Pendulum clock. Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

Answers

Answer 1

Answer:

The time period for this pendulum is 1.68 seconds

Explanation:

Solution

Given that:

The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum

Now,

In this case, the length becomes:

L= 80 - 15+5

L = 70 cm

The time period = T = 2π √L/g

T = 2* 3.14 *√0.7/9.8

= 1.68 seconds

Note: Kindly find an attached work to the part of the solution of the given question

Pendulum Clock. Your Friend Is Trying To Construct A Clock For A Craft Show And Asks You For Some Advice.

Related Questions


What is an independent variable?
A. A variable that is intentionally changed during an experiment
B. A variable that depends on the experimental variable
C. A variable that is not used in an experiment
D. A variable that is unknown during the experiment

Answers

Answer:

The answer is A

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent

Answer:

A.

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent.

Match these items.


1 . pls help


asteroids

between Mars and Jupiter

2 .

fission

ice, dust, frozen gases

3 .

energy

sun's atmosphere

4 .

fusion

ability to do work

5 .

corona

splitting atoms

6 .

comets

the combining of atomic nuclei to form one nucleus

Answers

Answer:

Here's your answer :

Asteroids - Between mars and JupiterFission - splitting atomsEnergy - Sun's atmosphereFusion - The combining of atomic nuclei to form one nucleusCorona - Ability to do workComets - Ice, dust, frozen gases

hope it helps!

Distributions of electric charges in a cell play a role in moving ions into and out of a cell. In this situation, the motion of the ion is affected by two forces: the electric force due to the non-uniform charge distribution in the cell membrane, and the resistive force (viscosity) due to colliding with the fluid molecules. In order to begin our analysis of this, let's consider a toy model in which the ion is moving in response to electric forces alone.

Charges in a cell membrane are distributed along the opposite sides of the membrane approximately uniformly. This leads to an (on the average) constant electric field inside the membrane. A simple model that gives this kind of field is two large parallel plates close together. The field between the plates is approximately constant pointing from the negative to the parallel plate. This results in a charge feeling a constant force anywhere between the plates (sort of like flat-earth gravity turned sideways). Outside of the plates the electric fields from the two plates cancel and there is no force.

2. The electric field between the plates (inside the membrane) is about 107 N/C and the thickness of the membrane is about 7 nm. Estimate:

2.1 The electric force on the ion when it is in the center of the channel.
F = N

Explain your reasoning.



2.2 The acceleration of the ion when it is in the center of the channel.
a = nm/s2
Explain your reasoning.



2.3 The magnitude of the change in the ion's potential energy as it crosses from one side of the plates to the other.
U = J
Explain your reasoning.



2.4 The kinetic energy the ion would gain as it crosses from one side of the plates to the other.
KE = J
Explain your reasoning.

Could you explain 2.3!

Answers

Answer:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.Explanation:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

What is atom?

Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.

Each atom is made up of a nucleus and one or more electrons that are linked to it. One or more protons and a significant number of neutrons make up the nucleus. Only the most prevalent type of hydrogen is neutron-free.

Atoms that are neutral or ionized make up every solid, liquid, gas, and form of plasma. Atoms are incredibly tiny, measuring typically 100 picometers across. The nucleus of an atom contains more than 99.94% of its mass.

Therefore, An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

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In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give your answer in units of Amps. An Amp is 1 Coulomb of charge flowing through a cross-sectional area of the wire per second - that's a lot of charge per second and will warm up a typical wire quite a bit! Most devices have circuits with larger resistors - kLaTeX: \OmegaΩ (103 LaTeX: \OmegaΩ) and MLaTeX: \OmegaΩ (106 LaTeX: \OmegaΩ) are common.

Answers

Answer:

The current pass the [tex]R_2[/tex] is  [tex]I = 0.25 A[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  [tex]V = 3V[/tex]

     The first resistance is  [tex]R_1 = 7 \Omega[/tex]

     The second resistance is  [tex]R_2 = 5 \Omega[/tex]

Since the resistors are connected in series their equivalent resistance is  

       [tex]R_{eq} = R_1 +R_2[/tex]

Substituting values

         [tex]R_{eq} = 7 + 5[/tex]

         [tex]R_{eq} = 12 \Omega[/tex]

Since the resistance are connected in serie the current passing through the circuit  is the same current passing through [tex]R_2[/tex] which is mathematically evaluated as

        [tex]I = \frac{V}{R_{eq}}[/tex]

Substituting values  

      [tex]I = \frac{3}{12}[/tex]

      [tex]I = 0.25 A[/tex]

The uniform slender bar of mass m and length l is released from rest in the vertical position and pivots on its square end about the corner at O. (a) If the bar is observed to slip when   30 , find the coefficient of static friction s between the bar and the corner. (b)If the end of the bar is notched so that it cannot slip, find the angle  at which contact between the bar and the corner ceases.

