write the adavandages and disadvanges of solar energy

Answer:

0.82 m

Explanation:

Given:

v₀ᵧ = 4 m/s

aᵧ = -9.8 m/s²

Find: Δy when vᵧ = 0 m/s

vᵧ² = v₀ᵧ² + 2aᵧΔy

(0 m/s)² = (4 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 0.82 m

The maximum height reached by the athlete is 0.82 meters.  The jump is not successful.

Answer:

Energy=3.1times 10^-17 J

Rest mass: 6.2 kg

Speed: 47.5 m/s

Wavelength: 2.659 times 10^-6

Momentum: 67.3 kg(m/s)

Explanation:

Answer:

[tex] \boxed{\huge {\sf A. \ 657.5 \ N}} [/tex]

Given:

Mass = 23.40 kg

Acceleration = 28.10 m/s

To Find:

Amount of force causing acceleration

Explanation:

[tex] \sf Force = Mass \times Acceleration \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 23.40 \times 28.10 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 657.54 \ N \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \approx 657.5 \ N [/tex]

Answer:

ω, the angular frequency of the source equals 377 rad/s

Explanation:

From the question, V(t) = V cosωt.

Now, ω = the angular frequency of the sinusoidal wave is given by

ω = 2πf where f = the frequency of the source = 60 Hz

So, the angular frequency of the source ,ω = 2π × the frequency of the source.

So, ω = 2πf

ω = 2π × 60 Hz

ω = 120π rad/s

ω = 376.99 rad/s

ω ≅ 377 rad/s

So, ω, the angular frequency of the source equals 377 rad/s

Answer:

5.10 m

Explanation:

Given that:

Initial velocity given to the ball, v = 10 m/s

Mass of the ball, m = 0.3 kg

Acceleration due to gravity, g = 9.8m/[tex]s^2[/tex]

To find: The height upto which the ball will go =  ?

Solution:

Initially, the ball will have kinetic energy and at top point the velocity will be zero, it will have only potential energy due to height.

Formula for kinetic energy and potential energy:

[tex]KE = \dfrac{1}{2}mv^2\\PE = mgh[/tex]

Applying conservation of energy principle:

[tex]\dfrac{1}{2}mv^2=mgh\\\Rightarrow \dfrac{1}{2}v^2=gh[/tex]

Putting the values of v and g to find h:

[tex]\Rightarrow \dfrac{1}{2}10^2=9.8\times h\\\Rightarrow h = \dfrac{50}{9.8}\\\Rightarrow h = 5.10\ m[/tex]

The ball will go 5.10 m high.

Answer:

Increases, increases

Explanation:

The current is directly proportional to the voltage and inversely proportional to the resistance. The implication of this is that, whenever the voltage is increased, the current increases simultaneously. On the other hand, if the resistance is increased, the current will decrease accordingly and vice versa.

Recall that power is given by P= V^2/R where;

P= power, V= voltage and R= resistance

We can see that power and resistance are inversely related hence decreasing the resistance increases the power output of the lightbulb.

Answer:

Increase, increase

Explanation:

Ohm's law states that R=V/I .

This can be rearranged as I = V/R , showing that current is inversely proportional to resistance. When resistance decreases, current increases (with constant voltage).

Power dissipated by a resistor is given by the equation P = RI^2 . Showing that as R decreases and increases I, the increase to I is exponential while the decrease in resistance is not, which leads to increased power dissipation.

Answer:

In  a parallel circuit, the blank

is the same for every leg in the circuit why?

Explanation:

Because the voltage is common across the elements of a parallel circuit, the voltage drops are all equal to each other, and the ap- plied voltage is equal to any one of thJ individual voltage drops. are also 60v. In a series circuit, the same current flows. through every component.

Answer:

voltage

Explanation:

a    p    e    x

Answer:

Of course, with a triangle with a 90° angle, a right triangle, you can simply use pythagoras theorem  (a2+b2=c2)  then SOH-CAH-TOA to solve for the angle  (θ=Cos−1(adjacenthypotenuse)) .

But what if it isn't a right triangle? The cosine rule works for all triangles, even if they don't have a 90° angle.

a2=b2+c2−2(b)(c)Cos(A)  

Where a is the missing side, and A is the angle opposite to side a.

