A 125-g coin is placed 8.0 cm from the axis of rotation of a horizontally rotating turntable as shown. The coefficient of static friction between the coin and the turntable is μs . The turntable makes exactly 1.0 revolution per second and the coin remains in place without slipping. [a] (8 pts) Draw a free-body force diagram for the coin. [b] (10pts)What is the frictional force acting on the coin? [c] (12pts)When the rotational speed is increased to 1.2 revolutions per second, the coin starts to slip. Calculate the coefficient of static friction μs between the coin and the turntable?

Answer:

X = 50 g

Explanation:

Please see attached photo for explanation.

From the attached photo,

Anticlock–wise moment = X × 20

Clockwise moment = 100 × 10

Anticlock–wise moment = clockwise moment

X × 20 = 100 × 10

X × 20 = 1000

Divide both side by 20

X = 1000 / 20

X = 50 g

Therefore, the value of X is 50 g

Answer:

Wavelength = 1.36 * 10^{-34} meters

Explanation:

Given the following data;

Mass = 0.113 kg

Velocity = 43 m/s

To find the wavelength, we would use the De Broglie's wave equation.

Mathematically, it is given by the formula;

[tex] Wavelength = \frac {h}{mv} [/tex]

Where;

h represents Planck’s constant.

m represents the mass of the particle.

v represents the velocity of the particle.

We know that Planck’s constant = 6.6262 * 10^{-34} Js

Substituting into the formula, we have;

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43} [/tex]

[tex] Wavelength = \frac {6.6262 * 10^{-34}}{4.859} [/tex]

Wavelength = 1.36 * 10^{-34} meters

Answer: b

Explanation:

The reading of the scale of the elevator  ascending with constant velocity is 150 N.

The reading of the scale on the elevaor is calculated by applying Newton's second law of motion;

R = m(a + g)

R = ma + mg

R = F + W

where;

At constant velocity, the acceleration of the object is zero (0).

R = 0 + 100 + 50

R = 150 N

Thus, the reading of the scale of the elevator  ascending with constant velocity is 150 N.

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Answer:

1 / i + 1 / o = 1 / f     thin lens equations

i = o f / (o - f)   rearranging

Lens 1:   object = 30 cm    f = 15.2 cm

i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm

o2 = 40.2 - 30/8 = 9.4 cm    distance of image 1 from lens 2

i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm

The final image is 24.6 cm to the left of lens 2

The first image is inverted

The second image is erect (as seen from the first image)

So the final image is inverted

M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69

It can be deduced that the final image's distance from Lens 2 will be 30.8 cm.

By using the Lens formula, the distance will be calculated thus:

1/v + 1/u = 1/15.2

1/v + 1/30.0 = 1/15.2

v = 30.8cm

In this case, the image formed will be to the right.

Lastly, the overall magnification of the lens pair will be:

M = (30.8/30.0)[(-28.4/40.7 - 30.8)]

M = -2.94.

In conclusion, the magnification is -2.94.

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Solution :

We know that the time period of oscillation of a spring mass system is given by :

[tex]$T = 2 \pi \sqrt{\frac{m}{k}}$[/tex]     , where m is mass and k is the spring constant

∴ [tex]$T_A = 2 \pi \sqrt{\frac{m}{k}}$[/tex]    .........(i)

  [tex]$T_B = 2 \pi \sqrt{\frac{2m}{k}}$[/tex] ..........(ii)

  [tex]$T_C = 2 \pi \sqrt{\frac{3m}{6k}} = 2 \pi \sqrt{\frac{m}{2k}}$[/tex]   ..........(iii)

  [tex]$T_D = 2 \pi \sqrt{\frac{m}{4k}}$[/tex]   ...............(iv)

Comparing the equations (i), (ii), (iii) and (iv)

We get

[tex]$T_B > T_A > T_C > T_D$[/tex]

So in increasing order of time period, we get

[tex]$T_D < T_C < T_A < T_B$[/tex]

Answer:

[tex]7.32\ \text{m/s}[/tex]

Explanation:

[tex]v_1[/tex] = Velocity at initial point = 0

[tex]P_1[/tex] = Pressure in tank = 120 kPa

[tex]P_2[/tex] = Pressure at outlet = 101 kPa

[tex]\rho[/tex] = Density of kerosene = [tex]750\ \text{kg/m}^3[/tex]

[tex]Z_1[/tex] = Tank height = 15 cm

[tex]Z_2[/tex] = Height of pipe exit = 0

[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

From Bernoulli's equation we have

[tex]\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+Z_2\\\Rightarrow \dfrac{P_1}{\rho g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}\\\Rightarrow v_2=\sqrt{2g(\dfrac{P_1}{\rho g}+Z_1-\dfrac{P_2}{\rho g})}\\\Rightarrow v_2=\sqrt{2\times 9.81(\dfrac{120\times 10^3}{750\times 9.81}+0.15-\dfrac{101\times 10^3}{750\times 9.81})}\\\Rightarrow v_2=7.32\ \text{m/s}[/tex]

The exit velocity from the tube is [tex]7.32\ \text{m/s}[/tex].

