The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lights?
Picture of three solar panels street light on a sunny day with blue background
A.
gravitational potential energy → vibrational energy → chemical potential energy
B.
radiant energy → chemical potential energy → motion energy
C.
radiant energy → electric energy → radiant energy
D.
sound energy → chemical potential energy → radiant energy
E.
gravitational potential energy → motion energy → radiant energy
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The sequence that shows the energy transformations taking place in these lights are radiant energy → electric energy → radiant energy.
What is law of conservation of conservation of energy?
The principle or law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.
Based on this law, the energy of a substance can be converted from one form to another
For example, energy can be converted as follows;
potential energy to kinetic energychemical energy to electric energyelectrical energy to sound energyetc,The sequence of energy that takes place on street lights powered by solar panels is given as follows;
Radiant energy (light energy from sun) to electrical energy (converted by photo voltaic cell of the panels) to radiant energy (light given by the street lights).
Thus, the sequence that shows the energy transformations taking place in these lights are radiant energy → electric energy → radiant energy.
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Answer:
C
Explanation:
I did the test
A steel cable on a bridge has a linear mass density of 15 kg/m. If the cable has been pulled taunt with a tension of 5536 N, what is the speed of a wave on it?
Please help me with this!
The balloon goes 25 m east in first 10 s.
Then the wind blows the balloon 35 m west in 8 s.
a
The balloon travels 25 m in 10s.
b
the balloon travels a distance 35 m in next 8 s.
c
thhe total distance travelled by the balloon is,
[tex]\begin{gathered} d=25+35 \\ =60\text{ m} \end{gathered}[/tex]d
Average speed in 10 s is,
[tex]\begin{gathered} s=\frac{25}{10} \\ =2.5\text{ m/s} \end{gathered}[/tex]e
The average speed in next 8 s is,
[tex]\begin{gathered} s_8=\frac{35}{8} \\ =4.37\text{ m/s} \end{gathered}[/tex]f
the average speed for the entire trip is,
[tex]\begin{gathered} s_{av}=\frac{25+35}{10+8} \\ =3.33\text{ m/s} \end{gathered}[/tex]g
displacement during the first 10 s is.
[tex]d_1=(25m)\hat{i}[/tex]h.
displacement during next 8 s is,
[tex]d_2=(-35m)\hat{i}[/tex]i
The total displacement is,
[tex]\begin{gathered} d_1+d_2=(25-35)\hat{i}_{} \\ =-(10m)\hat{i} \end{gathered}[/tex]j
the average velocity in 10 s is,
[tex]\begin{gathered} v_1=\frac{25m}{10}\hat{i} \\ =(2.5m)\hat{i} \end{gathered}[/tex]k
The average velocity in 8 s is,
[tex]\begin{gathered} v_2=\frac{-35\text{ m}}{8\text{ s}}\hat{i} \\ =-(4.37m)\hat{i} \end{gathered}[/tex]l
The average velocity entire the whole trip is,
[tex]\begin{gathered} v_{av}=\frac{d_1+d_2}{18} \\ =\frac{-10m\hat{i}}{18\text{ s}} \\ =-(0.55\hat{m/s)i} \end{gathered}[/tex]An archery bow is drawn a distance d = 0.39 m and loaded with an arrow of mass m = 0.088 kg. The bow acts as a spring with a spring constant of k = 195 N/m, and the arrow flies with negligible air resistance. To simplify your work, let the gravitational potential energy be zero at the initial height of the arrow. If the arrow is shot at an angle of θ = 45° above the horizontal, how high, in meters above the initial height, will the arrow be when it reaches its peak?
The maximum height reached by the arrows is determined as 8.6 m.
What is the initial speed of the arrow?The initial velocity of the arrow is calculated by applying the principle of conservation of energy as shown below;
K.E = U
where;
K.E is the kinetic energy of the arrowU is the elastic potential energy of the bow¹/₂mv² = ¹/₂kx²
mv² = kx²
v² = kx²/m
v = √(kx²/m)
where;
k is spring constant of the bowm is the mass of the arrowx is the extension of the bowv = √(195 x 0.39²/0.088)
v = 18.36 m/s
The maximum height reached by the arrow is calculated as follows;
H = (v² sin²θ) / (2g)
where;
θ is angle of projection of the arrowg is acceleration due to gravityH = (18.36² (sin45)²) / (2 x 9.8)
H = 8.6 m
Thus, the height of the arrow above the ground when it reaches its peak is 8.6 m.
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For an object spinning around a central point, what will happen if its distance from the center is decreased
Answer:
Its a acceleration will increase
Explanation:
The force required to keep an object in a circular motion is given by
[tex]F=\frac{mv^2}{R}[/tex]where v is the radial velocity and R is the radius of the object with mass m.
Now our question is what happens to the above equation as we decrease R?
We can see that as R decreases the quantity mv^2 /R increases (since R is getting smaller ).
Hence, we conclude that F increases. But what if F? it is the centripetal force.
Since centripetal force has increased, so has the quantity v^2 /R (called the acceleration ).
Meaning an increase in centripetal force implies an increase n acceleration.
Since in the answer choices we are not given the option to increase our centripetal force, the next best choice is to choose 'acceleration will increase. '
what energy is gotten from wind
Kinetic energy is gotten from wind which is converted into rotational energy.
