tex]
So
[tex] \mu_s * F_n = m * a [/tex]
=> [tex] \mu_s * 225 = 25 * 2.40 [/tex]
=> [tex] \mu_s = 0.27 [/tex]
Generally from the workdone equation we have that
[tex]KE_f - KE_i = W_f[/tex]
Here [tex]W_f[/tex] is the work done by friction which is mathematically represented as
[tex]W_f = m * g * \mu_k * s[/tex]
Here s is the distance covered by the bag
[tex]KE_f[/tex] is zero given that velocity at rest is zero
and
[tex]KE_i = \frac{1}{2} * m* v_i^2[/tex]
so
[tex] \frac{1}{2} * m* v_i^2 = m * g * \mu_k * s [/tex]
=> [tex] \frac{1}{2} * v_i^2 = g * \mu_k * s [/tex]
substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that
[tex] \frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s [/tex]
=> [tex] s = 0.948 \ m [/tex]
A physics student rolls a ball down a hill with an initial velocity of 2.5m/s. If it accelerates down the hill at a rate of 6.3m/s^2, How long will it take to reach the bottom of the hill, 150 meters away?
Please help!!!
I'll assume you mean the hill is 150 m along its slope, since that's the direction the ball rolls.
The distance x it rolls after time t is
x = (2.5 m/s) t + 1/2 (6.3 m/s²) t²
Set x = 150 m and solve for t :
150 m = (2.5 m/s) t + 1/2 (6.3 m/s²) t²
(3.15 m/s²) t² + (2.5 m/s) t - 150 m = 0
t = 1/2 [(-2.5 m/s) + √((2.5 m/s)² - 4 * (3.15 m/s²) * (-150 m))]
(Use the quadratic formula, and take the positive root)
t ≈ 6.52 s
A student pushes a 0.5 kg lab cart with a force of 3 Newtons. Determine the acceleration of the cart in
m/s2.
Answer:
The acceleration is 6 [tex]\frac{m}{s^2}[/tex]
Explanation:
Use Newton's second law to solve the problem:
[tex]Force = mass\,\,*\,\,acceleration\\3\,N = 0.5\,\,kg\,*\,a\\a = \frac{3}{0.5} \frac{m}{s^2} \\a=6\,\,\frac{m}{s^2}[/tex]
the acceleration of the cart given its mass and force exerted is 6m/s².
Given the data in the question;
Mass of lab cart; [tex]m = 0.5kg[/tex]
Force applied; [tex]F = 3N[/tex]
Acceleration of cart; [tex]a = \ ?[/tex]
To determine the acceleration of the cart, we use the expression from Newton's second law of Motion:
[tex]F = m* a[/tex]
Where F is force exerted, m is mass and a is acceleration
We substitute our given values into the equation
[tex]3N = 0.5kg \ * \ a\\\\3kgm/s^2 = 0.5kg\ *\ a\\\\a = \frac{3kgm/s^2}{0.5kg} \\\\a = 6m/s^2[/tex]
Therefore, the acceleration of the cart given its mass and force exerted is 6m/s².
Learn more: https://brainly.com/question/15839425
A .What is the difference between fossils fuels and nuclear fuel ?
B .Explain why it is important for us to find alternatives to fossil fuels to meet pur energy?
C .What are renewable energy sources?give three examples.
Answer:
A. Nuclear and fossil fuel-burning power plants differ mainly in where their energy comes from; a nuclear reactor produces heat from radioactive metals, and a fossil-fuel plant burns coal, oil or natural gas.
B. Environmental and economic benefits of using renewable energy include: Generating energy that produces no greenhouse gas emissions from fossil fuels and reduces some types of air pollution. Diversifying energy supply and reducing dependence on imported fuels
C. These are energy sources that are constantly being replenished, such as sunlight, wind, and water.
The most popular renewable energy sources currently are:
Solar energy.
Wind energy.
Hydro energy.
Tidal energy.
Geothermal energy.
Biomass energy.
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