An object that moves in uniform circular motion has a centripetal acceleration of 13 m/s^2 . If the radius of the motion is 0.02m, what is the frequency of motion?

When the string of the instrument is tightened then the length of the string decreases hence the pitch and frequency will increase so, option B is correct.

The frequency at which the sound waves that create a sound vibrate determines its pitch. High-frequency sound waves produce high-pitched noises, and low-frequency sound waves make low-pitched noises. The ability to discern between harsh and flat sounds is known as pitch.

A string will vibrate at a varied frequency depending on its length. Higher frequency and higher pitch are produced by shorter strings.

The pitch rises as the anxiety does as well. A string's length is also crucial. A string vibrates and makes music when it is supported at two points and pulled. The pitch of this string will, however, rise if the length is shortened.

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Answer:

0.967 m/s

1855 N

[tex]46.375\ \text{MPa}[/tex]

Explanation:

v = Final velocity

u = Initial velocity = 0

s = Displacement = 4.77 cm

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

From the kinematic equations

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]

The velocity of the object at the moment of impact is 0.967 m/s

Now

[tex]\Delta v[/tex] = Change in velocity = 1 m/s

t = Time taken = 20 ms

m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]

[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]

The force associated with a single step of the person is 1855 N

A = Area = [tex]0.4\ \text{cm}^2[/tex]

Stress is given by

[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]

The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]

The speed of object during falling is 0.967 m/s.

(A)  The magnitude of force  associated with this single step for a person is 1855 N.

(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

Given data:

The height of fall is, h = 4.77 cm = 0.0477 m.

The magnitude of downward velocity is, v' = 1.0 m/s.

The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].

The mass of person is, m = 74.2 kg.

The magnitude of force is, F' = 1880 N.

The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].

The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,

[tex]v^{2}=u^{2}+2gh[/tex]

Solving as,

[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]

Thus, the speed of object during falling is 0.967 m/s.

(A)

Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.

Apply the expression of average force as,

[tex]F =\dfrac{m'v'}{t}[/tex]

Solving as,

[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]

Thus, the magnitude of force  associated with this single step for a person is 1855 N.

(B)

The expression for the stress is given as,

[tex]\sigma = \dfrac{F'}{A}[/tex]

Solving as,

[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]

Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

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Answer:

Speed = 100 km/h

Explanation:

Given:

Speedometer reading = 100 kph for one lap

Assume;

Time taken to complete one lap = 1 hour

Computation:

Speed =  Distance / Time

Speed =  100 / 1

Speed = 100 km/h

Answer:

Q' = 320 J

Explanation:

The arrangements are given in attachment. The Fourier's Law of Heat Conduction states that:

Q = KAΔT/L

where,

Q = Heat Transferred

K = Constant (Conduction Coefficient)

A = Surface Area of Heat Transfer

ΔT = Difference of Temperature between two surfaces

L = Length between surfaces

For Arrangement AL

Q = 80 J

Therefore,

80 = KAΔT/L   ------------- equation (1)

Now, for arrangement B:

A' = 2 A (As, the rods are now connected in parallel with each other)

L' = L/2

Therefore,

Q' = K(2A)ΔT/(L/2)

Q' = 4 KAΔT/L

using equation (1)

Q' = 4(80 J)

Q' = 320 J

Answer:Acceleration implies any change in the velocity of the object with respect to time. Velocity is nothing but the rate of change of displacement. On the other hand, acceleration is the rate of change of velocity with respect to time.

Explanation:

Answer:

Nope

Explanation:

Answer:abc defg hijk lmnop qrs tuv wx y and z

Explanation: now i know my abc's

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

Since we are finding the new volume

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

From the question we have

[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]

We have the final answer as

Hope this helps you

Answer:

1.45

Explanation:

Refractive index of the liquid is given as;

Refractive index = [tex]\frac{speed of light in vacuum}{speed of light in liquid}[/tex]

But,

speed = [tex]\frac{distance}{time}[/tex]

Since a certain light of specific wavelength was used during the same time, let the time be represented by t.

So that;

speed of light in vacuum = [tex]\frac{3500}{t}[/tex]

speed of light in the liquid = [tex]\frac{2350}{t}[/tex]

Refractive index = [tex]\frac{3500}{t}[/tex] ÷ [tex]\frac{2350}{t}[/tex]

                            = [tex]\frac{3500}{t}[/tex] x [tex]\frac{t}{2350}[/tex]

Refractive index = [tex]\frac{3500}{2350}[/tex]

                          = 1.4536

                          = 1.45

The refractive index of the liquid is 1.45.

