The length of stereocilia actually vary from 10 to 50 micrometers. Again, assuming that they behave like simple pendula, over what frequency range of sound waves would they resonate. (The actual frequency range of human hearing is 20 Hz - 20,000 Hz, so there certainly must be other mechanisms involved in determining the frequency of these being the pendular model is rather oversimplified.)

a. About 70 Hz -160 Hz.
b. About 440 Hz - 1000 Hz.
c. About 20 Hz - 50 Hz.
d. About 0.07 to 0.16 Hz.

Answer:

Explanation:

this is like rubbing a balloon on your head to make your hair stand up.  Do that to the can.    The balloon is filled , ofc,  and then just rub the balloon on the can.  This will charge the can with static electricity.   :P

Answer:

Explanation:

Physics is the branch of science that studies the natural world and its rules and orders. It is one of the oldest sciences as ancient people study the stars and astronomy is considered part of Physics.

Answer:

w = w₀ / 2    the angular velocity is half the initial value.

Explanation:

We can analyze this exercise as if we added another disk to obtain a disk with twice the mass, for which if the system is two disks, the angular tidal wave is conserved

initial instant.

          L₀ = I₀ w₀

final moment

          L_f = I w

the moment is preserved

          L₀ = L_f

          I₀ w₀ = I w

the moment of inertia of a disk is

         I = ½ m R²

we substitute

          ½ m R² w₀ = ½ (2m) R² w

          w = w₀ / 2

for the case of a disk with twice the mass, the angular velocity is half the initial value.

Answer:

Any food made from wheat, rice, oats, cornmeal, barley, or another cereal grain is a grain product. Anything else is not

Explanation:

Answer:

B. Containing charged regions

Explanation:

The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions

So in the given case, the option b would be contributed to the molecules that have intermolecular forces

hence, the option b is correct

Answer:

Explanation:

The equation of the capacitance of the capacitor can be represented as:

[tex]C = \dfrac{\varepsilon_oA}{d}[/tex]

Also, the electric field between the plates  can be expressed as:

[tex]E = \dfrac{\sigma}{\varepsilon_o}[/tex]

[tex]= \dfrac{Q}{A \varepsilon _o} \ \ (surface \ charge\ density \ \sigma =\dfrac{Q}{\varepsilon_o})[/tex]

However;

when pushed to a distance d/2, the new capacitance of the capacitor is:

[tex]C = \dfrac{\varepsilon _oA}{(d/2))}[/tex]

[tex]=\dfrac{2 \varepsilon_oA}{d}[/tex]

[tex]=2C \\ \\ Q' = CV \\ \\ = 2CV \\ \\ =2Q[/tex]

SImilarly, the new electric field between the plates is:

[tex]E' = \dfrac{\sigma'}{\varepsilon_o} \\ \\ = \dfrac{Q'}{A \varepsilon_o} \\ \\ = \dfrac{2Q}{A \varepsilon_o} \\ \\ =2E[/tex]

For Battery disconnected:

The electric field between the plates doesn't rely upon the distance between the plates yet relies upon the magnitude of the charge. At the point when the battery is detached, the charge on the capacitor stays as before, so does the electric field.  

Therefore, the magnitude of the electric field when the battery is associated is twice however much the magnitude of the electric field when the battery is separated and disconnected.

Hence, the ratio is :

[tex]\dfrac{E_{connected}}{E_{disconnected}} =\dfrac{2E}{E} \\ \\ \dfrac{E_{connected}}{E_{disconnected}} = 2[/tex]

Hence, the ratio is = 2

Answer:

A)    t = 10.56 s, B)  x = 235 m, C) v = 25.2 m / s

Explanation:

A) We can solve this problem using kinematics expressions.

The distance traveled by the truck is

       x_c = v_c t

Distance traveled by the car.

The car must travel the distance that separates them from the truck x₀=25.0.   Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2  4.5 = 9 m, therefore the total length traveled by the car is

          x_t = x₁ + x₂ + x₃

          x_t = 26.5 + 20.7 +9 = 56.2 m

the distance traveled by the car when it returns to the lane is

         x_c + x_t = x₀ + v₀ t + ½ a t²

when the car passes the car the distance traveled by the two vehicles is the same, we substitute

         v_c  t + x_t = x₀ + v₀ t + ½ a t²

         ½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0

we substitute the values

         ½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =

         0.28 t² -31.2 = 0

         t = [tex]\sqrt{ \frac{31.2}{0.28} }[/tex]

         t = 10.56 s

This is the time it takes for the car to pass the truck and back into the lane.

