If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​

If R = 20 , What Is The Equivalent Resistance Between Points A And B In The Figure?

Answers

Answer 1

Answer:

Option C. 70 Ω

Explanation:

Data obtained from the question include:

Resistor (R) = 20 Ω

From diagram given ABOVE, we observed the following

1. R and R are in parallel connections.

2. 2R and 2R are in parallel connections.

3. 4R and 4R are in parallel connections.

Next, we shall determine the equivalent resistance in each case.

This is illustrated below:

1. Determination of the equivalent resistance for R and R parallel connections.

R = 20 Ω

Equivalent R = (R×R) /(R+R)

Equivalent R = (20 × 20) /(20 + 20)

Equivalent R = 400/40

Equivalent R = 10 Ω

2. Determination of the equivalent resistance for 2R and 2R parallel connections.

R = 20 Ω

2R = 2 × 20 = 40 Ω

Equivalent 2R = (2R×2R) /(2R+2R)

Equivalent 2R = (40 × 40) /(40 + 40)

Equivalent 2R = 1600/80

Equivalent 2R = 20 Ω

3. Determination of the equivalent resistance for 4R and 4R parallel connections.

R = 20 Ω

4R = 4 × 20 = 80 Ω

Equivalent 4R = (4R×4R) /(4R+4R)

Equivalent 4R = (80 × 80) /(80 + 80)

Equivalent 4R = 6400/160

Equivalent 4R = 40 Ω

Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:

Equivalent of R = 10 Ω

Equivalent of 2R = 20 Ω

Equivalent of 4R = 40 Ω

Equivalent =?

Equivalent = Equivalent of (R + 2R + 4R)

Equivalent = 10 + 20 + 40

Equivalent = 70 Ω

Therefore, the equivalent resistance between point A and B is 70 Ω.


Related Questions

When you squeeze together the coils of a spring and then release them, you are creating a ____ wave? transverse compressional water seismic

Answers

Answer:

Hey there!

Because you are compressing the spring, these are compressional waves.

Let me know if this helps :)

If we use to construct the latches on the windows and doors, then the magnetism will keep thee latches secure.

a. True
b. False

Answers

Answer:

a. True

Explanation:

I'll assume the question is about magnetic latches and locks.

Magnetic door locks use an electromagnetic force to stop doors from opening, so they are ideal for security. There are two main types of electric locking devices. Locking devices can either be a fail-secure locking device that remains locked when power is lost, or a fail-safe locking device that is unlocked when de-energized. An electromagnetic lock creates a magnetic field when energized or powered up, this causes an electromagnet and armature plate to become attracted to each other strongly enough to keep a door from opening.

Which of the following elements tend to be radioactiv
Check all that apply.
A. Ne
B. Am
C. U
D. K

Answers

A . Ne

B. Am

D. K

Hopes this helps!!

The 60.0 g mass, attached to a light spring with a 40.0 N/m force constant, vibrates with an amplitude of 5.00 cm on a horizontal, frictionless plane. find (a) the total energy of the pulsating System, (b) the speed of mass when the displacement is 2.00 cm. When the displacement is 2.50 cm, (c) the kinetic energy and (d) the potential energy

Answers

Explanation:

(a) The total energy is the elastic potential energy at maximum displacement.

E = ½ kx²

E = ½ (40.0 N/m) (0.0500 m)²

E = 0.05 J

(b) At this displacement, there is both elastic potential energy and kinetic energy.

E = EE + KE

0.05 J = ½ kx² + ½ mv²

0.05 J = ½ (40.0 N/m) (0.0200 m)² + ½ (0.060 kg) v²

v = 1.18 m/s

(c) E = EE + KE

0.05 J = ½ kx² + ½ mv²

0.05 J = ½ (40.0 N/m) (0.0250 m)² + KE

KE = 0.0375 J

(d) EE = ½ kx²

EE = ½ (40.0 N/m) (0.0250 m)²

EE = 0.0125 J

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por qué? b) Con m 0.500 kg, k 150 N/m y A 0.150 m, calcule la magnitud de la fuerza sobre la masa y la aceleración de la masa en x 0, 0.050 m y 0.150 m.