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47  m

Explanation:

U = 40m/s

V = 30m/s

T = 5 sec

A = ?

[tex]a = \frac{u - v}{t}[/tex]

[tex]a = \frac{40 - 30}{5}[/tex]

[tex]a = \frac{10}{5}[/tex]

[tex]a = 2[/tex]

since it's decreasing in speed, The acceleration will be " - 2.0ms^-2 " or " - 2.0m/s^2 "

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Answer:

I'm pretty sure its A

Explanation:

because its a reflection- Hope you get a good grade!

Answer:B

Explanation: The book is thinner making magnets attraction stronger, making the paper clip harder to move

Answer:

The total torque is 27.5 Nm

Explanation:

Given;

5.0-N force directed upward is applied 4.0 m to the left of the pivot,

5.0-N force directed upward is applied 1.5 m to the right of the pivot,

Taking the moment about the pivot, the total torque is given by;

τ = Fr

where;

F is the appllied force

r is the radius of the force arm

τ = (5 N x 4 m) + (5 N x 1.5 m)

τ = 27.5 Nm

Therefore, the total torque is 27.5 Nm

Answer:

The workdone is  [tex]W = 1.712 *10^{-20 } \ J[/tex]  

Explanation:

From the question we are told that

    The potential difference is  [tex]V = 107 mV = 107 *10^{-3} \ V[/tex]

Generally the charge on  [tex]Na^{+}[/tex] is  [tex]Q_{Na^{+}} = 1.60 *10^{-19 } \ C[/tex]

 Generally the workdone is mathematically represented as

         [tex]W = Q_{Na^{+}}V[/tex]

=>     [tex]W = 1.60 *10^{-19 } * 107 *10^{-3}[/tex]    

=>     [tex]W = 1.712 *10^{-20 } \ J[/tex]    

Answer:

I = 1.05x10⁻³ A

Explanation:

By definition, an electric current is the rate of charge flow at a given time:

[tex] I = \frac{q}{t} [/tex]

Where:

q: is the electrons charge = 1.602x10⁻¹⁹ C

t: is the time

In a circular motion, the time is given by:

[tex] t = T = \frac{2\pi}{\omega} = \frac{2\pi}{v/r} = \frac{2\pi r}{v} [/tex]

Where:

ω: is the angular speed = v/r

v: is the speed = 2.19x10⁶ m/s

r: is the radius = 5.29x10⁻¹¹ m

[tex] t = \frac{2\pi r}{v} = \frac{2\pi 5.29 \cdot 10^{-11} m}{2.19 \cdot 10^{6} m/s} = 1.52 \cdot 10^{-16} s [/tex]

Now, the effective current is:

[tex] I = \frac{q}{t} = \frac{1.602 \cdot 10^{-19} C}{ 1.52 \cdot 10^{-16} s} = 1.05 \cdot 10^{-3} A [/tex]  

Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.

I hope it helps you!                                

Answer:

a) they both will appear to have the same loudness

Explanation:

When we talk about how loud a sound is , we describe it in decibels (dB). The value of this unit on measurement in this question is what we are going to build our answer on.

Their sound intensity level is the same at 60db even though each of these separate sounds have different frequencies of 40Hz and 100Hz respectively.

The both of them will therefore have the same loudness actually as their sound intensity level has no difference bat 60db each.

Answer:

C. lenses refract light; mirrors do not

This question involves the concepts of reflection and refraction.

The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".

When it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the  bending of light rays, while passing through a medium, without any rebound or absorption.

Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.

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Answer:

9.8m/s^2 down  (option C)

Explanation:

The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.

Answer:

(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰

(a)(ii) The maximum torque is 0.132 Nm

(b) The orientation of the coil is 45⁰

Explanation:

Given;

diameter of the circular wire, d = 8.6 cm = 0.086 m

radius of the wire, r = d /2 = 0.043 m

number of turns, N = 15 turns

magnetic field, B = 0.56 T

The torque on the wire is given by;

τ = NIABsinθ

where;

θ is the orientation of the wire

(a) maximum torque occurs when the orientation of the wire is at 90⁰

The maximum torque is given by;

τ = NIABsin(90⁰)

τ = NIAB

τ = (15)(2.7)(π x 0.043²)(0.56)

τ = 0.132 Nm

(b)

71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm

[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]

Answer:

biking to school

Explanation:

plato

Answer:

this is a projectile

Answer:

D. projectile

Explanation:

Since it's dropped from a rest that means that it's velocity at the beginning is 0.

