iven a 36.0 V battery and 14.0 Ω and 84.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series.

Answers

Answer 1

Answer:

0.367A = Current of both resistors

For resistor 1: 1.89W; For resistor 2: 11.3W

Explanation:

When the resistors are connected in series, the equivalent resistance is the sum of both resistors, that is:

R = 14.0Ω + 84.0Ω = 98.0Ω

Using Ohm's law, we can find the current of the circuit (Is the same for both resistors):

V = RI

V / R = I

36.0V / 98.0Ω = I

0.367A = Current of both resistors

Power is defined as:

P = I²*R

For resistor 1:

P = 0.367A²*14.0Ω = 1.89W

For resistor 1:

P = 0.367A²*84.0Ω = 11.3W


Related Questions

When the adjustable mirror on the Michelson interferometer is moved 20 wavelengths, how many fringe pattern shifts would be counted

Answers

Answer:

The number of  fringe pattern shift is   m  = 40

Explanation:

 From the question we are told that

      The  Michelson interferometer is moved 20 wavelengths i.e  [tex]20 \lambda[/tex]

Generally the distance which the  Michelson interferometer is moved is mathematically represented as

         [tex]d = \frac{m * \lambda}{2}[/tex]

Here [tex]m[/tex] is the number of  fringe pattern shift

     So

         [tex]20 \lambda = \frac{m * \lambda}{2}[/tex]

         [tex]40 \lambda = m * \lambda[/tex]

        m  = 40

A woman was told in 2020 that she had exactly 15 years to live. If she travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:

Answers

Answer:

2035

Explanation:

The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035

As a skydiver falls, his potential energy ___ and his kinetic energy __​
increases,increases
increases,decreases
decreases,increases
decreases, decreases

Answers

Answer:

Hey there!

PE=mgh, so as height decreases, so does the potential energy.

KE=mv^2, so as velocity increases, kinetic energy increases.

Thus, the correct answer would be Decreases, Increases.

Let me know if this helps :)

A loop of wire is at the edge of a region of space containing a uniform magnetic field B. The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dA/dt=−c, with c>0.Required:a. The induced emf in the loop is measuredto be V. What is the magnitude B of the magnetic field that the loop was in?b. For the case of a square loop of sidelength L being pulled out of the magneticfield with constant speed v, What is the rate of change of area c= -dA/dt

Answers

Answer:

The question is not clear enough. So i have attached a copy of the correct question.

A) B = V/c

B) c = Lv

Explanation:

A) we know that formula for magnetic flux is;

Φ = BA

Where B is magnetic field and A is area

Now,

Let's differentiate with B being a constant;

dΦ/dt = B•dA/dt

From faradays law, the EMF induced is given as;

E = -dΦ/dt

However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.

Thus, V = -dΦ/dt

Thus, we can now say that;

-V = B•dA/dt

Now from the question, we are told that dA/dt = - c

Thus;

-V = B•-c

So, V = Bc

Thus, B = V/c

B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:

Thus;

E =LvB or can be written as; V = LvB

Where;

V is EMF

L is length of bar

v is velocity

From the first solution, we saw that;

V = Bc

Thus, equating both of the equations, we have;

Bc = LvB

B will cancel out to give;

c = Lv

Explanation:

A hot cup of coffee is placed on a table. Which will happen because of conduction? Answer options with 4 options A. The temperature of the coffee will decrease while the temperature of the table decreases. B. The temperature of the coffee will increase while the temperature of the table increases. C. The temperature of the coffee will decrease while the temperature of the table increases. D. The temperature of the coffee will increase while the temperature of the table decreases.

Answers

Answer:

C.