Answer:

[tex]from \: the \: wave \: equation \\ velocity = frequency \times wavelength \\ 340 = 230 \times \lambda \\ \lambda = \frac{340}{230} \\ \lambda = 1.5 \: m[/tex]

Answer:

Wavelength = [tex]1.36\times 10^{-34}\ m[/tex]

Explanation:

Given that,

The mass of a ball, m = 0.113 kg

Velocity of the ball, v = 43 m/s

We need to find the wavelength of the ball. It can be calculated by applying the De-Broglie concept. So,

[tex]\lambda=\dfrac{h}{mv}[/tex]

Put all the values,

[tex]\lambda=\dfrac{6.63\times 10^{-34}}{0.113\times 43}\\\\=1.36\times 10^{-34}\ m[/tex]

So, the wavelength of the ball is equal to [tex]1.36\times 10^{-34}\ m[/tex].

Explanation is[tex]^{}[/tex] in a file

bit.[tex]^{}[/tex]ly/3tZxaCQ

Answer:

351.28 N

Explanation:

Let F be the force on the object and f be the frictional force. The component of the force acting in the horizontal direction causing the object to move is FcosФ where Ф is the angle between F and the horizontal = 43.0°. The frictional force on the packing crate f = μN where μ = coefficient of kinetic friction = 0.35 and N = normal force = W = weight of the packing crate = mg where m = mass of crate = 74.9 kg and g = acceleration due to gravity = 9.8 m/s².  So, f = μN = μW = μmg

So, the net force on the packing crate is

FcosФ - f = ma

FcosФ - μmg = ma

Since the crate moves at constant speed, its acceleration a = 0

So, FcosФ - μmg = ma

FcosФ - μmg = m(0)

FcosФ - μmg = 0

FcosФ = μmg

F = μmg/cosФ

Substituting the values of the variables into the equation, we have

F = μmg/cosФ

F = 0.35 × 74.9 kg × 9.8 m/s²/cos43.0°

F = 256.907 kg-m/s²/0.73135

F = 351.28 kg-m/s²

F = 351.28 N

Answer:

please give me brainlist and follow

Explanation:

Given:

The speed of the train ,which the boy is observing(t) = 18 km/hr

[ Now , to convert km/hr into m/s we have to multiply by 5/18 ]

So,

speed of the observed train = 18× 5/18

☞ 05 m/s

Also, The speed of the train in which the boy is sitting is(s) = 36 km/hr

Speed in m/s will be = 36 × 5/18

☞ 10 m/s

Also, The Length of the Train = 90m

Therefore:.

The speed of the train is given (expressed) by

= > speed \: = \frac{distance}{time}=>speed=

time

distance

= > time \: = \frac{distance}{Velocity }=>time=

Velocity

distance

Where,

velocity = Realtive velocity of both trains

Distance = Length of the train

So, The above mentioned formulae can also be written as :

= > time = \frac{length}{relative \: velocity}=>time=

relativevelocity

length

Here,

Relative velocity will be :

☞. s -(-t)

☞. 10 -(-5)

☞.15 m/s = Relative velocity

Now , Finally putting the values in the formulae respectively

■ Time = 90/15

■ Time = 6 sec [Answer]

Answer:

Potential energy only occurs when an object is in motion.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Kinetic energy (K.E): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

b. Potential energy (P.E): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Additionally, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Mechanical energy = P.E + K.E

Answer:

The angular speed of the ball in radians per second is 5.55 rad/s.

Explanation:

Given;

mass of the ball, m = 1.8 kg

number of the ball's rotation per minute, n = 53 RPM

The angular speed of the ball in radians per second is calculated as follows;

[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]

Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.

Explanation:

Net force on the car= mass of the car × acceleration

F=1×10^3×4.5

=4.5×10^3 N

Answer:

Pair 1

Explanation:

The gravitational force greater between the objects in Pair 1 because in Pair 2 the objects are far apart and in pair 1 the object are more closer to each other.

So, Gravitational force directly proportional to distance

Thus, Increase in distance = Increased in gravitational force

-TheUnknownScientist

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

[tex]I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2[/tex]

The moment of inertia of the rod about its center is calculated as follows;

[tex]I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2[/tex]

The initial rotational kinetic energy of the disk and rod;

[tex]K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J[/tex]

The final rotational kinetic energy of the disk-rod system is calculated as follows;

[tex]K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J[/tex]

The loss in rotational kinetic energy due to the collision is calculated as follows;

[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J[/tex]

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

Answer:

One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point. The zero point on the Celcius scale was defined as the freezing point of water, which means that there are higher and lower temperatures around it.