B) the distance traveled is

        x = v₀ t + ½ a t²

        x = 19.3 10.56 + ½ 0.560 10.56²

        x = 235 m

C) the final velocity is

         v = v₀ + a t

         v = 19.3 + 0.560 10.56

         v = 25.2 m / s

To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

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Answer:

λ = a

Explanation:

This is a diffraction exercise that is described by the expression

          a sin θ = m λ

         sin θ  = m λ/ a

the first zero of the diffraction occurs for m = 1

        sin θ  = λ / a

angles are generally very small and are measured in radians

         sin θ  = θ  = y / x

we substitute

         [tex]\frac{y}{x} = \frac{\lambda}{a}[/tex]

the width of the central maximum is twice the distance to zero

         w = 2y

in the exercise indicate that this width is equal to twice the distance to the screen (2x)

          W = 2x

           2y = 2x

we substitute

          1 = λ/ a

          λ = a

we see that the width of the slit is equal to the wavelength used.

Answer:

v(t)=3t²+3, and a(t)=6t

v(0)=3cm/sec, and a(0)=0cm/sec²

v(1)=6cm/sec, and a(1)=6cm/sec²

Explanation:

Find velocity and acceleration functions ( v(t) and a(t) )

(Relevant background: Let's say f(t) is a function of y with respect to x. Think of the derivative of this function, f'(t) , as representing the rate at which y changes with respect to x)

s(t) is a function of distance with respect to time. Therefore, s'(t) - the derivative of this - represents the rate at which distance changes with time, which is just the definition of velocity. So we can say velocity v(t) = s'(t).

s(t)=t³+3t+3

s'(t)=v(t)=3t²+3

Similarly, if v(t) is a function of speed with respect to time, then v'(t) represents the rate at which speed changes with time, which is acceleration. So we can say that acceleration a(t)=v'(t)=s''(t)

v(t)=3t²+3

v'(t)=a(t)=6t

Find velocity and acceleration at t=0 and t=1

t=0

v(t)=3t²+3

v(0)=3(0²)+3

v(0)=3 cm/sec

a(t)=6t

a(0)=6(0)

a(0)=0 cm/sec²

t=1

v(t)=3t²+3

v(1)=3(1²)+3

v(1)=3+3

v(t)=6 cm/sec

a(t)=6t

a(1)=6(1)

a(1)=6 cm/sec²

The function ​s(t)=−t3+3t+3 gives the distance from a starting point at time t of a particle moving along a line.

v(t)=3t²+3, and a(t)=6t

v(0)=3cm/sec, and a(0)=0cm/sec²

v(1)=6cm/sec, and a(1)=6cm/sec²

Relevant background: Let's say f(t) is a function of y with respect to x. Think of the derivative of this function, f'(t) , as representing the rate at which y changes with respect to x.

s(t) is a function of distance with respect to time. Therefore, s'(t) - the derivative of this - represents the rate at which distance changes with time, which is just the definition of velocity. So we can say velocity v(t) = s'(t).

s(t)=t³+3t+3

s'(t)=v(t)=3t²+3

Similarly, if v(t) is a function of speed with respect to time, then v'(t) represents the rate at which speed changes with time, which is acceleration. So we can say that acceleration a(t)=v'(t)=s''(t)

v(t)=3t²+3

v'(t)=a(t)=6t

Find velocity and acceleration at t=0 and t=1

t=0

v(t)=3t²+3

v(0)=3(0²)+3

v(0)=3 cm/sec

a(t)=6t

a(0)=6(0)

a(0)=0 cm/sec²

t=1

v(t)=3t²+3

v(1)=3(1²)+3

v(1)=3+3

v(t)=6 cm/sec

a(t)=6t

a(1)=6(1)

a(1)=6 cm/sec²

Therefore, The function ​s(t)=−t3+3t+3 gives the distance from a starting point at time t of a particle moving along a line.

v(t)=3t²+3, and a(t)=6t

v(0)=3cm/sec, and a(0)=0cm/sec²

v(1)=6cm/sec, and a(1)=6cm/sec²

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Weathering is the breaking down or dissolving of rocks and minerals on Earths surface.

Erosion is the process by which the surface of the Earth gets worn down. Erosion can be caused by natural elements such as wind and glacial ice

Deposition is the dropping of sediment by wind, water, ice, or gravity.

Answer:

Weathering: refers to the process of breaking down and disintegrating rocks, minerals, soils, as well as several other materials.

Erosion: refers to the process of wearing down the surface of the earth due to glacial ice, wind and other natural elements.

Deposition: refers to the geological process, of sediments and soil, added to landforms due to wind, ice and other natural elements to build up layers od sediments.

Explanation:

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          [tex]\frac{1}{f_1} = \frac{1}{q}[/tex]

           q = f₁

2) for an object located at p = 25 cm

            [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]

        [tex]\frac{1}{f_2}[/tex] = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = [tex]\frac{1}{f}[/tex]

where the focal length is in meters

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

Answer:

K = 1800 kJ

Explanation:

Given that,

The speed of the object, v = 30 m/s

Mass of the object, m = 4000 kg

We need to find the kinetic energy of the object. The formula for the kinetic energy is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 4000\times 30^2\\\\K=1800000\\\\or\\\\K=1800\ kJ[/tex]

So, the required kinetic energy is equal to 1800 kJ.