Explanation:

C. The temperature of the coffee will decrease while the temperature of the table increases.

Are Quantum Physics, Quantum mechanics,Quantum Engagement same?
or, Do they branch of each others ​

Answers

Answer:

The topic of quantum entanglement is at the heart of the disparity between classical and quantum physics: entanglement is a primary feature of quantum mechanics lacking in classical mechanics. ... In the case of entangled particles, such a measurement will affect the entangled system as a whole

Explanation:

Answer:

quantum entanglement is thought to be one of the trickiest concepts in science, but the core issues are simple. And once understood, entanglement opens up a richer understanding of concepts such as the “many worlds” of quantum theory.

Explanation:

Which one of the following lists gives the correct order of the electromagnetic spectrum from low to high frequencies?
A) radio waves, infrared, microwaves, ultraviolet, visible, x-rays, gamma rays
B) radio waves, ultraviolet, x-rays, microwaves, infrared, visible, gamma rays
C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays
D) radio waves, microwaves, visible, x-rays, infrared, ultraviolet, gamma rays
E) radio waves, infrared, x-rays, microwaves, ultraviolet, visible, gamma rays

Answers

Answer:

C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays

Explanation:

radio waves have lowest  energy , lowest  frequency and highest  wavelength

gamma rays  have highest  energy , highest  frequency and least  wavelength

Answer: C

Explanation:

How much time will elapse if a radioisotope with a half-life of 88 seconds decays to one-sixteenth of its original mass?

Answers

Answer:

352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.

Explanation:

The decay of radioisotopes are represented by the following ordinary differential equation:

[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]

Where:

[tex]t[/tex] - Time, measured in seconds.

[tex]\tau[/tex] - Time constant, measured in seconds.

[tex]m[/tex] - Mass of the radioisotope, measured in grams.

The solution of this expression is:

[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.

The ratio of current mass to initial mass is:

[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]

The time constant is now calculated in terms of half-life:

[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]

Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.

Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:

[tex]\tau = \frac{88\,s}{\ln 2}[/tex]

[tex]\tau \approx 126.957\,s[/tex]

Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:

[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]

[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]

[tex]t \approx 352\,s[/tex]

352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion

Answers

Answer:

C. planetary accretion

Explanation:

Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.

Answer:

[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]

Explanation:

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.

Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.

A stationary coil is in a magnetic field that is changing with time. Does the emf induced in the coil depend

Answers

Answer:

Explanation:

The e.m.f induced in the coil depend on the following :

(a) No. of turns in the coil

(b) Cross-sectional Area of the coil

(c) Magnitude of Magnetic field

(d) Angular velocity of the coil

6. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v?

Answers

Answer:

   t = t₀ / 2

Explanation:

In this exercise we must use Newton's second law

          F = m a

          a = F / m

now we can use kinematics

  as in object part of rest (v₀ = 0)

        v =a t₀

        t₀ = v / a

these results are with the first experiment

now repeat the experiment, but F = 2F₀

           a = 2F₀ / m = 2 a₀

          v = 2 a₀ t

          t = v / 2a₀

          t = t₀ / 2

The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].

Given the following data:

Initial speed = 0 m/s (since the object is at rest)Final speed = VTime = [tex]\Delta t[/tex]Speed = V

To find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:

Mathematically, Newton's Second Law of Motion is given by this formula;

[tex]F = \frac{M(V-U)}{t}[/tex]

Where:

F is the force.V is the final velocity.U is the initial velocity.t is the time.

Substituting the given parameters into the formula, we have;

[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]

When the experiment is repeated, the magnitude of the force is doubled:

[tex]F = 2F[/tex]

Now, we can find the time interval that is required to reach the same final speed (V):

[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]

Substituting the value of F, we have:

[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]

Read more here: https://brainly.com/question/24029674

A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 105 m/s to pass straight through without deflection? A. 70 V B. 140 V C. 450 V D. 1,400 V E. 2,800 V

Answers

Answer:

C. 450v

Explanation:

Using

Voltage= B*distance of separation*velocity

3mm x 0.3T x 5E5m/s

= 450v

What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×10-34 Js.)

Answers

Answer:

9.82 × [tex]10^{-35}[/tex] Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = [tex]\frac{h}{mv}[/tex]

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = [tex]\frac{h}{mv}[/tex]

  = [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]

 = [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]

 = 9.8222 × [tex]10^{-35}[/tex]

The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.