Answers

Answer:

A) 0.188

B) 53.1 ⁰

Explanation:

taking moment about 0

∑ Mo = Lo∝

mg 1/2 sin∅ = 1/3 m L^2∝

note ∝ = w[tex]\frac{dw}{d}[/tex]∅

forces acting along t-direction ( ASSUMED t direction)

∑ Ft = Ma(t) = mr∝

mg sin ∅ - F = m* 1/2 * 3g/2l sin∅

therefore F = mg/4 sin∅

forces acting along n - direction ( ASSUMED n direction)

∑ Fn = ma(n) = mr([tex]w^{2}[/tex])

= mg cos∅ - N = m*1/2*3g/1 ( 1 - cos∅ )

hence N = mg/2 ( 5cos∅ -3 )

A ) Angle given = 30⁰c find coefficient of static friction

∪ = F/N

  = [tex]\frac{\frac{mg}{4}sin30 }{\frac{mg}{2}(5cos30 -3) }[/tex]  = 0.188

B) when there is no slip

N = O

   = 5 cos ∅ -3 =0

   therefore cos ∅ = 3/5  hence ∅ = 53.1⁰

Which term defines the distance from crest to crest

Answers

Answer:

The horizontal distance between two adjacent crests or troughs is known as the wavelength.

Answer: Wavelength

Explanation:

From crest to crest, it is one full wavelength

Under electrostatic conditions, the electric field just outside the surface of any charged conductor

A. is always zero because the electric field is zero inside conductors
B. can have non zero components perpendicular to and parallel to the surface of the conductor
C. is always perpendicular to the surface of the conductor
D. is always parallel to the surface
E. is perpendicular to the surface of the conductor only if it is a sphere, a cylinder, or a flat sheet.

Answers

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly  distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

Which of these parameters is directly related to sound frequency?

Answers

Answer:Velocity

Explanation:

Velocity is directly proportional to the frequency of a wave.

Velocity=frequency x wavelength

Which of the following BEST summarizes the relationship between groups and culture and critical thinking?

Answers

Answer:

Groups and culture helps in influencing our values,ethics and beliefs. This influence should always be questioned through the process of thinking critically.

This best summarizes the relationship between groups and culture and critical thinking.

An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0.260t2, where t is in seconds and θ is in radians. When t = 2.40 s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?

Answers

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

[tex]\theta=0.260t^2[/tex]

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]

Then, you evaluate for t=2.40 s:

[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]

(b) The linear velocity is calculated by using the following formula:

[tex]v=\omega r[/tex]

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]

(c) The tangential acceleration is:

[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]

(d) The radial acceleration is:

[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]

plzzz help will mark the brainliest

Answers

Ciara is winging....etc
The answer is : 0.60 N, toward the center of the circle


A satellite....etc
The Answer is : 7400 m/s


What is the .....etc
The Answer is : 2.60 m/s

which one of the following statements is true? A.in an elastic collision,only momentum is conserved B. in any collision,both momentum & kinetic energy are conserved C.in an inelastic collision,both momentum & kinetic energy are conserved D.in an elastic collision,only kinetic energy is conserved ​

Answers

Answer:

option C is correct

................