Answer:

Ratio of kinetic energy of object A to B = 1:5

Explanation:

Given:

Mass of object A = 200 kg

Mass of object B = 1,000 kg

Find:

Ratio of kinetic energy of object A to B

Computation:

Kinetic energy = (1/2)(m)(v²)

Kinetic energy of object A = (1/2)(200)(v²)

Kinetic energy of object A = (100)(v²)

Kinetic energy of object B = (1/2)(1,000)(v²)

Kinetic energy of object B = (500)(v²)

Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B

Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)

Ratio of kinetic energy of object A to B = 100 / 500

Ratio of kinetic energy of object A to B = 1/5

Ratio of kinetic energy of object A to B = 1:5

Answer:

No, the mass will never come to rest

Explanation:

It is so because even at arbitrarily small distance it will experience some amount of force (irrespective of how small the value of force is).

This does not allow the mass to become stationary or in a equilibrium state as it is still subject to some amount of force.

Hence, the the mass will never come to rest

Answer:

I=R×V

360×120 V

=43,200 A

Answer:

A.)0.33 A

Explanation:

i just took the quiz 2021

Answer:

8.33 meters/sec.

time = 30 sec. 30 sec. = 8.33 meters/sec.

I could make a poem for you if you actually gave the words...... what 10 words do i need to incorporate???☹︎

Answer:

Normal force=mg

Explanation:

The reaction force of weight is the normal force.

in order to find the normal for we need to write all the forces and set it equal to the net force:

N-mg=ma (since it is a constant speed the a=0)

N=mg

Answer:

sudden cardiac arrest

Explanation:

Answer:

a) 5.23599

b) 0.314159

c) 1.8326

Explanation:

Answer:

a. 5.236 rad

b. 0.314 rad

c. 1.833 rad

Explanation:

300° × π/180 = 5.236rad

18° × π/180 = 0.3142rad

105° × π/180 = 1.833rad

Answer:

F = 812.25 N

Explanation:

Given (convert to SI units):

m = 57 g = 0.057 kg

[tex]v_{1}[/tex] = -25 [tex]\frac{m}{s}[/tex]

[tex]v_{2}[/tex] = 32 [tex]\frac{m}{s}[/tex]

t = 4 ms = 0.004 s

Find acceleration:

a = [tex]\frac{v_{2}-v_{1} }{t}[/tex] = [tex]\frac{32-(-25)}{0.004}[/tex] = 14250

Find force:

F = ma = 0.057(14250) = 812.25 N

Answer:

Yes, it would make it back up.

Explanation:

If it has 100,000 Joules of gravitational potential energy at the top of the hill, by the time the cart gets to the bottom, it will become PE = 0, KE = 90,000 since 10% of 100,000 is 10,000. The cart only requires 80,000J to climb back up so it should easily do so.

I didn't quite understand if the 10% energy loss is total, or every time it goes up or down, but it isn't a problem because 10% of 90,000 is 9,000, which means it would have 81,000J of energy on the way back up IF it loses energy due to friction on the way back up also.

The only physical law you need to prove this is the Law of Conservation of Energy: no energy is lost, only transformed; 10% of the energy becomes heat, the rest remains mechanical energy, which is the reason why the reasoning above works.

Answer:

0.1 kg/s.

Explanation:

The density of water, d = 1000 kg/m³

Volume, V = 2 L

Time, t = 20 s

We need to find the mass flow rate from the faucet. We know that the density of an object is given by :

[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]

So, the mass flow rate is equal to 0.1 kg/s.

Answer:

the speed of the fetal heart wall at the instant is 0.0325 m/s

Explanation:

Given the data in the question;

f₀ = 2.50 MHz = 1.80 × 10⁶ Hz

f[tex]_B[/tex]  = 78 beat per sec ( Hz )

V = 1500 m/s

the speed of the fetal heart wall at the instant this measurement is made = ?

now, if v is the magnitude of hart wall speed, V is speed of sound and f[tex]_h[/tex] the frequency the heart receives( and reflects), f₀ is original frequency and f, is reflected back;

so

heart is moving observer and device is stationary source

f[tex]_h[/tex] = ((V + v)/v )f₀

Heart is moving source and device is stationary observer

f' = (V/(V-v ))f[tex]_h[/tex]

Beats

f[tex]_B[/tex] = f₀ - f' = f₀ - f₀( V+v / V-v ) = f₀( 2v / V-v )

so we solve for v

v = V( f[tex]_B[/tex] / ( 2f₀ + f[tex]_B[/tex] )

so we substitute

v = 1500 ( 78 / ( (2×1.80 × 10⁶) + 78 )

v = 1500 ( 78 / ( 3,600,000 + 78 )

v = 1500 ( 78 / 3,600,078 )

v = 1500 ( 2.1666 × 10⁻⁵ )

v = 0.0325 m/s

Therefore,  the speed of the fetal heart wall at the instant is 0.0325 m/s

Answer:

Right

Explanation:

Coil move right yes