A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up from chest height to full arm extension, a distance of 0.53 m .
Part A: How much work does the weightlifter do to lift the barbell one time?
Part B: If the weightlifter does 23 repetitions a day, what total energy does she expend on lifting, assuming a typical efficiency for energy use by the body?
Part C: How many 500 Calorie donuts can she eat a day to supply that energy?

Answers

Answer:

A) Workdone = 223.57 N-m

B) 22357 J of energy

C) Number of donuts = 10.7 donuts

Explanation:

A) The work done is calculated from the formula;. Work done = Force × Distance

We are given;

Mass; m = 43 kg

Distance = 0.53 m

Force(weight) = mg = 43 × 9.81

Thus;

Work done = 43 × 9.81 × 0.53

Workdone = 223.57 N-m

B) We are told she does 23 repetitions a day.

Thus, we assume 23% efficiency.

So, Work = Energy

Thus;

At 100% efficiency;

Energy = (223.57/100%) × 23 repetitions = 5142.11 J

Now, since she is only 23% efficient, she will expend; 5142.11/0.23 J = 22357 J of energy to do 5390 J of work.

C) from conversions; 4.18 J = 1 calorie

Thus;

22357 J ÷ 4.18 J/cal = 5348.565 calories

We how many 500 calorie donuts she can eat in a day to supply that energy.

Thus;

Number of donuts = 5348.565 cal ÷ 500 cal /donut

Number of donuts = 10.7 donuts

Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How far away (in m) is the planet Venus if the echo time is 900 s? m (b) What is the echo time (in µs) for a car 80.0 m from a Highway Patrol radar unit? µs (c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 11.5 m? ns

Answers

Answer:

a) 1.35 x 10^11 m

b) 0.53 µs

c) 8 ns

Explanation:

Radar involves the use of radio wave which has speed c = 3 x 10^8 m/s

a) for 900 s,

The distance for a round trip = v x t

==>  (3 x 10^8) x 900 =  2.7 x 10^11 m

The distance of Venus is half this round trip distance = (2.7 x 10^11)/2 = 1.35 x 10^11 m

b) for a 80.0 m distance of the car from the radar source, the radar will travel a total distance of

d = 2 x 80 = 160 m

the time taken = d/c = 160/(3 x 10^8) = 5.3 x 10^-7 s = 0.53 µs

c) accuracy in distance Δd = 11.5 m

Δt = accuracy in time = Δd/c = 11.5/(3 x 10^8) = 3.8 x 10^-8 = 38 ns

Light of wavelength 550 nm is incident on a slit having a width of 0.200 mm. The viewing screen is 1.90 m from the slit. Find the width of the central bright fringe

Answers

Answer:

The width of Center bright fringe is 10.2mm

Explanation:

Given that if

Y/ L << 1 then

Sin theta will be approx Y/L

So sin theta approx Y/L = lamda/a

Y= a x lambda/a

By substituting

1.9x 10^ -3m x 550*10^-9/ 0.2 x 10^-3m

= 5.2mm

But

Change in y = 2y = 10.4mm

What is the separation in meters between two slits for which 594 nm orange light has its first maximum at an angle of 32.8°?

Answers

Answer:

1.1micro meter

Explanation:

Given that

Constructive interference is

ma = alpha x sin theta

Alpha = 1 x 594 x10^ -9/ sin 32.8°

= 1.1 x 10^ -6m

Explanation:

Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?

Answers

Answer: Only Tech B is correct.

Explanation:

First, tech A is wrong.

The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.

Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.

A locomotive is pulling three train cars along a level track with a force of 100,000N. The car next to the locomotive has a mass of 80,000kg, next one, 50,000kg, and the last one, 70,000 kg. you can neglect the friction on the cars being pulled.
A) what if the magnitude of the force between that the 80,000-kg car exerts on the 50,000-kg car?
B) what is the magnitude of the force that the 50,000-kg car exerts on the 70,000-kg car?