Answers

Answer:

a) the correct answer is 1 , b) x=0   F=0, a=0

x= 0.050    F= -7.5 N,  a= -15 m/s²

x= 0.150     F= 22.5 N,  a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

          F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values ​​are from the equilibrium position of the spring.

Let's calculate the force

x = 0

      F = 0

x = 0.050

      F = - 150 0.050

      F = - 7.50 N

x = 0.150

      F = - 150 0.150

      F = - 22.5 N

let's use Newton's second law to find the acceleration

      F = m a

      a = F / m

x = 0 m

      a = 0

x = 0.050 m

      a = -7.50 / 0.50

      a = - 15 m / s²

x = 0.150 m

      a = - 22.5 / 0.50

      a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte  la fuerza es dada por la fuerza de Hoke,  

          F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es  1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza  

x=0  

              F= 0

x=0.050  

              F = - 150 0.050

              F= - 7.50 N

x= 0.150  

                F= - 150 0.150

                F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

          F = m a

          a = F/m

x =0  m

        a = 0

x= 0.050 m

         a = -7.50/ 0.50

          a =- 15 m/s²

x= 0.150 m

          a= - 22.5 / 0.50

          a= - 45 m/s²

Water flowing through a 1.6-cm-diameter pipe can fill a 500 L bathtub in 6.5 min.What is the speed of the water in the pipe?

Answers

Answer:

v = 6.38 m/s

Explanation:

discharge Q = 500 L / 6.5 min

flow through pipe is 1.6 cm x (1 m/100 cm ) = 0.016 m

using the flow rate formula Q = A * v

where A = area, v = velocity

the speed of water in the pipe = v = Q / A

      500 L        1 m³          1 min           π (0.016 m)²

v = ----------   x  --------  x ------------    ÷  --------------------

      6.5 min       10³ L       60 sec                 4

v = 6.38 m/s

The speed of the water is "12.95 m/s".

According to the question,

Volume,

V = 500 L

           = [tex]500\times 10^{-3} \ m^3[/tex]

Time,

t = 3.2 min

          = [tex]3.2\times 60[/tex]

          = [tex]192 \ s[/tex]

Diameter,

d = 1.6 cm

           = 0.016 m

As we know,

Area of cross-section of pipe will be:

→ [tex]A = \pi (\frac{d}{2} )^2[/tex]

      [tex]= 3.14\times (\frac{0.016}{2} )^2[/tex]

      [tex]= 2.01\times 10^{-4} \ m^2[/tex]

Flow rate is:

→ [tex]\frac{V}{t} = Av[/tex]

   [tex]v = \frac{V}{At}[/tex]

      [tex]= \frac{500\times 10^{-3}}{2.01\times 10^{-4}\times 192}[/tex]

      [tex]= 12.95 \ m/s[/tex]

Thus the above approach is right.          

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Which of the following is not an example of matter?
A. sand
B. heat from a fire,
C. helium in a balloon
D. fog paint on a canvas

Answers

Heat from fire is the correct answer

Answer:

B. heat from fire

Explanation:

It is not an example of matter because heat is energy and not matter

Select True or False for the following statements about Heisenberg's Uncertainty Principle.
A) It is not possible to measure simultaneously the x and z positions of a particle exactly.
B) It is possible to measure simultaneously the x and y momentum components of a particle exactly.
C) It is possible to measure simultaneously the y position and the y momentum component of a particle exactly.

Answers

Answer:

A and B are true C is false

Explanation:

Because it states that if we know everything about where a particle is located (the uncertainty of position is small), we know nothing about its momentum (the uncertainty of momentum is large), and vice versa.

So in A we can know the positions of two objects

In B we can measure the momentum at two different places

While in C we cannot measure both the position and momentum of y accurately

Drag and drop each description into the correct category
They copy the original sound
Analog signals
Digital signals
They change sound into numbers.
They reduce unwanted noise.
They may pick up unwanted noise.
They flow continuously.
They do not flow continuously
Intro
Done

Answers

Explanation:

Analog signals :

Analog signals are continuous signal. These are like wave signals that change with time period.The main disadvantage of analog signal is that it contains noise. .They flow continuously.