Answer:

Teaching is the process of attending to people's needs, experiences and feelings, and making specific interventions to help them learn particular things.

Explanation:

Answer:

In education, teaching is the concerted sharing of knowledge and experience, which is usually organized within a discipline and, more generally, the provision of stimulus to the psychological and intellectual growth of a person by another person or artifact.

Answer:

F1 = K Q1 Q2 / R1^2

F2 = K Q1 / 2 * Q2 / (2 R1)^2

F2 / F1 = 1/2 / 4 = 1/8

The new force is 5N   (1/2 due to charge and 1/4 due to distance)

Answer: True!

Explanation: The force is proportional to the square of the distance between 2 point masses

Answer:

The correct answer is "3899.92 N".

Explanation:

The given values are:

Force,

[tex]F_{app}=58 N[/tex]

Ratio,

[tex]\frac{R_2}{R_1}=8.2[/tex]

As we know,

Area, [tex]A=\pi r^2[/tex]

or,

⇒  [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]

On substituting the value of "A", we get

⇒  [tex]\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}[/tex]

⇒  [tex]\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}[/tex]

On applying cross-multiplication, we get

⇒  [tex]F_2=F_1\times (\frac{r_2}{r_1} )^2[/tex]

On substituting the given values, we get

⇒       [tex]=58\times (8.2)^2[/tex]

⇒       [tex]=58\times 67.2[/tex]

⇒       [tex]=3899.92 \ N[/tex]

Answer:

23. Option B. Hertz

29. Refraction.

Explanation:

23. Determination of the unit of measurement of frequency.

Frequency is simply defined as the number of oscillation made in one second. Mathematically, frequency can be represented as:

Frequency = 1/period

f = 1/T

Period is measured in seconds.

Thus, the unit of frequency becomes

f = 1/T

f = 1/s = s¯¹ (Hertz)

Therefore, the unit of frequency is Hertz.

29. When a wave enters a new medium, the speed of the wave is uttered. This leads to the bending of the wave. When this occurs, we can say refraction has taken place.

Answer:

the  wavelength of the interfering waves for this mode is 2.4168

Explanation:

The computation of the wavelength of the interfering waves for this mode is shown below:

Given that

The length is 7.25m

And, the number of nodes is 6

So, the wavelength should be

= 2 × 7.25m ÷ 6

= 2.4168 m

Hence, the  wavelength of the interfering waves for this mode is 2.4168

The same would be considered

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

Answer:

dorsiflexion

Explanation:

Hope this helps

In the model 1 AU = 100 meters (the length of a football field) , then mars would be 150 meters far from the school, therefore the correct answer is option C.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.

As given in the problem the  Earth Science students are making a human-scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters,

As given in the table marks is 1.5 AU far from Mars,

1 AU = 100 meters

1.5 AU = 1.5 × 100

           = 150 meters

Thus, mars would be 150 meters far from the school, therefore the correct answer is option C.

Learn more about the unit of measurement here, refer to the link ;

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Answer:

271.095 m

Explanation:

✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)

✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)

✓ Let the distance travelled by the sound = s

✓duration of Time that exist between heard of sounds = 0.70s

But we know that

Time = (Distance / Speed)

✓Time it takes the sound to travel through air= s/vi = s/343

✓Time it takes the sound to travel through concrete= s/vc = s/3000

✓ (s/343) - (s/3000) = 0.70

Finding LCM and simplify

[(3000s - 343s)]/1029000 = 0.70

2657s /1029000 = 0.70

Making " s" subject of the formula

s= (1029000 × 0.70)/2657

s=720300/ 2657

s= 271.095 m

Hence, The impact took place at a distance of 271.095 m away from the person.

Answer:

a) 1.092 m/s

b) 0.33 m

c) 0.25 m

Explanation:

To start with, from the formula of wave, we know that

v = f λ, where

v = velocity of wave

f = frequency of the wave

λ = wavelength of the wave

Again, on another hand, we know that

T = 1/f, where T = period of the wave

From the question, we are given that

t = 2.7 s

d = 0.66 m

λ = 5.9 m

Period, T = 2 * t

Period, T = 2 * 2.7

Period, T = 5.4 s

If T = 1/f, then f = 1/T, thus

Frequency, f = 1/5.4

Frequency, f = 0.185 hz

Remember, v = f λ

v = 0.185 * 5.9

v = 1.092 m/s

Amplitude, A = d/2

Amplitude, A = 0.66/2

Amplitude, A = 0.33 m

If the other distance travelled by the boat is 0.5, then Amplitude is

A = 0.5/2

A = 0.25 m