Answers

Answer:

a) 60000 N

b) 35000 N

Explanation:

Force from locomotive = 100000 N

mass of first car = 80000 kg

mass of second car = 50000 kg

mass of third car = 70000 kg

friction is neglected in this system

Total mass of the cars = 80000 + 50000 + 70000  = 200000 kg

All the car in the system will accelerate at the same rate since they are pulled by the same force

We know that force F = ma

where

a is the acceleration of the cars

m is the total mass in the system

from this we can say that

a = F/m

a = 100000/200000 = 0.5 m/s^2

a) The total mass involved in this case = mass of the last two cars after the 80000 kg car =  50000 + 70000 = 120000 kg

therefore force exerted F = ma

F = 0.5 x 120000 = 60000 N

b) The total mass in this case = mass of the third car only = 70000 kg

F = ma

F = 70000 x 0.5 = 35000 N

Select the situation for which the torque is the smallest.

a. A 200 kg piece of silver is placed at the end of a 2.5 m tree branch.
b. A 20 kg piece of marble is placed at the end of a 25 m construction crane arm.
c. A 8 kg quartz rock is placed at the end of a 62.5 m thin titanium rod.
d. The torque is the same for two cases.
e. The torque is the same for all cases.

Answers

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

a)

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

τ = 4900 N.m

b)

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

τ = 4900 N.m

c)

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

τ = 4900 N.m

Hence, the correct answer will be:

e. The torque is the same for all cases.

The A block, with negligible dimensions and weight P, is supported by the coordinate point (1.1/2) of the parabolic fixed grounded surface, from equation y = x^2/2 If the block is about to slide, what is the coefficient of friction between it and the surface; determine the force F tangent to the surface, which must be applied to the block to start the upward movement.

Answers

Answer:

μ = 1

F = P√2

Explanation:

The parabola equation is: y = ½ x².

The slope of the tangent is dy/dx = x.

The angle between the tangent and the x-axis is θ = tan⁻¹(x).

At x = 1, θ = 45°.

Draw a free body diagram of the block.  There are three forces:

Weight force P pulling down,

Normal force N pushing perpendicular to the surface,

and friction force Nμ pushing up tangential to the surface.

Sum of forces in the perpendicular direction:

∑F = ma

N − P cos 45° = 0

N = P cos 45°

Sum of forces in the tangential direction:

∑F = ma

Nμ − P sin 45° = 0

Nμ = P sin 45°

μ = P sin 45° / N

μ = tan 45°

μ = 1

Draw a new free body diagram.  This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.

Sum of forces in the tangential direction:

∑F = ma

F − Nμ − P sin 45° = 0

F = Nμ + P sin 45°

F = (P cos 45°) μ + P sin 45°

F = P√2

A wire carries current in the plane of this paper toward the top of the page. The wire experiences a ma netic force toward the right edge of the page. The direction of the magnetic field causing this force is:
A. in the plane of the page and toward the left edge
B. in the plane of the page and toward the bottom edge
C. upward out of the page
D. downward into the page

Answers

Answer:

D) True. In this case the thumb goes up the page, the fingers are extended out of the page and the palm points to the left

Explanation:

The magnetic force on a conductor is given by

        F = i L x B

bold letters indicate vectors. We can write this expression in the form of magnitudes

         F = i L B sin θ

The direction of the force can be found by the rule of the right hand, the thumb points in the direction of the current, the fingers extended in the direction of the magnetic field and the palm gives the direction of the force

Let's apply this expression to the case presented.

A) False. In this case the force is out of the page and is in contradiction with the real force

B) False. In this case the force is zero since the displacement of the current and the field would be parallel

C) False. In this case the force is to the left

D) True. In this case the thumb goes up the page, the fingers are extended out of the page and the palm points to the left


Somebody please help it’s urgent!!!!