With this rule, we can find a missing angle or a missing side to any type of triangle, assuming you have the needed variables.

Explanation:

The box acted upon by the force of 5 N will move on a frictionless surface, with the condition S = P + Q.

The effect of a push or a pull on the body is known as force. The main types of forces include friction forces, nuclear forces, and gravitational forces. For instance, when a hand strikes a wall, the wall exerts a force on the hand as well as the hand exerting a force on the wall. Newton was given several laws to comprehend force.

Given: with a triangle with a 90° angle, a right triangle, you can simply use Pythagoras theorem  (a2+b2=c2)  then SOH - CAH - TOA to solve for the angle  (θ=Cos−1(adjacent hypotenuse)).

The cosine rule works for all triangles, even if they don't have a 90° angle.

a2=b2+c2−2(b)(c)Cos(A)  

A is the angle opposite to side a,

With this rule, we can find a missing angle or a missing side to any type of triangle, assuming you have the needed variables.

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Answer:

516m/s^2

Explanation:

Given the following :

Height of aircraft = 10000m

Acceleration due to gravity (g) = 10m/s^2

Angle of projection (θ) = 60°

Height of aircraft = maximum height

Maximum height of a projectile:

H = (u^2sin^2θ) / 2g

Where H = height

u = initial velocity

10000 = [(usinθ)^2] / 2g

10000 = [(u * sin60°) ^2] / 2*10

10000 = (0.866 * u)^2 / 20

20 * 10000 = 0.749956 * u^2

200000 = 0.749956u^2

u^2 = (200000/0.749956)

u^2 = 266,682.31

u = √266,682.31

u = 516.41292

Initial velocity (u) = 516m/s^2

Answer:

Explanation:

Three screens are placed one behind the other and light falls on all of them from a torch . In this case a mirror can be only the last screen . In case of first or second screen being a mirror , light would not have reached the last screen because the nature of mirror is to reflect the light back.

Hence , the only possibility is that , the last screen is mirror.

Answer:

Explanation:

According to Hooke's law "provided the elastic limit of a material is not exceeded the extension e is directly proportional to the applied force"

Given data

mass m= 85 kg

spring constant k= 250 N/m

assuming g= 9.81 m/s^2

extension e= ?

We know that

[tex]F= ke \\mg= ke[/tex]

solving for the extension e

[tex]85*9.81= 250*e\\\e= \frac{833.85}{250} \\\e= 3.33m[/tex]

to get how far bellow the bridge the student falls before coming to stop, we have to add the extension e to the length of the bungee cord.

=12+3.33= 15.33 m

The student fall  below the bridge(12+3.33 = 15.33 m ) before coming to a stop. Where 12 is the length of bungee cord.

According to Hooke's law in the elastic limit of a material the the strain in the spring is directly proportional to the applied force or stress.  

Formula

[tex]\bold {F_s = -kx}[/tex]

Where,

= spring force

k = spring constant =  250 N/m

x = spring stretch or compression = ?

The force,

F = mg

[tex]\bold {F = 85 \times 9.18}\\\\\bold {F = 833.85 }[/tex]

So,

[tex]\bold {x = \dfrac {F_s }{k} }\\\\\bold {x = \dfrac {833.85} {250} }\\\\\bold {x = 3.33}[/tex]  

Therefore, the student fall  below the bridge(12+3.33 = 15.33 m ) before coming to a stop. Where 12 is the length of bungee cord.

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Answer:

Archimedes Principle states that "any body completely or partially submerged in water is acted upon by an upthrust force which is equal to the magnitude of Weight of the body."

Answer:

We know that due to upthrust ,objects tend to lose weight when immersed in liquid.

This law is applicable for the gaseous medium too.The weight of the displaced liquid =pgv where p is density of liquid ,g is acceleration due to gravity and v is volume of displaced liquid.

Hope this helps...

Good luck on your assignment..