Answer:

C- in an inelastic collision, both momentum & kinetic energy are conserved

Explanation:

Took the test

In order to get going fast, eagles will use a technique called stooping, in which they dive nearly straight down and tuck in their wings to reduce their surface area. While stooping, a 6- kg golden eagle can reach speeds of up to 53 m/s . While golden eagles are not very vocal, they sometimes make a weak, high-pitched sound. Suppose that while traveling at maximum speed, a golden eagle heads directly towards a pigeon while emitting a sound at 1.1 kHz. The emitted sound has a sound intensity level of 30 dB when heard at a distance of 5 m .A) Model this stooping golden eagle as an object moving at terminal velocity. The eagle’s drag coefficient is 0.5 and the density of air is 1.2 kg/m 3 . What is the effective cross-sectional area of the eagle’s body while stooping?B) What is the doppler-shifted frequency that the pigeon will hear coming from the eagle?C) Consider the moment when the pigeon is 5 m away from the eagle. At the pigeon’s position, what is the intensity (in W/m^2 ) of the sound the eagle makes?D) The golden eagle slams into the 250- g pigeon, which is initially moving at 10 m/s in the opposite direction (toward the eagle). The eagle grabs the pigeon in its talons, and they move off together in a perfectly inelastic collision. How fast do they move after the collision?

Answers

Answer:

Check the explanation

Explanation:

Part A

F = CA

this drag force balances the weight = 6X 9.8

so

6X9.8 = 0.5 X A X0.5 X 1.2 X 532

A= 0.069 m2

Part B

here the sorce is moving and the observer is at rest

so f= f(- 1 - 1

f = 1.1X10 343 343 – 53

f' = 1.3 KHz

Part C:

given the intensity = 30 dB

we know that I dB = 10 log (I(W/m2))

so we get I (W/m2) = 1000

Part D : The catch

Given that U1 = 53 M1 = 6 kg

U2 =-10 M2=0.25

V1=V2

now conserving momentum

6 X 53 -0.25 X10 =(6+0.25)V

V= 50.48 m/sec

Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance ddd is


|F|=K|QQ′|d2|F|=K|QQ′|d2,


where K=14πϵ0K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2)ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.


Consider two point charges located on the x axis: one charge, q1q1q_1 = -15.0 nCnC , is located at x1x1x_1 = -1.660 mm ; the second charge, q2q2q_2 = 34.5 nCnC , is at the origin (x=0.0000)(x=0.0000).


What is the net force exerted by these two charges on a third charge q3q3q_3 = 47.0 nCnC placed between q1q1q_1 and q2q2q_2 at x3x3x_3 = -1.240 mm ?


Your answer may be positive or negative, depending on the direction of the force.

Answers

Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is true about this situation?

Answers

Answer:

Answer: the true statement form the given statements is “the athletes is not doing any work because he does not move weight”

Explanation:

The athlete isn’t doing any work because he doesn’t move the weight is the correct statement.

What is Work? Work is the energy transferred to or from an object via the application of force along a displacement.Work = Force x Displacement.

How to solve this Problem?The weight of an object given is 50kgsThe time of holding an object given is 10 secondsWe need to justify the statements

Here ,

There is no displacement that means displacement is zero.If displacement is zero then work done will also be zero

Hence there is no work done by the athlete

Therefore ,The athlete isn’t doing any work because he doesn’t move the weight is the correct statement

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This is a measure of quantity of matter

Answers

Answer:

Mass

Explanation:

Mass is the measure of amount of matter contained within any substance and hence mass determines the weight. Unit of mass is kilogram as per ISI system of units.  

Mass is measured through a balance. The more is the mass of an object, the more the balance tilts towards the object side.  

Weight is equal to product of mass and the gravitational constant i.e 9.8m/s^2

Is mercury (the planet) rocky or gaseous(meaning relating to or having the characteristics of a gas.)

Answers

Answer:

Mercury is rocky

Explanation:

Answer:

Rocky

Explanation:

It has no atmosphere so it cannot hold gas.