Digital signals :

Digital signals do not flow continuously. It has finite number of values. They form in the form of numbers. They reduce unwanted noise. It is the advantage of digital signals.

Answer:

Analog -

They flow continuously

They may pick up unwanted noise

They copy the original sound

Digital -

They do not flow continuously

They reduce unwanted noise.

They change sound into numbers.

Mark Brainliest Please

(b) A piece of wood of volume 0.6 m² floats in water. Find the volume
exposed. What force is required to immerse it completely under water?
(Density of wood = 600 kg/m3, water = 1000 kg/m3)
[8]​

Answers

Answer:

Explanation:

Let the volume below water be v . Then

buoyant force = v d g where d is density of water , g is acceleration due to gravity

= v x 1000 x g

weight of wood piece = volume x density of wood x g

= .6 x 600 x g

for equilibrium while floating

buoyant force = weight

= v x 1000 x g  =  .6 x 600 x g

v = .36 m²

volume above water or volume exposed = .6 - .36

= .24 m²

When immersed completely ,

buoyant force = .6 x 1000 x 9.8

= 5880 N

weight of wood

=  .6 x 600 x g

= 3528 N

buoyant force is more than the weight . In order to equalise them for floating with full volume in water

weight required = 5880 - 3528

= 2352 N.

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.30 m before stopping. How far does the lighter fragment slide

Answers

Answer:

the distance d traveled by the lighter fragment is 58.1 m.

Explanation:

mass of the lighter fragment = m

the lighter fragment traveled a distance = ?

mass of the heavier fragment = 7m

the distance covered by the heavier fragment = 8.30 m

The two particles will be given the same amount of energy from the explosion. This energy is used to do work by the two fragments.

work done by heavier fragment w = mgd

where m is the mass

g is acceleration due to gravity

d is the distance traveled.

substituting, the work done by the heavier fragment is

w = 7m x g x 8.3 = 58.1mg

The same way, the lighter fragment does work of

w = mgd

equating the two work done since they are given the same amount of energy from the explosion, we have

58.1mg = mgd

mg cancels out, we have

the distance d traveled by the lighter fragment d = 58.1 m

Gauss’s law applies to:_____________
a. lines.
b. flat surfaces.
c. spheres only.
d. closed surfaces.

Answers

Answer:

D. Closed Surfaces

The gaussian surface could be a sphere and can be applied to flat surfaces, but it's always a closed surface.

the lenses in a students eyes have arefractive power of 52. 0 diopters when she is able to focus on the board if the distance between the ey lens and the retina is 2.00 cm find how

Answers

Answer:

     p = 49.95 cm

This is the distance from the student to the stepped, in order to be able to reach this distance they must sit in the first row

Explanation:

In medicine it is very common to express the potential visual corrections that is

         P = 1 / f

where P is the power and f the focal length in meters

In this exercises give the power, let's find the focal length

        f = 1 / p

        f = 1/52

        f = 0.01923 m = 1.923 cm

For geometrical optics calculations the most used equation is the constructor equation

        1 / f = 1 / p + 1 / q

Where p and q are the distance to the object and the image, respectively

To be able to see an object clearly, its image must be on the retina,

       q = 2.00 cm

find the distance to the object

        1 / p = 1 / f - 1 / q

        1 / p = 1 / 1,923 - 1/2.00

        1 / p = 0.02002

         p = 49.95 cm

This is the distance from the student to the stepped, in order to be able to reach this distance they must sit in the first row

The coil has a radius of 8 cm and 110 turns. While moving the magnet closer towards the coil, at a certain point in time, the voltage meter shows -310 mV. What is the magnitude of the rate of change of the flux through the coil?