In the tug of war game, none of the teams won. What can you conclude about the forces of the two teams ? Write all the evidence to support your answer.

Answers

Answer:

Explanation:

We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.

hope this helps

plz mark as brainliest!!!!!!!

If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.

Answers

Answer:

D. A convex lens in air

Explanation:

This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens

Calcular la resistencia de una varilla de grafito de 170 cm de longitud y 60 mm2. Resistividad grafito 3,5 10-5 Ωm

Answers

Answer:

R = 0.992 Ω

Explanation:

En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.

Matemáticamente,

Resistencia = (resistividad * longitud) / Área De la pregunta;

Resistividad = 3,5 * 10 ^ -5 Ωm

longitud = 170 cm = 1,7 m

Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2

Conectando estos valores a la ecuación anterior, tenemos;

Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =

(3.5 * 1.7) / 6 = 0.992 Ω

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation. With what period does it now oscillate?

Answers

Answer:

The new time period is  [tex]T_2 = 3.8 \ s[/tex]

Explanation:

From the question we are told that

  The period of oscillation is  [tex]T = 5 \ s[/tex]

   The  new  length is  [tex]l_2 = 0.76 \ m[/tex]

Let assume the original length was [tex]l_1 = 1 m[/tex]

Generally the time period is mathematically represented as

         [tex]T = 2 \pi \sqrt{ \frac{ I }{ mgh } }[/tex]

Now  I is the moment of inertia of the stick which is mathematically represented as

           [tex]I = \frac{m * l^2 }{12 }[/tex]

So

        [tex]T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }[/tex]

Looking at the above equation we see that

        [tex]T \ \ \ \alpha \ \ \ l[/tex]

=>    [tex]\frac{ T_2 }{T_1} = \frac{l_2}{l_1}[/tex]

=>    [tex]\frac{ T_2}{5} = \frac{0.76}{1}[/tex]

=>     [tex]T_2 = 3.8 \ s[/tex]

Find the focal length of contact lenses that would allow a nearsighted person with a 130 cmcm far point to focus on the stars at night.

Answers

Answer:

130cm

Explanation:

The lens equation is expressed as;

1/f = 1/u+1/v where;

f is the focal length of the lens

u is the object distance

v is the image distance

Since the near sighted person wants focus the starts at nigt, the stars at night are the images located that infinity. Hence the image distance v = ∞.

The object distance u = 130cm

Substituting the given parameters in the formula to get the focal length f

[tex]\frac{1}{f} = \frac{1}{\infty} + \frac{1}{130} \\\\As \ x \ tends \ to \ \infty, \, \frac{a}{x} \ tends \ to \ 0 \ where\ 'a' \ is \ a\ constant \\\\} \\\\[/tex]

[tex]\frac{1}{f} = 0+ \frac{1}{130}\\\\[/tex]

[tex]\frac{1}{f} =\frac{1}{130}\\cross\ multiply\\\\f = 130*1\\\\f = 130cm[/tex]

Hence the focal length of contact lenses that would allow a nearsighted person with a 130 cm far point to focus on the stars at night is 130cm

To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.0 m/s. How much time does it take for the glove to return to the pitcher

Answers

Answer:

The glove takes 1.02s to return to the pitchers hand.

Explanation:

Given;

initial velocity the pitcher's glove, u = 5 m/s

Apply kinematic equation

s = ut - ¹/₂gt²

where;

g is acceleration due to gravity = 9.8 m/s²

t is the time takes the glove to return to the pitchers hand

s is the displacement of the glove, which will be equal to zero when the glove returns to the pitchers hand. (s = 0)

0 = ut - ¹/₂gt²

ut = ¹/₂gt²

u = ¹/₂gt

gt = 2u

t = (2u) / g

t = (2 x 5) / 9.8

t = 1.02 s

Therefore, the glove takes 1.02s to return to the pitchers hand.

A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.

In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?

The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.