A cobalt-60 source with activity 2.60×10-4 Ci is embedded in a tumor that has
mas 0.20 kg. The source emits gamma photons with average energy 1.25 MeV.
Half the photons are absorbed in the tumor, and half escape.
i. What energy is delivered to the tumor per second? [4 marks]
ii. What absorbed dose, in rad, is delivered per second? [2 marks]
iii. What equivalent dose, in rem, is delivered per second if the RBE for
these gamma rays is 0.70? [2 marks]
Page 6 of 7
iv. What exposure time is required for an equivalent dose of 200 rem? [2
marks]
B. A laser with power output of 2.0 mW at a wavelength of 400 nm is projected
onto a Calcium metal. The binding energy is 2.31 eV.
i. How many electrons per second are ejected? [6 marks]
ii. What power is carried away by the electrons? [4 marks]
C. A hypodermic needle of diameter 1.19 mm and length 50 mm is used to
withdraw blood from a patient? How long would it take for 500 ml of blood to be
taken? Assume a blood viscosity of 0.0027 Pa.s and a pressure in the vein of
1,900 Pa. [10 marks]
D. A person with lymphoma receives a dose of 35 gray in the form of gamma
radiation during a course of radiotherapy. Most of this dose is absorbed in 18
grams of cancerous lymphatic tissue.
i. How much energy is absorbed by the cancerous tissue? [2 marks]
ii. If this treatment consists of five 15-minute sessions per week over the
course of 5 weeks and just one percent of the gamma photons in the
gamma ray beam are absorbed, what is the power of the gamma ray
beam? [4 marks]
iii. If the gamma ray beam consists of just 0.5 percent of the photons
emitted by the gamma source, each of which has an energy of 0.03
MeV, what is the activity, in Curies, of the gamma ray source? [4 marks]
E. A water heater that is connected across the terminals of a 15.0 V power supply
is able to heat 250 ml of water from room temperature of 25°C to boiling point
in 45.0 secs. What is the resistance of the heater? The density of water is 1,000
kg/m2 and the specific heat capacity of water is 4,200 J/kg/°C. [10 marks]

Answers

Answer:

A i. E = 9.62 × 10⁻⁷ J/s

ii. The absorbed dose is 4.81 × 10⁻⁶ Gy

iii. The equivalent dose is  3.37 × 10⁻⁴ rem/s

iv.  t = 593471.81 seconds

B. i. 4.025 × 10¹⁵/s

ii. 0.512 mW

C. 7218092.2 seconds

D. i. 6.3 × 10⁻¹ J

ii. 1.4 × 10⁻² W

iii. 1.57 × 10³ Curie

E. 0.129 Ω

Explanation:

The given parameters are;

Mass of tumor = 0.20 kg

Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci

Photon energy = 1.25 MeV

(i) The energy, E, delivered to the tumor is given by the relation;

[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]

[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]

E = 9.62 × 10⁻⁷ J/s

(ii) The equation for absorbed dose is given as follows;

Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg

Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy

1 Gray = 100 rad

4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s

(iii) Equivalent dose, H, is  given by the relation;

H = D × Radiation factor, [tex]w_R[/tex]

∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s

(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;

[tex]\dot{H} = \dfrac{H}{t}[/tex]

Therefore;

[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]

∴ t = 6.9 days

B. The number of electrons ejected is given by the relation;

[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]

[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]

(ii) The power carried by the electron

The energy carried away by the electrons is given by the relation;

[tex]KE_e = hv - \Phi[/tex]

[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]

[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]

Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW

C. The given parameters are;

d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m

l = 50 mm = 5 × 10⁻³ m

V = 500 ml = 5 × 10⁻⁴ m³

η = 0.0027 Pa

p = 1,900 Pa.

[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]

[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]

[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]

t = 7218092.2 seconds

D) i. Energy absorbed is given by the relation;

E = m×D

Where:

D = 35 Gray = 35 J/kg

m = 18 g = 18 × 10⁻³ kg

∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J

ii. Total time for treatment = 15 × 5 = 75 minutes

Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J

Power = Energy(in Joules)/Time (in seconds)

∴ Power = 63/(75×60) = 1.4 × 10⁻² W

iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;

[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]

1 MeV = 1.60218 × 10⁻¹³ J

0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon

Therefore, the number of disintegration per second = 0.28 J/s ÷  4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second

1 Curie = 3.7 × 10¹⁰  disintegrations per second

Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie

= 1.57 × 10³ Curie

E. The parameters given are;

Density of water = 1000 kg/m³

Volume of water = 250 ml = 0.00025 m³

Initial temperature, T₁, = 25°C

Final temperature, T₂, = 100°C

Change in temperature, ΔT = 100 - 25 = 75°

Specific heat capacity of the water = 4200 J/kg/°C

Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg

∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J

Time to heat the water = 45.0 sec

Therefore, power = Energy/time = 78750/45 = 1750 W

The formula for electrical power = I²R =VI = V²/R

Therefore, where V = 15.0 V, we have;

15²/R = 1750

R = 15²/1750 = 0.129 Ω.

The resistance of the heater = 0.129 Ω.

