1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alina1380 [7]
3 years ago
5

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

e bottom.
a) From what height should a circular hoop of radius R bereleased on the same slope in order to equal the sphere's speed atthe bottom?
b) Can a circular hoop of different diameter be released froma height of 30 cm and match the sphere's speed at the bottom? If,so what is the diameter? If not, why not?
Physics
1 answer:
photoshop1234 [79]3 years ago
8 0

Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

You might be interested in
A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of
Vikentia [17]

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)

mv^2 = mv_t^2 + 2mgR(1 + cos \theta)

we know that,

N - mgcos \theta = \dfrac{mv^2}{R}

N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}

N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)

N  = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)

N  = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)

N = 26.59 N

3 0
3 years ago
A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’
scZoUnD [109]

A charged particle moving in a magnetic field experiences a force equal to:

\vec{F}=q\vec{v}\times \vec{B}

Thus, the magnitude of the force that the proton experiences is given by:

F=qvBsin\theta

The magnetic field is perpendicular to the proton's velocity, therefore, we have \theta=90^\circ. Replacing the given values, we obtain:

F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N

3 0
3 years ago
Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typica
kogti [31]

Answer:

The break even cost is $0.0063825

Explanation:

Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.

Wattage = W

Cost per kilo watt hour = C

Number of hours per year = H

Price per bulb/CFL = P

Discount rate = 11%

Life of bulb = 2 years

Price of bulb = $0.39

Wattage consumption of bulb per hours = 60

Life of CFL = 24 years

Price of CFL = $3.10

Wattage consumption of CFL per hour = 15

Calculate the Equated Annual Cost (EAC) of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)

PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods

PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)

<u>Calculate the EAC of bulb</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

EAC = {-0.39 – (51.37570 x C)}/ 1.712523, <em>consider this equation 1</em>

<u>Calculate the EAC of CFL</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, <em>consider this equation 2</em>

<u>Equate 1 and 2 to find the amount of C</u>

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

C = $0.0063825

Thus, the break- even cost per kilo – watt is $0.0063825

3 0
3 years ago
A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy.
statuscvo [17]
Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.

According to the first law of thermodynamics, 
ΔE = Q + W
178 = 658 + W 
∴ W = 178-658 = -480 J

Minus sign indicates that work is done by the system. 
7 0
3 years ago
Read 2 more answers
A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con
erastova [34]

Let the key is free falling, therefore from equation of motion

h = ut +\frac{1}{2}gt^2..

Take initial velocity, u=0, so

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2.

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

d= v \times t

From above substituting t,

d = v \times \sqrt{\frac{2h}{g} }.

Now substituting all the given values and g = 9.8 m/s^2, we get

d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m.

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0
3 years ago
Other questions:
  • Define mechanical clock
    13·1 answer
  • A model rocket is launched at an angle of 70° above the x axis, with an initial velocity of 40 m/s. How high will the rocket be
    11·2 answers
  • You put a new battery into your MP3 player and listen to a song. What are the energy transformations taking place in this system
    11·1 answer
  • At take off, a plane flies 100 km north before turning to fly 200 km east. How far is its destination from where the plane took
    10·2 answers
  • (1 point) A horizontal clothesline is tied between 2 poles, 20 meters apart. When a mass of 5 kilograms is tied to the middle of
    15·1 answer
  • Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi
    15·1 answer
  • The 64.5-kg climber in is supported in the “chimney” by the friction forces exerted on his shoes and back. The static coefficien
    8·1 answer
  • A neighborhood transformer on the top of a utility pole transforms 12.0 kV 60.0 Hz alternating voltage down to 120.0 V to be use
    13·1 answer
  • An element’s __________ is its row in the periodic table.
    8·1 answer
  • Which is the only element in group 1 on the periodic table that forms covalent bonds?
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!