Answer:

This question is incomplete.

Explanation:

This question is incomplete. However, it should be noted that the voltage, V, across resistors in parallel is the same (although there currents are not the same). Thus, if a voltage has been provided, it remains the same but if not provided, you can solve for it using the formulas below

V = IR

where V is the voltage. I is the current and R is the resistance

R in parallel can be calculated as R = 1/R₁ + 1/R₂ + 1/R₃ + ......

Answer:

Diameter of the piston would be 0.71 m (71.1 cm)

Explanation:

From the principle of pressure;

[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]

Let [tex]F_{1}[/tex] = 2903.57 lb, [tex]F_{2}[/tex] = 24.41 lbs, [tex]r_{2}[/tex] = 3.26 cm = 0.0326 m.

[tex]A_{2}[/tex] = [tex]\pi r^{2}[/tex]

    = [tex]\frac{22}{7}[/tex] x [tex](0.0326)^{2}[/tex]

    = 0.00334 [tex]m^{2}[/tex]

So that:

[tex]\frac{2903.57}{A_{1} }[/tex] = [tex]\frac{24.41}{0.00334}[/tex]

[tex]A_{1}[/tex] = [tex]\frac{2903.57*0.00334}{24.41}[/tex]

    = 0.3973

[tex]A_{1}[/tex] = 0.4 [tex]m^{2}[/tex]

The radius of the piston can be determined by:

[tex]A_{1}[/tex] = [tex]\pi r^{2}[/tex]

0.3973 = [tex]\frac{22}{7}[/tex] x [tex]r^{2}[/tex]

[tex]r^{2}[/tex] = [tex]\frac{0.3973*7}{22}[/tex]

   = 0.1264

r = [tex]\sqrt{0.1264}[/tex]

 = 0.3555

r = 0.36 m

Diameter of the piston = 2 x r

                                     = 2 x 0.3555

                                     = 0.711

Diameter of the piston would be 0.71 m (71.1 cm).

Answer:

Explanation:

The giant star must have at least once been the more massive star and then subsequently transferred some of its mass to its companion, the other star.

The two stars would be around the same age, so the more massive one would have turned into a giant first before the other one did or even had a chance to

Answer:

So, at the depth of 24 cm below the surface of the glycerine the pressure is  2970 Pa. Hence, this is the required solution.

Explanation:

Given that,

Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.

The density of glycerine,  

We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :

h = 0.24 meters

or

h = 24 cm

Answer:

a

[tex]\lambda = 3.68 *10^{-36} \ m[/tex]

b

[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Explanation:

From the question we are told that

   The mass of the person is  [tex]m = 180 \ kg[/tex]

    The speed of the person is  [tex]v = 1 \ m/s[/tex]

    The energy of the proton is  [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]

Generally the de Broglie wavelength is mathematically represented as

      [tex]\lambda = \frac{h}{m * v }[/tex]

Here  h is the Planck constant with the value

      [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

So  

     [tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]

=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]

Generally the energy of the proton is mathematically represented as

         [tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]

Here [tex]m_p[/tex]  is the mass of proton with value  [tex]m_p = 1.67 *10^{-27} \ kg[/tex]

=>     [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]

=>   [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]

=>   [tex]v = 3.09529 *10^{7} \ m/s[/tex]

So

        [tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]

so    [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]

=>     [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Answer:

The value is  [tex]b = 0.00026 \ kg / m[/tex]

Explanation:

From the question we are told that

   The mass of the  Ping-Pong is  [tex]m = 2.3 \ g = 0.0023 \ kg[/tex]

    The terminal  speed is  [tex]v = 9.3 \ m/s[/tex]

    The drag force is  [tex]bv^2[/tex]

Generally the resultant force on the Ping- Pong is mathematically represented as

      [tex]F = mg - bv^2[/tex]

when  terminal velocity is  attained  , the resultant force is zero  so

      [tex]0 = mg - bv^2[/tex]

=>   [tex]b = \frac{m * g}{v^2}[/tex]

=>    [tex]b = \frac{0.0023 * 9.8}{ 9.3 ^2}[/tex]

=>    [tex]b = 0.00026 \ kg / m[/tex]