Air is matter which backs best support the statement

Answers

Answer: A. Balloons can be filled with air.

C. Air has mass.

Explanation:

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Balloons are able to be filled with air and air has mass.

A pipe branches symmetrically into two legs of length L, and the whole system rotates with angular speed ω around its axis of symmetry. Each branch is inclined at angle α to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate Q. The pipe diameter, D, is much smaller than L. Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration ω_ ?

Answers

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on the right has a current of 2 A. We can say that:

A. The left wire attracts the right wire and exerts twice the force as the right wire does.
B. The left wire attracts the right wire and exerts half the force as the right wire does.
C. The left wire attracts the right wire and exerts as much force as the right wire does.
D. The left wire repels the right wire and exerts twice the force as the right wire does.
E. The left wire repels the right wire and exerts half the force as the right wire does.
F. The left wire repels the right wire and exerts as much force as the right wire does.

Answers

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]

Next, you use the formula for the magnetic force produced by the wires:

[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]

Hence, due to this result you have that:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is the coefficient of kinetic friction?
A block is given an initial speed of 3m/s up a 25° incline. Coefficient of friction

Answers

Answer:

a)  μ = 0.475 , b)   μ = 0.433

Explanation:

a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

X axis

     Wₓ - fr = m a

the friction force has the expression

     fr = μ N

y Axis

     N - [tex]W_{y}[/tex] = 0

let's use trigonometry for the components the weight

     sin 27 = Wₓ / W

     Wₓ = W sin 27

     cos 27 = W_{y} / W

     W_{y} = W cos 27

     N = W cos 27

     W sin 27 - μ W cos 27 = m a

     mg sin 27 - μ mg cos 27 = m a

      μ = (g sin 27 - a) / (g cos 27)

      very = tan 27 - a / g sec 27

      μ = 0.510 - 0.0344

      μ = 0.475

b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

         μ = tan 25 - 0.3 / 9.8 sec 25

         μ = 0.466 -0.03378

         μ = 0.433

Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-degrees above the forward direction .Find the magnitude and direction(relative to forward direction of the resultant force that these forces exert on the body)​

Answers

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

[tex]\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N[/tex]

Thus, the net force over the body is:

[tex]F=(913.14N)\hat{i}+(274.87N)\hat{j}[/tex]

Next, you calculate the magnitude of the force:

[tex]F=\sqrt{(913.14N)+(274.87N)^2}=953.61N[/tex]

and the direction is:

[tex]\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°[/tex]

Which of the following is not true about of the use of MRI in medicine?
1) It produces no negative side-effects on the human body
2) It produces high resolution images of soft tissues
3) It is very cheap
4) It requires very strong magnetic fields

Answers

Answer:

3) False. It is expensive since it requires sophisticated equipment and very low temperatures

Explanation:

Nuclear magnetic resonance imaging measurements consist of magnetic resonance imaging to analyze tissues by the transition of the unpaired electron at carbon 13, giving information on the structure and composition of tissues. This information is processed in computers and transformed into images.

So the physical measurement is the MRN

Now we can analyze the statements in the problem

1) True by itself a magnetic measurement is non-invasive

2) True. Measuring carbon transitions has information about the soft tissue of the body

3) False. It is expensive since it requires sophisticated equipment and very low temperatures

4) Right. The applied magnetic field is high to be able to induce carbon transaction

A particular coil has 100 turns and a diameter of 6.0 m. When it's time for a measurement, a 4.5 A current is turned on. The large diameter of the coil means that the field in the water flowing directly above the center of the coil is approximately equal to the field in the center of the coil. The field is directed downward and the water is flowing east. The water is flowing above the center of the coil at 1.5 m/s .

What is the magnitude of the field at the center of the coil?

Answers

Answer:

The magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

Explanation:

Information from the question:

Number of turns of the coil = 100 turns

The diameter of the coil = 6 m

The radius of the coil = diameter / 2 = 3 m

The coil current = 2.5 A

Formula : The Magnetic field at the center of the coil =

                                  k * number of turns * current / 2 * radius

Therefore, The Magnetic field at the center of the coil=

                                 (4 * [tex]\pi[/tex] * 10 ^ -7 * 100 * 2.5 ) / (2 * 3)

The Magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

I need some help!!!!!!!!!