Answer: s(5) = -200,  v(5) = -90

              s(10) = -900, v(10) = -190

Explanation:

Position: s(t)

Velocity:  s'(t) = v(t)                ⇒      [tex]s(t)=\int {v(t)} \, dt[/tex]

Acceleration     v'(t) = a(t)      ⇒      [tex]v(t)=\int {a(t)} \, dt[/tex]

We are given that acceleration a(t) = -20   and velocity v(t) = 10

[tex]v(t)=\int {a(t)} \, dt\\\\v(t)=\int{-20}\, dt\\\\v(t)=-20t + C \\\\v(t)=10\quad \longrightarrow \quad C=10\\\\v(t)=-20t+10[/tex]

[tex]s(t)=\int {v(t)} \, dt\\\\s(t)=\int {(-20t+10)} \, dt\\\\s(t)=-10t^2+10t\\\\[/tex]

(a) Input t = 5 into the s(t) and v(t) equations

   s(5) = -10(5)² + 10(5)           v(5) = -20(5) + 10

          =  -250  +  50                     = -100   +  10

          =       -200                          =        -90

(b) Input t = 10 into the s(t) and v(t) equations

     s(10) = -10(10)² + 10(10)           v(10) = -20(10) + 10

            =  -1000  +  100                     = -200   +  10

            =       -900                          =        -190

Answer:

yeah..

as we all know that the cheapest fuel is petrol because it is the last product of petroleum.

hope it helps...

Answer:

a) 50 rad/s

b) 39.8 rev

Explanation:

Given:

r = 0.36 m

v₀ = 0 m/s

a = 1.8 m/s

t = 10 s

a) Find: ω

v = at + v₀

v = (1.8 m/s) (10 s) + (0 m/s)

v = 18 m/s

ω = (18 m/s) / (0.36 m)

ω = 50 rad/s

b) Find: Δθ

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1.8 m/s) (10 s)²

Δx = 90 m

Δθ = (90 m) / (2π × 0.36 m)

Δθ = 39.8 rev

Answer: 5000

Explanation: pressure: force / area

SI unit of force: newton

SI unit of area: metre²

Force exerted: 10 N

Area on which it is exerted : 20 cm²

Since area has to be in m² , we convert cm² to m² by dividing it by 10000( m:100cm, 1m²: 10000 cm²)

Area=20/10000: 0.002 m²

So, pressure= 10/0.002=5000 pa( SI unit of pressure is pascal)

The pressure is the force exerted by unit area.  The force on the brick by its weight is 10 N and the surface area is 20cm². Then, the pressure in the given area will be 5000 Pa.

Pressure on an object is the force exerted per unit area of that object. Pressure is dependent on the volume and temperature as well. It is commonly expressed in the unit of atm. Other units of pressure are bar, pascal, torr, mmHg etc.

The normal force acting on the brick is the force by its own weight.

normal force = 10 N

surface area = 20 cm²

20 cm² /(100)² = 0.002 m²

Pressure = force/ area

p = 10 N/0.002 m²

  = 5000 N/m²  = 5000 pa

Therefore, the pressure exerted on the brick is 5000 Pa

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Answer:

2943 J

Explanation:

Potential energy = mass x acceleration due to gravity x height in meters

(P.E. = mgh)

Substitute the given numbers: (I take acceleration due to gravity as 9.81 m s^-2)

PE = 40 x 9.81 x (0.15x50)

PE = 2943 J

Answer:

Explanation:

slit separation d = .3 x 10⁻³ m

screen distance D = 4.35 m

fringe width = 6.4 x 10⁻² / 9

= .711 x 10⁻² m

a ) Interference pattern on the screen is created due to interference of two light waves coming from two slits and falling in screen . When they fall on different points on the screen they travel different distance . This is called path difference . When their path difference is equal to their wave length or its integral multiple , they reinforce each other and at that point the brightness increases . On the other hand when at a point the path difference is equal to odd multiple of half wavelength , they kill or destroy each other and darkness is found at that point . In this way dark band and bright band appear on the screen . This is called interference pattern .

b )

Fringe width  = λ D / d

Putting the values

.711 x 10⁻² = λ x 4.35 /  0.3 x 10⁻³

λ = .049 x 10⁻⁵

= 490 x  10⁻⁹

= 490 nm

The colour of this wavelength will be blue .