Answers

Answer:

The magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s

Explanation:

Given;

radius of coil, r = 8 cm = 0.08 m

number of turns of the coil, N = 110 turns

the induced emf through the coil, E = -310mV

The induced emf through the coil is given by;

[tex]E = - N\frac{d \phi}{dt} \\\\\frac{d \phi}{dt} = \frac{E}{-N}[/tex]

Where;

dФ/dt is the magnitude of the rate of change of the flux through the coil

[tex]\frac{d\phi}{dt} = \frac{-0.31}{110} \\\\\frac{d\phi}{dt} =2.82*10^{-3} \ T.m^2/s[/tex]

Therefore, the magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s

If you wish to detect details of the size of atoms (about 1 ✕ 10−10 m) with electromagnetic radiation, it must have a wavelength of about this size.
(a) What is its frequency?
(b) What type of electromagnetic radiation might this be?

Answers

Answer:

a) 3×10^18 Hz

b) The electromagnetic wave is an x-ray

Explanation:

We know that the speed of an electromagnetic wave is given by;

c= λf

Where;

c= speed of electromagnetic waves = 3×10^8 ms-1

f= frequency of electromagnetic waves= the unknown

λ = wavelength of the electromagnetic wave= 1 ×10^-10 m

Hence;

f = c/λ= 3×10^8/ 1×10^-10

f= 3×10^18 Hz

b) The electromagnetic wave is an x-ray.

(a) The frequency of the electromagnetic radiation is [tex]3\times 10^{18} \ Hz[/tex]

(b)  The electromagnetic radiation is ultra violet ray.

The given parameters;

distance of the atom, λ = 1 x 10⁻¹⁰ m

The frequency of the electromagnetic radiation is calculated as follows;

[tex]c =f \lambda \\\\f = \frac{c}{\lambda} \\\\f = \frac{3\times 10^8}{1\times 10^{-10}} \\\\f = 3\times 10^{18} \ Hz[/tex]

The frequency range of ultra violet ray is between 10¹⁴ to 10¹⁸ m.

Thus, the electromagnetic radiation is ultra violet ray.

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Assuming two hypothetical maps that each cover a standard 8.5 by 11-inch sheet of paper, the larger-scale map would cover a larger geographic area than the smaller-scale map.
a) true
b) false

Answers

Answer:

b) false

Explanation:

The scale of a map is the ratio of a distance on the map to the corresponding distance on the ground. Scaling allow us to capture a large geographical area on a reduced platform while still retaining the relative sizes and positioning of places on the map to their real life sizes and positioning. If both maps cover a standard 8.5 by 11-inch sheet of paper, then the map with the smaller ratio will have the bigger geographical area.

To understand better, let us assume two geographical areas A and B. A is bigger than B. If we were to put them both on the same area of map paper, then we'll have to scale up the smaller geographical area B so as to fit into the map paper. This means that the geographical area with the smaller area B will have the larger scale on the map.

An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
A: What is the force exerted on the electron?
B: How strong is the electric field at the electron’s location?
C: How much work would be done on the electron if it was moved so that it’s .001m away from the sphere?
D: Now replace the electron with a positron (q=+1.602×10-19C). Explain in your own words how that would affect the results in parts A, B, and C.

Answers

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

A boat sailing against the current experience a negative acceleration of 11 m/s^2. If the boat's initial velocity is 44m/s upstream, how long until it comes to a stop?

Answers

Answer:

4 s

Explanation:

Given:

v₀ = 44 m/s

v = 0 m/s

a = -11 m/s²

Find: t

v = at + v₀

0 m/s = (-11 m/s²) t + 44 m/s

t = 4 s

If the boat's initial velocity is 44m/s upstream, the time it takes to stop is found to be 4 seconds.

What is Acceleration?

Acceleration may be defined as the rate of change of velocity with respect to time. It is a vector quantity as it has both magnitude and direction. It is also the second derivative of position and the first derivative of velocity with respect to time.

According to the question,

v₀ = 44 m/s

v = 0 m/s

a = -11 m/s²

Now, you have to calculate the time, t.

So, you calculate the time, t with the help of the following formula:

v = at + v₀

       0 m/s = (-11 m/s²) t + 44 m/s

        t = 4 s.

Therefore, if the boat's initial velocity is 44m/s upstream, the time it takes to stop is found to be 4 seconds.