Answers

Answer:

The Object will immediately begin moving toward the left

Explanation:

Because the force of thirteen is greater than ten and applied to the opposite side

Find the frequency of the 4th harmonic waves on a violin string that is 48.0cm long with a mass of 0.300 grams
and is under a tension of 4.00N. ​

Answers

Answer:

The frequency of the 4th harmonic of the string is 481.13 Hz.

Explanation:

When a stretch string fixed at both ends is set into vibration, it produces its lowest sound of possible note called the fundamental frequency.  Under certain conditions on the string, higher frequencies called harmonics or overtones can be produced.

The frequency of the forth harmonic is the third overtone of the string and can be determined by:

          f = [tex]\frac{2}{L}[/tex][tex]\sqrt{\frac{T}{m} }[/tex]

Given that; L = 48.0 cm = 0.48 m,

                 m = 0.3 g = 0.0003 Kg,

                 T = 4.0 N,

         f = [tex]\frac{2}{0.48}[/tex][tex]\sqrt{\frac{4}{0.0003} }[/tex]

         f = 4.1667 × 115.4701

           = 481.1252

        f = 481.13 Hz

The frequency of the 4th harmonic of the string is 481.13 Hz.

The current in the wires of a circuit is 60 milliamps. If the resistance of the circuit were doubled (with no change in voltage), then it’s new current would be _____ milliamps

Answers

Answer:30

Explanation:

Current=60 milliamps

Current=(voltage)/(resistance)

60=(voltage)/(resistance)

Doubling the resistance means multiplying both sides by 1/2

60x1/2=(voltage)/(resistance) x 1/2

30=(voltage)/2(resistance)

Therefore the resistance would be 30 milliamp if we double the resistance

The figure shows a crane whose weight is 12.5 kN and center of gravity in G. (a) If the crane needs to suspend the 2.5kN drum, determine the reactions on the wheel in A and B when the boom is in the position shown.(b) Considering the same situation illustrated, what is the value of the maximum weight that the crane can suspend without tipping over?

Answers

Answer:

(a) Ra = 9.25 kN; Rb = 5.75 kN

(b) 26.7 kN

Explanation:

(a) Draw a free-body diagram of the crane.  There are four forces:

Reaction Ra pushing up at A,

Reaction Rb pushing up at B,

Weight force 12.5 kN pulling down at G,

and weight force 2.5 kN pulling down at F.

Sum of moments about B in the counterclockwise direction:

∑τ = Iα

-Ra (0.66 m + 0.42 m + 2.52 m) + 12.5 kN (2.52 m + 0.42 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0

-Ra (3.6 m) + 12.5 kN (2.94 m) − 2.5 kN (1.38 m) = 0

Ra = 9.25 kN

Sum of moments about A in the counterclockwise direction:

∑τ = Iα

Rb (0.66 m + 0.42 m + 2.52 m) − 12.5 kN (0.66 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° + 0.66 m + 0.42 m) = 0

Rb (3.6 m) − 12.5 kN (0.66 m) − 2.5 kN (4.98 m) = 0

Rb = 5.75 kN

Alternatively, you can use sum of the forces in the y direction as your second equation.

∑F = ma

Ra + Rb − 12.5 kN − 2.5 kN = 0

Ra + Rb = 15 kN

9.25 kN + Rb = 15 kN

Rb = 5.75 kN

However, you must be careful.  If you make a mistake in the first equation, it will carry over to this equation.

(b) At the maximum weight, Ra = 0.

Sum of the moments about B in the counterclockwise direction:

∑τ = Iα

12.5 kN (2.52 m + 0.42 m) − F ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0

12.5 kN (2.94 m) − F (1.38 m) = 0

F = 26.7 kN

Which of the following statements are true? a. Kinematics is the science that studies forces and motion of particles and bodies. b. Speed is a vector quantity. c. The units of velocity are length divided by time. d. The term deceleration is commonly used to describe a negative acceleration.

Answers

Answer:

true. b, c and d

Explanation:

Let's review each statement separately  

a) False. The kinematics studies the position, speed and acceleration of the bodies, but not what causes these changes

b) True. Velocity is the displacement between time, displacement is a vector, and time is a scalar, so the division between them gives a vector

c) True. speed is the displacement that is a length between time, so its unit is length / time

d) true  desaceleration = - aceleration

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