Derive an expression for Root Mean Square (RMS) cur-rent through the capacitor as a function of the applied voltage amplitude. Recall that the RMS value of a sinewave with unity amplitude is 0.707.

Answers

Answer 1

Answer:

hello your question is incomplete attached below is the complete question

answer: Irms = 0.707*2[tex]\pi FcVo\\[/tex]

Explanation:

To derive the expression for the root mean square ( RMS ) current through the capacitor as a function of the applied voltage amplitude

RMS value of a sine wave with unity amplitude = 0.707

when the dc component is removed the i.e. the RMS signal = standard deviation ( signal )

attached below is the derivation

Derive An Expression For Root Mean Square (RMS) Cur-rent Through The Capacitor As A Function Of The Applied
Derive An Expression For Root Mean Square (RMS) Cur-rent Through The Capacitor As A Function Of The Applied

Related Questions


Does changing the height of point C affect the speed of the coaster car at point D?​

Answers

Without friction, NO.

The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.

The speed of the coaster car at point D will be affected if  the height of point C is changed.

Potencial Energy:

It is the enrgy in a body due to the position of differnt part of the object or system.

As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.

Therefore, the speed of the coaster car at point D will be affected if  the height of point C is changed.

To know more about  speed of the coaster car,

https://brainly.com/question/9178285

A 40 kg boy standing on a skateboard throws a 2 kg ball 20 m/s to the left.
a. What is the ball's momentum?
O 10 kg m/s
O 20 kg mis
O 40 kg m/s
O 1 kg m/s

Answers

Answer:

40 kg m/s

Explanation:

Given the following data;

Mass of boy = 40kg

Mass of ball = 2kg

Velocity = 20m/s

To find the momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum =mass * velocity

Substituting into the equation, we have

Momentum = 2 * 20

Momentum = 40 kg m/s

Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I predicted. This is because I used the wrong mass for the puck when I did all my calculations. I accidentally used the mass of the smaller puck rather than the mass of the larger puck in my video." Is this a plausible explanation? Would the using the wrong mass for the puck during the calculations mean the puck would reach a greater height? Explain your reasoning.

Answers

Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the Principle of Conservation of Energy, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".

When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.

So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.

The representation is as followed:

[tex]K_{i}+P_{i}=K_{f}+P_{f}[/tex]

[tex]K_{i}+0=0+P_{f}[/tex]

[tex]\frac{1}{2}mv^{2} = mgh[/tex]

As we noticed, mass of the object can be cancelled from the equation, making height be:

[tex]h=\frac{v^{2}}{2g}[/tex]

So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.

A box has a mass of 150 kg. If a net force of 3,000 N acts
on the box, what is the box’s acceleration?

Answers

a = m/f
a = 150/3000
a = 15/300
a = 0.05m/s/s

True or False. The larger a waves wavelength, the more energy it carries. (1 Point) True O False O Maybe​

Answers

Explanation:

False, it is oppisite the shorter the wavelength the more energy it carries.

A iron block with a mass of 4.8 kg initially slides over a rough horizontal surface with a speed of 1.2 m/s. Friction slows the block to rest. While slowing to rest, 85.0% of the kinetic energy of the block is absorbed by the block itself as internal energy. What is the temperature increase of the block

Answers

Answer:

Explanation:

Kinetic energy of block will be converted into heat energy by friction .

Heat energy produced = 1/2 m v²

= .5 x 4.8 x 1.2²

= 3.456 J

85% of energy is converted into heat energy , so heat energy produced

= .85 x 3.456 = 2.9376 J .