Answers

Answer:

a)   Δt = 24.96 s , b)  τ = 0.078 N m

Explanation:

This is a rotational kinematics exercise

        θ = w₀ t - ½ α t²

Let's reduce the magnitudes the SI system

       θ = 60 rev (2π rad / 1 rev) = 376.99 rad

       w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s

       

      α = (w₀ t - θ) 2 / t²

let's calculate the annular acceleration

      α = (43.98 10 - 376.99) 2/10²

      α = 1,258 rad / s²

Let's find the time it takes to reach zero angular velocity (w = 0)

        w = w₀ - alf t

         t = (w₀ - 0) / α

         t = 43.98 / 1.258

         t = 34.96 s

this is the total time, the time remaining is

         Δt = t-10

         Δt = 24.96 s

To find the braking torque, we use Newton's law for angular motion

        τ = I α

the moment of inertia of a circular ring is

       I = M r²

we substitute

         τ = M r² α

we calculate

        τ = 0.625  0.315²  1.258

        τ = 0.078 N m

The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Given data:

The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex]   (rps means rotation per second).

The time interval is, t' = 10.0 s.

The number of rotations made by wheel is, n = 60.0.

The mass of bike wheel is, m = 0.625 kg.

The radius of wheel is, r = 0.315 m.

The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,

[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]

Here, [tex]\theta[/tex] is the angular displacement, and its value is,

[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]

And, angular speed is,

[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]

Solving as,

[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]

Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.

[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]

Then total time is,

T = t - t'

T = 35.18 - 10

T = 25.18 s

Now, use the standard formula to obtain the value of braking torque as,

[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]

Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Learn more about the rotational motion here:

https://brainly.com/question/1388042

A stone is dropped from the upper observation deck of a tower, 50 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 9 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

Answers

Answer:

A. Using displacement =Ut + 1/2gt²

=> 0 + 1/2 (-9.8)t²

= -4.9t²

So

h(t) = 50+ displacement

= 50 - 4.9t²

B. To reach the ground

h(t) = 0

So

50-4.9t²= 0

t = √ (50/4.9)

= 3.2s

C. Using

V = u+ gt

U= 0

V= - 9.8(3.2)

= 31.4m/s

D. If u = -9m/s

Then s = ut + 1/2gt²

5t- 1/2gt²

But distance from the ground is

=.> 50-5t- 4.8t²= 0

So t solving the quadratic equation

t= 3.58s

(a) The distance of the stone above the ground level at time t is [tex]h(t) = 50 - 4.9t^2[/tex]

(b) The time taken for the stone to strike the ground is 3.19 s.

(c) The velocity of the stone when it strikes the ground is 31.4 m/s.

(d) The time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.

The given parameters;

height above the ground, h₀ = 50 m

The distance of the stone above the ground level at time t is calculated as;

[tex]h(t) = h_0 - ut - \frac{1}{2} gt^2\\\\h(t) = 50 - 0 -0.5\times 9.8t^2\\\\h(t) = 50 - 4.9t^2[/tex]

The time taken for the stone to strike the ground is calculated as;

[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 50}{9.8} } \\\\t = 3.19 \ s[/tex]

The velocity of the stone when it strikes the ground is calculated as;

[tex]v =u + gt\\\\v = 0 + 3.2 \times 9.8\\\\v = 31.4 \ m/s[/tex]

The time taken for the stone to reach the ground when thrown at speed of 9 m/s is calculated as;

[tex]50 = 9t + \frac{1}{2} (9.8)t^2\\\\50 = 9t + 4.9t^2\\\\4.9t^2 + 9t - 50 = 0\\\\a = 4.9 \, \ b = 9, \ \ c = -50\\\\solve \ the \ quadratic \ equation\ using \ formula \ method\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-9 \ \ + /- \ \ \sqrt{(9)^2 - 4(4.9 \times -50)} }{2(4.9)} \\\\t = 2.41 \ s \ \ or \ \ - 4.24 \ s[/tex]

Thus, the time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.

Learn more here:https://brainly.com/question/9527588

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