Answer:

Ohm's Law:

[tex]V = I R[/tex]     ------------------(1)

Where V is voltage, I is current and R is resistance.

Whereas,

P = I V   --------------------(2)

Where P is power, I is current and V is Voltage.

Putting (1) into (2)

=> P = I (I R)    [∴ V = I R]

=> P = I²R    (Derived!)

Answer:

Distance

Explanation:

Distance is the base quantity (measured in lengths) while speed, Area and Force are all derived quantities derived from base quantities.

Answer:

balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure

Explanation:

Este ejercicio es referente a la mecánica de fluidos, usemos la expresión para la presión  

       P = ρ g h

En es el caso del balón  usemos la presión en la pared extrema, llamemos P la presión por el gas en el interior y P_ext la presión atmosférica del lugar

        cuando se llena el valor en una ciudad de baja altura la presión atmosférica es mas alta

          P_int1 < P_ext1

por lo cual la pared del balón no se mantiene rígida.

Cuando el balón es trasladado a una ciudad con mayor altura sobre el nivel del mar la presión exterior disminuye

       P_ext2 = ρ g h₂ < P_ext1

en promedio la presión disminuye con la altura  en 0,029 atm cada 250 m

por lo tanto como la cantidad de gas es constante en el interior la presión interior del globo es constante y en esta ciudades se hace igual o un poco mayor que la presión atmosférica, en consecuencia la pared del globo esta rígida

        P_int2 >P_ext2

Traslate

This exercise is related to fluid mechanics, let's use the expression for pressure

       P = ρ g h

In the case of the balloon, let's use the pressure on the extreme wall, let's call P the pressure for the gas inside and P_ext the atmospheric pressure of the place

        when the value is filled in a low-lying city the atmospheric pressure is higher

          P_int1 <P_ext1

therefore the wall of the ball does not remain rigid.

When the ball is transferred to a city with higher altitude above sea level, the external pressure decreases

       P_ext2 = ρ g h <P_ext1

on average the pressure decreases with height by 0.029 atm every 250 m

therefore as balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure, therefore the wall of the

        Pint 2> Pe

Answer:

The ratio of the time period of object 1 to the time period of object 2 is [tex]\frac{1}{4}[/tex]. (T₁ : T₂) = 1 : 4

Explanation:

Let suppose that both objects are moving along the circular paths at constant speed, such that period of rotation of each object is represented by the following formula:

[tex]\omega = \frac{2\pi}{T}[/tex]

Where:

[tex]\omega[/tex] - Angular speed, measured in radians per second.

[tex]T[/tex] - Period, measured in seconds.

The period is now cleared:

[tex]T = \frac{2\pi}{\omega}[/tex]

Angular speed ([tex]\omega[/tex]) and linear speed ([tex]v[/tex]) are related to each other by this formula:

[tex]\omega = \frac{v}{R}[/tex]

Where [tex]R[/tex] is the radius of rotation, measured in meters.

The angular speed can be replaced and the resultant expression is obtained:

[tex]T = \frac{2\pi\cdot R}{v}[/tex]

Which means that time period is directly proportional to linear speed and directly proportional to radius of rotation. Then, the following relationship is constructed and described below:

[tex]\frac{T_{2}}{T_{1}} = \left(\frac{v_{1}}{v_{2}}\right)\cdot \left(\frac{R_{2}}{R_{1}} \right)[/tex]

Where:

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Time periods of objects 1 and 2, measured in seconds.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Linear speed of objects 1 and 2, measured in meters per second.

[tex]R_{1}[/tex], [tex]R_{2}[/tex] - Radius of rotation of objects 1 and 2, measured in meters.

If [tex]\frac{v_{1}}{v_{2}} = 2[/tex], [tex]R_{1} = 10\,m[/tex] and [tex]R_{2} = 20\,m[/tex], the ratio of time periods is:

[tex]\frac{T_{2}}{T_{1}} = 2\cdot \left(\frac{20\,m}{10\,m} \right)[/tex]

[tex]\frac{T_{2}}{T_{1}} = 4[/tex]

[tex]\frac{T_{1}}{T_{2}} = \frac{1}{4}[/tex]

The ratio of the time period of object 1 to the time period of object 2 is [tex]\frac{1}{4}[/tex].