If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature

Q = msΔt

specific heat of iron s = 462 J / kg C

Putting the values ,

2.9376 = 4.8 x 462 x Δt

Δt = 13.24 x 10⁻⁴ ⁰C.

greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but
3 leaps of the hound are equal to 5 leaps of the hare. Compare the speed of the
hound and the hare,

need full solution:-​

Answers

[tex]{\large{\bold{\rm{\underline{Given \; that}}}}}[/tex]

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

[tex]{\large{\bold{\rm{\underline{To\; find}}}}}[/tex]

★ The speed of the hound and the hare

[tex]{\large{\bold{\rm{\underline{Solution}}}}}[/tex]

★ The speed of the hound and the hare = 25:18

[tex]{\large{\bold{\rm{\underline{Full \; Solution}}}}}[/tex]

[tex]\dashrightarrow[/tex]  As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

 So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

 But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

 Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

(5/3)a × 5

25/3a metres

 Now the distance travelled by the hare in it's 6 leaps..!

6a metres

 Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

25/3a = 6a

25/3 = 6

25:18

A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through a distance of 1.0 m before leaving the skater's hand at a speed of 7.0 m/s. Assume the skater and the ball to be point-like and the surface to be frictionless and ignore air resistance. Use a vertical y-axis with the positive direction pointing up and a horizontal x-axis with the positive direction pointing to the right. What will happen to the skater and the ball after the ball is thrown

Answers

Answer:

[tex]a_2=24.5\ \text{m/s}^2[/tex] towards right

[tex]a_1=2.1875\ \text{m/s}^2[/tex] towards left

Explanation:

[tex]m_1[/tex] = Mass of skater = 56 kg

[tex]m_2[/tex] = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2[/tex]

So, the ball will move towards right with a magnitude of acceleration [tex]a_2=24.5\ \text{m/s}^2[/tex].

The force on the ball will be

[tex]F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}[/tex]

The force on the ball is [tex]122.5\ \text{N}[/tex]

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

[tex]-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2[/tex]

So, the skater will move towards left with a magnitude of acceleration [tex]2.1875\ \text{m/s}^2[/tex]

After the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².

The given parameters

Mass of the skater, m1 = 56 kgMass of the ball, m2 = 5 kgDistance traveled by the ball, d = 1 mSpeed of the ball, v = 7 m/s

The acceleration of the ball as it leaves the skaters hand is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\ v^2 = 0 + 2as\\\\ a = \frac{v^2}{2s} \\\\ a = \frac{7^2}{2(1)} \\\\ a = 24.5 \ m/s^2[/tex]

The force experienced by the ball is calculated as follows;

[tex]F = ma\\\\ F_b = 5 \times 24.5\\\\ F_b = 122.5 \ N[/tex]

The force experienced by the skater is equal in magnitude to that of the ball but opposite in direction.

[tex]F_s = -F_b\\\\ F_s = -122.5 \ N[/tex]

The acceleration of the skater after the ball was thrown is calculated as follows;

[tex]F = ma \\\\ a = \frac{F}{m} \\\\ a = \frac{-122.5 }{56} \\\\ a =-2.19 \ m/s^2[/tex]

Thus, after the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².

Learn more about Newton's third law of motion here: https://brainly.com/question/25998091

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 4.00 kg particle such that the center of mass of the three-particle system has the coordinates (-0.500 m, -0.700 m)

Answers

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ([tex]\vec r_{cm}[/tex]), measured in meters, is defined by this weighted average:

[tex]\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}[/tex] (1)

Where:

[tex]m_{i}[/tex] - Mass of the i-th particle, measured in kilograms.

[tex]\vec r_{i}[/tex] - Location of the i-th particle with respect to origin, measured in meters.

If we know that [tex]\vec r_{cm} = (-0.500\,m,-0.700\,m)[/tex], [tex]m_{1} = 1\,kg[/tex], [tex]\vec r_{1} = (-1.20\,m, 0.500\,m)[/tex], [tex]m_{2} = 4.50\,kg[/tex], [tex]\vec r_{2} = (0.600\,m, -0.750\,m)[/tex] and [tex]m_{3} = 4\,kg[/tex], then the coordinates of the third particle are:

[tex](-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}[/tex]

[tex](-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})[/tex]

[tex](4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)[/tex]

[tex](x_{3},y_{3}) = (-1.562\,m,-0.944\,m)[/tex]

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

"45 meters north" is an example of

Answers

Answer:

Displacement

Explanation:

The quantity 45m north is a typical example of displacement.

Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.

When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacement

In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
A.
BC shows zero acceleration, and AB shows negative acceleration.
OB.
AB shows zero acceleration, and CD shows negative acceleration.
O C.
BC shows zero acceleration, and CD shows negative acceleration.
D.
AB shows zero acceleration, and BC shows negative acceleration.

Answers

Answer:

c

Explanation:

Answer:

c

Explanation:

A constant force FA is applied to an object of mass M, initially at rest. The object moves in the horizontal x-direction, and the force is applied in the same direction. After the force has been applied, the object has a speed of vf. Which mathematical routines can be used to determine the time in which the force is applied to the object of mas

Answers

Answer:

 t = [tex]\frac{ v \ F}{ m}[/tex]

Explanation:

The question is a bit strange, for this exercise we must use the mathematical relationship of Newton's second law to find the acceleration of the body

          F = m a

          a = F / m        (1)

with this acceleration the mathematical relations of kinematics of accelerated motion must be used

          v = v₀ + a t

with the body starting from rest its initial velocity is zero

          v = a t

          t = v / a          (2)

if we substitute the equation 1 in 2

          t = [tex]\frac{ v \ F}{ m}[/tex]

this is the final mathematical expression that allows to find the time based on the data of the problem

A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that is approximated as 0.018 for all speeds. The vehicle is tested unloaded and achieves the theoretical minimum stop in 180 ft (from brake application). The initial speed was 60 mi/h. Ignoring aerodynamic resistance, what is the unloaded braking efficiency

Answers

Answer:

the unloaded braking efficiency is 84.6 %

Explanation:

Given the data in the question;

by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula

S = (Y[tex]_{b}[/tex]( V₁² - V₂²)) / ( 2g( ηbμ + [tex]f_{rl}[/tex] ± sin∅[tex]_{g}[/tex]))

now given that the tracked is levelled, ∅[tex]_{g}[/tex] = 0, also Y[tex]_{b}[/tex] = 1.04 for level or flat road

Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s

now, we substitute in our values to get the braking efficiency;

180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))

180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)

180ft = 8053.76 / ( 0.5152ηb + 1.1592)

180( 0.5152ηb + 1.1592)  = 8053.76

( 0.5152ηb + 1.1592) = 8053.76 /180

0.5152ηb + 1.1592 = 44.7431

0.5152ηb = 44.7431 - 1.1592

0.5152ηb = 43.5839

ηb = 43.5839 / 0.5152

ηb = 84.596 ≈ 84.6 %

Therefore,  the unloaded braking efficiency is 84.6 %

A boat is cruising in a straight line at a constant speed of 2.5 m/s when it is shifted into neutral. After coasting 14 mm the engine is engaged again, and the boat resumes cruising at the reduced constant speed of 1.5 m/s. Assuming constant acceleration while coasting,

a. How much time did it take for the boat to coast the 12 m?
b. What was the boat’s acceleration while it was coasting?
c. What was the speed of the boat when it had coasted for 6.0 m? Explain.

Answers

Answer:

Explanation:

a ) distance s = 12 m .

constant speed = 1.5 m/s

time taken to coast 12 m

= distance / speed

= 12 / 1.5 = 8 s

b ) initial velocity u = 2.5 m /s

final velocity v = 1.5 m /s

displacement s = 14 m

acceleration a = ?

v² = u² + 2 as

1.5² = 2.5² + 2 a x 14

a = - .1428 m /s²

= - 14.28 cm / s²

c )

v = ?

u = 2.5 m /s

s = 6 m

a = -   .1428 m /s²

v² = u² - 2 as

= 2.5² - 2 x 6 x .1428

= 6.25 - 1.71

= 4.54

v = 2.13 m /s .

g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament

Answers

Answer:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

What is a work out time setting? (Gym)

Answers

Answer:

It determines how long you do a certain workout.