To learn more about Acceleration, refer to the link:

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g Let the orbital radius of a planet be R and let the orbital period of the planet be T. What quantity is constant for all planets orbiting the sun, assuming circular orbits?

Answers

Explanation:

Kepler's third law gives the relationship between the orbital radius and the orbital period of the planet. Its mathematical form is given by :

[tex]T^2=\dfrac{4\pi ^2}{GM}a^3[/tex]

Here,

G is gravitational constant

M is mass of sun

It means that the mass of Sun is constant for all planets orbiting the sun, assuming circular orbits.

Luz, who is skydiving, is traveling at terminal velocity with her body parallel to the ground. She then changes her body position to feet first toward the ground. What happens to her motion? She will continue to fall at the same terminal velocity because gravity has not changed. She will slow down because the air resistance will increase and be greater than gravity. She will speed up because air resistance will decrease and be less than gravity. She will begin to fall in free fall because she will have no air resistance acting on her.

Answers

Answer:

She will speed up because air resistance will decrease and be less than gravity.

Explanation:

While Luz was falling initially, she was experiencing a gravitational force downwards, and air resistance that arise from the drag force on her body

Her gravitational force downwards is constant, and she fall down with a net force of

[tex]F_{net}[/tex] = [tex]F_{g}[/tex] - [tex]F_{d}[/tex]

where [tex]F_{net}[/tex] is the net force on Luz downwards

[tex]F_{g}[/tex] is the gravitational force on Luz

[tex]F_{d}[/tex] is the drag force on Luz

The drag force on on Luz is proportional to her attack surface area.  When Luz changes her body position from her frontal area, parallel to the ground to falling with her feet first, she reduces the area available for drag force from her whole frontal area to just about a little more than the areas of the sole of her feet. This action reduces the drag force due to air resistance on her body.

Answer:

The correct answer to the question is the third statement.

Explanation:

She will speed up because air resistance will decrease and be less than gravity. Changing position during skydiving is one of the factors affecting the speed. Since the cross section area is smaller than the first position, she experiences lesser air resistance, which causes her to speed up.

How much work did the movers do (horizontally) pushing a 46.0-kgkg crate 10.5 mm across a rough floor without acceleration, if the effective coefficient of friction was 0.50

Answers

Answer:

7.1 J

Explanation:

From the question,

Work done by the mover  = work done in pushing the crate + work done against friction

W = W'+Wf................. Equation 1

W = mgd+mgμd............ Equation 2

W = mgd(1+μ)................ Equation 3

Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of  friction.

Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5

constant: g = 9.8 m/s²

Substitute these values into equation 3

W = 46×9.8×0.0105(1+0.5)

W = 7.1 J

A flatbed truck is carrying a 20-kg crate up a sloping road. The coefficient of static friction between the crate and the bed is 0.40, and the coefficient of kinetic friction is 0.30. What is the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place

Answers

Answer:

The angle is  [tex]\theta =21.8 ^o[/tex]

Explanation:

From the question we are told that

   The mass of the crate is  [tex]m_c = 20 \ kg[/tex]

   The coefficient of static friction is  [tex]\mu_s = 0.40[/tex]

   The  coefficient of kinetic friction is  [tex]\mu_k = 0.30[/tex]

Generally  for the the crate not to slip , the static frictional  must be equal to the force driving the truck

   i.e

           [tex]F_f = F[/tex]

Now since we are considering a slope that static frictional force is mathematically represented as

            [tex]F_f = mg * cos(\theta) * \mu_s[/tex]

While the force driving the truck is mathematically represented as

          [tex]F = mg * sin (\theta )[/tex]

Here  mg is the weight of the crate  so

         So

               [tex]mg * cos (\theta ) \mu_s = mg * sin (\theta )[/tex]

=>             [tex]\frac{sin (\theta )}{cos (\theta)} = \mu_s[/tex]

=>          [tex]\theta = tan ^{-1} [\mu_s ][/tex]

=>          [tex]\theta = tan ^{-1} [0.40 ][/tex]

=>          [tex]\theta =21.8 ^o[/tex]