How energy is produced from the wind?The wind is used to produce electricity using the kinetic energy created by air in motion wind turbines convert the kinetic energy in the wind into mechanical power. This mechanical ability can be used for particular tasks (such as grinding grain or forcing water) or can be converted into electricity by a generator. into electricity. In present wind turbines, wind rotates the rotor blades, which change kinetic energy into rotational energy. Wind turbines labor on an easy principle: in lieu of using electricity to make wind like a fan wind turbines use the wind to make electricity. The wind turns the rotter-like blades of a turbine around a rotor, which spins a generator, which produced electricity.
So we can conclude that Wind rotates the rotor blades that convert kinetic energy into rotational energy.
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What is the initial velocity of an automobile acquiring a final velocity of 32 m/s with an acceleration of 4.0m/s ²
Answer:
Explanation:
Given:
V = 32 m/s
a = 4.0 m/s²
__________
V₀ - ?
V = V₀ + a*t
V₀ = V - a*t = 32 - 4*t
Time is not set according to the condition of the problem!
There's not enough given information t o answer the question. It depends on how long the car has been accelerating.
it could be 28 m/s 1 second ago.
it could be 16 m/s 4 seconds ago.
it could be 10 m/s 5.5 seconds ago.
etc.
i'll take a wild guess and speculate that the question actually tells how long the car has been accelerating, but you didn't copy that part.
Two blocks of mass M₁ and M₂ are connected by a massless
string that passes over a massless pulley as shown in the
figure. M₁ has a mass of 3.75 kg and rests on an incline of
0₁ = 63.5°. M2 rests on an incline of 0₂ = 15.5°. Find the
mass of block M₂ so that the system is in equilibrium (i.e.,
not accelerating). All surfaces are frictionless
The correct answer is 58.58 Kg. (Mass of M_2)
What is mass string and friction system?
A spring-mass system in simple calculation can be described as a spring system where a block is hung or attached at the free end of the spring. If the surface is frictionless so µ = 0 (we can assume)
To just begin to slide up the friction will be kinetic friction
Applying free body diagram on blocks (as diagram is not given in question so assumption is the basis on given data only)
Given, M_1 = 3.75 Kg., M_2 =?
O_1 = 63.5◦ and O_2 = 15.5◦, g = 9.8 m/s2
So, if we require to keep the system in equilibrium position
Then we can write an equation as follows:
M_1x g x Sin63.5◦ = M_2 x g x Cos 15.5◦ (To be in Equilibrium)
63.5 x 9.8 x 0.89101 = M_2 x 9.8 x 0.9659
M_2 = 63.5 x 0.89101 / 0.9659
M_2 = 58.58 Kg. (Mass of M_2)
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As a torque activity, your Physics TA sets up the arrangement decribed below. A uniform rod of mass mr = 158 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the r1 = 10.0 cm and r2 = 90.0 cm mark, passed over pulleys, and masses of m1 = 281 g and m2 = 177 g are attached. Your TA asks you to determine the following: (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). r3 = Fp = F = (b) Let's now remove the mass m3 and determine the new mass m4 you would need to suspend from the rod at the position r4 = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 = Fp = F = (c) Let's now remove the mass m4 and determine the mass m5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location r5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 = r5 =
a) Recall, the net torque on the rod must be zero. Thus,
Σt = 0
where
t represents torque
Thus,
t1 + t2 - tr - t3 = 0
t = rF
where
F = force
r = distance
r1F1 + r2F2 - rrFr - r3F3 = 0
r3F3 = r1F1 + r2F2 - rrFr
r3 = (r1F1 + r2F2 - rrFr)/F3
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
Thus,
r3 = (r1m1g + r2m2g - rrmrg)/m3g
g cancels out
r3 = (r1m1 + r2m2 - rrmr)/m3
From the information given,
r1 = 10 cm = 10/100 = 0.1 m
r2 = 90 cm = 90/100 = 0.9 m
rr = 100/2 = 50 cm = 50/100 = 0.5 m
m1 = 281 g = 281/1000 = 0.281 kg
m2 = 177g = 0.177 kg
mr = 158g = 0.158 kg
m3 = 200g = 0.2kg
By substituting these values into the equation,
r3 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
r3 = 0.542 m
The force exerted by the pin, Fp = mg
g = 9.8
Fp = (m3 - mr - m1 - m2)g
Fp = (0.2 + 0.158 - 0.281 - 0.177)9.8
Fp = - 0.981
Taking the absolute value,
IFpI = 0.981 N
F = - 90 degrees
b) r1F1 + r2F2 - rrFr - r4F4 = 0
r4F4 = r1F1 + r2F2 - rrFr = 0
F4 = (r1F1 + r2F2 - rrFr)/r4
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
F4 = T4 = m4g
Thus,
m4g = (r1m1g + r2m2g - rrmrg)/r4
m4g = (r1m1 + r2m2 - rrmr)/r4
r4 = 0.2
By substituting these values into the equation,
m4 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
m4 = 0.542 kg
The force exerted by pin is
Fp = (m4 + mr - m1 - m2(g
Fp = (0.542 + 0.158 - 0.281 - 0.177)9.8
Fp = 2.37 N
Fp = 2.37 N
F = 90 degrees
c) When the pin does not exert a force,
Fp = 0
F1 + F2 - Fr = F5
m1 + m2 - mr = m5
m5 = 0.281 + 0.177 - 0.158
m5 = 0.3 kg
Since the net torque on the rod is zero,
t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
r5 = r1F1 + r2F2 - ffFr)/F5
r5 = (r1m1 + r2m2 - rrmr)/m5
r5 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.3
r5 = 0.36