A beaker is filled with water to the rim. Gently you place a toy duck in the beaker and some of the water spills out. The duck floats. The weight of the beaker with the duck floating in it is Group of answer choices less than its weight before adding the duck. the same as its weight before adding the duck. greater or less than its weight before adding the duck, depending on the weight of the duck. greater than its weight before adding the duck.

Answers

Answer:

Explanation:

The same as its weight before adding the duck

Lenz’s Law allows us to find _______.

the direction of the induced current.
the magnitude of the induced emf.
the direction of the induced emf.
the magnitude of the induced current.

Answers

Answer:

Explanation:

a

If you carry a box across the floor, you don't do work. Why not?

Answers

The force exerted by the person is an upward force equal to the weight of the box, and that force is perpendicular to the motion. If there is no motion in the direction of the force, then no work is done by that force. Yet you certainly feel like you are doing work if you carry a heavy box.

In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force. Energy is required to do work and the basic SI unit of energy is the joule, the amount of energy required to exert a force of 1 Newton through a distance of 1 meter (1 joule = 1 newton meter).*picture 1*

For the special case of a constant force, the work may be calculated by multiplying the distance times the component of force which acts in the direction of motion.*picture 2*

In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force.*picture 3*

For a constant force F which moves an object in a straight line from x1 to x2 , the work done by the force can be visualized as the area enclosed under the force line below*picture 4*

For the more general case of a variable force F(x) which is a function of x, the work is still the area under the force curve, and the work expression becomes an integral. *picture 5*

credits-goo.gle,hyper physics

It would be impossible to build a microscope that could use visible light to see the molecular structure of a crystal because. It would be impossible to build a microscope that could use visible light to see the molecular structure of a crystal because. lenses cannot be ground with fine enough precision. diffraction limits the resolving power to about the size of the wavelength of the light used. lenses with enough magnification cannot be made. lenses cannot be placed in the correct place with enough precision. More than one of the above is correct.

Answers

Answer:

the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used

Explanation:

To be able to solve two structures with a light source, the Rayleigh criterion must be met that stable the two structures are solved when the first minimum of diffraction at one point is in the code of the first maximum of the other point

Using this criterion we can find an expression for the first minimization of the diffraction spectrum m = 1

     sin θ tea = λ / a

now the structure of the comatose has a separation of around 1 nm and the wavelength of visible light ranges from 400 to 700 nm, when substituting we find

           sin θ = 400/1 10

           sin θ  = 400

           sin θ  = 700/1

           sin θ  = 700

These values ​​are neither impossible since the sin function is bounded between -1 to 1, so we cannot see the diffraction

When reviewing the different statements, the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used:

It would be impossible to build a microscope that could use visible light to

see the molecular structure of a crystal because diffraction limits the

resolving power to about the size of the wavelength of the light used.

Diffraction occurs when waves encounter an obstacle which leads to it

bending around it. Visible light when diffracted results in the overlapping of

the patterns thereby formation of images as one instead of in different parts.

This thereby leads to the resolving power being limited to the size of the

wavelength of light used and images won't be able to be identified

separately.

Read more about diffraction here https://brainly.com/question/16749356

Two trains, each having a speed of 22 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 51 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide

Answers

Answer:

the total distance the bird travels before the trains collide is 69.54 km

Explanation:

Given the data in the question,

dA = 22km/h × t

dB = 51km - 22km/h × t

now we find the time, when the two trains collide

dA = Db

22km/h × t =  51km - 22km/h × t

44km/h × t =  51km

t = 51km / 44km/h

t = 1.159 hrs

so the bird can fly back and forth for 1.159 hrs  before the train collide.

hence, distance travelled by the bird in total will be;

d = v × t

we substitute

d = 60 km/h × 1.159 h

d = 69.54 km

Therefore, the total distance the bird travels before the trains collide is 69.54 km

A magnifying glass uses a convex lens of focal length 6.25 cm. When it is held 5.20 cm in front of an object, what is the image distance?
(Mind your minus signs)
(Unit=cm)

Answers

Answer:

The answer is -30.95.