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, [tex]t_t[/tex] of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + [tex]t_t[/tex] + t₁ =6 + 4 + 1.6 = 11.6 seconds.

Willie's total time to cover the race is 11.6 seconds.

A velocity-time graph is a graph that is used to represent a way in which the motion of an object can be represented with the velocity change on the vertical (y)-axis and change in time on the horizontal (x)-axis.

From the given information:

Using the following first equation of motion:

The time during the first acceleration can be computed as:

v = u + at

12 = 0 + (2 × t)

t = 12/2

t = 6 seconds

Using the fourth equation of motion, the distance covered in the first acceleration is determined as follows:

[tex]\mathbf{v^2 = u^2 + 2as}[/tex]

[tex]\mathbf{12^2 = (0)^2 + 2(2) \times s}[/tex]

144 = 4s

s = 144/4

s = 36 meters

However, he maintained a speed of 12.0 m/s till he is 16.0 m away from the finish line when he starts to decelerate.

Hence, at his highest speed;

Recall that:

[tex]Time = \dfrac{distance}{velocity}[/tex]

Thus, the time for running at his highest speed is:

[tex]\mathbf{time = \dfrac{48 m}{12 m/s}}[/tex]

time = 4 seconds

His deceleration from his highest speed can be computed as:

[tex]\mathbf{v^2 = u^ 2 + 2as}[/tex]

here:

[tex]\mathbf{v^2 - u^2=2as}[/tex]

[tex]\mathbf{a = \dfrac{v^2 - u^2}{2s}}[/tex]

[tex]\mathbf{a = \dfrac{8^2 - 12^2}{2(16)}}[/tex]

[tex]\mathbf{a = \dfrac{64 - 144}{32}}[/tex]

a = -2.5 m/s²

Finally, using the first equation of motion, the total time can be computed as:

[tex]\mathbf{v_1 = u_1 + a_1t_1}[/tex]

[tex]\mathbf{8= 12 + (-2.5)t_1}[/tex]

[tex]\mathbf{t_1 = \dfrac{(8- 12)}{ (-2.5)}}[/tex]

[tex]\mathbf{t_1 = \dfrac{(-4)}{ (-2.5)}}[/tex]

[tex]\mathbf{t_1 = 1.6\ seconds}[/tex]

Thus, the total time = (6+4+1.6) seconds

The total time = 11.6 seconds

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Answer:

18 m

Explanation:

Given:

v₀ = 6 m/s

v = 12 m/s

a = 3 m/s²

Find: Δx

v² = v₀² + 2aΔx

(12 m/s)² = (6 m/s)² + 2 (3 m/s²) Δx

Δx = 18 m

Answer:

 c = 5 m

Explanation:

this exercise you want to divide the rectangular room into two triangular rooms

the area of ​​triangles is

           A = ½ base height

           A = ½ 4 3

          A = 6 m²

the length of the curtain can be found using the Pythagorean theorem

           c² = b² + a²

           c = √ (4² + 3²)

           c = 5 m

this is the length of the curtain

Answer:

Kinetic Energy

Answer:

Kinetic

Explanation:

got it on edge

Answer:

Option B

Explanation:

Kinetic Energy is the energy possessed by the body due to "motion".

The answer is:

Answer:

The graph shows direct proportion  because it is a straight line

through the origin.

Explanation:

Answer:

MY friend has already described the purpose of projectile motion so I will quickly go through the uses of each equation ...

Explanation:

TIME OF FLIGHT = it is given as 2Usin tita/g...it is the total time taken to and fro...it is 2x of the time taken ....

TIME taken ..t= Usin tita / g.....is the time taken to reach the maximum height which is 1/2 the TOTALTIME OF FLIGHT GIVEN ABOVE ..

MAXIMUM HEIGHT: the maximum height is the height attained by the projectile when projected ...it is calculate using the formula = U^2 sin^2 tita / 2g

Range =Search Results

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An object launched into projectile motion will have an initial launch angle anywhere from 0 to 90 degrees. The range of an object, given the initial launch angle and initial velocity is found with: R=v2isin2θig R = v i 2 sin ⁡ 2 θ i g .