A diagram show an illustration is on the first uploaded image

Pls answer 9 Through 12

Answers

Answer:

physics,chemistry,biology,astronomy and earht sciences

principles of science are integrity of knowledge honesty,collegiality(cooperation between colleagues) objectivity and openness

examples of science are actually branches of science so those are biology,mathematics,chemistry and physics etc

science is importent to provide our basic needs and to improve our living standard.It makes our life much easier by providing different technologies.It helps in the diagnosing and treatment of a disease

Two monatomic ideal gases are in thermal equilibrium with each other. Gas A is composed of molecules with mass m while gas B is composed of molecules with mass 4m. The ratio of the average molecular kinetic energy KA / KB is

Answers

Answer: [tex]\frac{K_{A}}{K_{B}}[/tex] = 1

Explanation: Average molecular kinetic energy for monatomic idela gases is given by:

[tex]E=\frac{3}{2}RT[/tex]

where

R is gas constant of ideal gas

T is temperature

Which means kinetic energy depnends only on temperature.

Since gas A and gas B are at the same temperature, kinetic energy will be the same. Therefore:

[tex]\frac{K_{A}}{K_{B}}[/tex] = 1

Ratio of this average molecular kinetic energy is 1.

A 817 kg car has four 8.91 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles

Answers

Answer:

0.0107

Explanation:

We know that

The rotational kinetic energy due to four wheel is

1/2ဃ²I x 4

So

1/4mR²(v/R)² = mv²

But kinetic energy along straight path of the car is 1/2mv²

=> 1/2( 817)v ²

Kc= 408.5v²

So The fraction of total kinetic energy that is due to rotation of the wheel about their axis

Is Kw/Kw+Kc

and Kw = 1/2* 8.91v²= 4.45v²

So 4.45v²/ 4.45v²+ 408.5v²

= 0.0107 as fraction of total kinetic energy

If an ancient campfire were analyzed, was found to have only about one-eighth the carbon-14 that is normally found in living things, how long ago was that campfire extinguished? Answer in units of y.

Answers

Answer:

17047.54 years or 17048 y

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2=half-life of the carbon-14 =5670 years

t= time elapsed

Ao= activity of living C-14

A= activity if the sample under study

But A= 1/8 Ao

Hence;

0.693/5670= 2.303/t log(Ao/1/8Ao)

1.22×10^-4 = 2.303/t log 8

1.22×10^-4 = 2.0798/t

t= 2.0798/1.22×10^-4

t= 17047.54 years

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the bob's mass is doubled, approximately what will the pendulum's new period be?
A: T/2
B: T
C: sqrt(2)*T
D: 2T

Answers

Answer:

B. T

Explanation:

This is because the mass of the Bob does no relate to the period as given by the relationship

T = 2π(√L/g) so double mass is still T

Answer:

The correct option is (B).

Explanation:

The time period of a pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is the length of the pendulum

g is acceleration due to gravity

It is very clear that the time period is independent of the mass of the bob. If the mass of the bob is doubled, the time period of the pendulum remains the same i.e. T. Hence, the correct option is (B).

Allowing all but one allows for accurate results

Answers

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From what maximum height could a 75 kg person jump and land rigidly upright on both feet without breaking their legs

Answers

Answer:

h is 0.27metres without breaking their leg

Explanation:

A body jumping from any heights has potential energy mgh and maximum energy which can be converted by human body from the jump is 200J derived from experiment so using

200J = mgh

h = 200/ 9.8* 75

h= 0.27 m

Explanation:

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron would the proton have to be? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C, mproton = 1.67 × 10-27 kg, melectron = 9.11 × 10-31 kg)

Answers

Answer:

r = 5.08 m

Explanation:

The electric force of attraction or repulsion is given by :

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.

So, the force from the proton is balanced by the mass of the electron.

[tex]\dfrac{kq_pq_e}{r^2}=mg[/tex]

r is distance

[tex]r=\sqrt{\dfrac{kq_pq_e}{mg}} \\\\r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times 9.8}} \\\\r=5.08\ m[/tex]

So, proton have to be at a distance of 5.08 meters above the electron.

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