Explanation:

Use the lens equation: 1/focal length = 1/object distance + 1/image distance. The answer comes out to -30.95. This is correct on Acellus.

When it is held 5.20 cm in front of an object, the image distance will be "-30.95 cm". To understand the calcultaion, check below.

Convex lens

According to the question,

Object distance, u = -5.20 cm

Focal length, f = 6.25 cm

By using the Lens formula, we get

→ [tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u}[/tex]

or,

→ [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]

By substituting the values, we get

     [tex]= \frac{1}{6.25} - \frac{1}{5.20}[/tex]

  [tex]\frac{1}{v} = -\frac{21}{650}[/tex]

By applying cross-multiplication,

   v = -30.95 cm

Thus the above answer is correct.  

Find out more information about Convex lens here:

https://brainly.com/question/13102253

A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8.11 m below its highest point?

Answers

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

Match each letter to the description


A woman walks in a

straight line with the sun

to her right at six o'clock in

the morning. Toward

which pole does she walk?


A line of latitude.


A man walks in a straight

line with the sun to his

right at six o'clock in the

evening. Toward which

pole does he walk?


A line of longitude.

Answers

A woman walks in a  straight line with the sun to her right at six o'clock in the morning.

The sun rises East of her, so the woman is walking toward the North pole.

A man walks in a straight line with the sun to his right at six o'clock in the  evening.

The sun sets West of him, so the man is walking toward the South pole.

The woman and the man are both walking along lines of constant longitude.

2.19 The drag characteristics of a blimp traveling at 4 m/s are to be studied by experiments in a water tunnel. The prototype is 20 m in diameter and 110 m long. The model is one-twentieth scale. What velocity must the model have for dynamic similarity

Answers

Answer:

[tex]Vm=0.894m/s[/tex]

Explanation:

From the question we are told that

Velocity if travel [tex]v=4m/s[/tex]

Diameter of  prototype [tex]d_1=20m[/tex] and [tex]d_2=110m[/tex]

Scale ratio=[tex]\frac{1}{20}[/tex]

Generally Velocity of of the model using Froud's model is mathematically given as

[tex]Fm=Fp[/tex]

[tex]\frac{Vm}{\sqrt{Lmg}} =\frac{Vp}{\sqrt{Lpg}}[/tex]

[tex]Vm=Vp*\frac{Vp}{\sqrt{Lpg} }[/tex]

[tex]Vm=4*\frac{1}{\sqrt{20}}[/tex]

[tex]Vm=0.894m/s[/tex]

Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment

Answers

Answer:

a delta

Explanation:

The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.

A delta is a depositional landform where a smaller body of water enters into a larger one.

The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.

So, this feature is a delta.  

A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.

Answers

Answer:

49 Ns

Explanation:

Given data

Force= 14N

time = 3.5seconds

Applying the expression for impulse

P= Ft

substitute

P=14*3.5

P=49 Ns

Hence the impulse is 49 Ns

If a person weighs 140 lb'on Earth, their mass in kilograms is

Answers

Answer:

70 kg

Explanation:

divide it by 2

Hope this helped!

Answer:

63.502932 Kilograms

Explanation:

a lamp with flux 1400 lm, has an intensity of

a.) 1400 cd
b.) 100 cd
c.) 1000 cd
d.) 111 cd

Answers

I think it’s c sorry if I’m wrong

Leticia leaves the grocery store And walks 150 M’s to parking lot then she turns 90° to the right and walks an additional 70 M’s to her car what is the magnitude of displacement of her car from the grocery store at exit

Answers

Hi my name is licia and thx you so much for the free points

Answer:

Explanation:

Its 165.5m

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