Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly atoms are assembled in this protein synthesis.

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.

Given data:

Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].

Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].

Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].

The expression for the angular frequency of spring-mass system is,

[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex]  ......................................................(1)

Here, k is the spring constant.

Angular frequency is also expressed as,

[tex]\omega = 2 \pi f[/tex] .........................................................(2)

here, f is the linear frequency of spring-mass system.

And linear frequency is,

[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]

Then substitute equation (2) in equation (1) as,

[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]

Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.

Learn more about spring-mass system here:

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Answer:

1008.57kg/m3

Explanation:

Now the mass of fresh water is 250×1000 /1000000 = 0.25kg

Now the mass of salt water is

100×1030 /1000000 = 0.103kg

Note Density = mass / volume

Mass = volume × density

Note that converting from cm3 to m3 we divide by 1000000

Total mass = 0.25kg +0.103kg= 0.353kg.

Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3

Hence the density of the mixture= total mass / total volume

0.353kg/35 × 10^{-5}m3=1008.57kg/m3

Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters )  is the airplane above the ground 13 seconds after takeoff?

Answer:

The height  is [tex]H = 318.5 \ m[/tex]

Explanation:

From the question we are told that

   The speed at which the plane takes off is  [tex]u = 49 \ m/s[/tex]

      The angle at which it takes off is  [tex]\theta = 30 ^o[/tex]

        The time taken is [tex]t = 13 s[/tex]

The vertical distance traveled is  mathematically represented as

          [tex]H = u sin \theta t[/tex]

Substituting values  

         [tex]H = (49) * sin (30) *13[/tex]

        [tex]H = 318.5 \ m[/tex]

Answer:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Explanation:

You have three forces F1, F2 an F3 that produce the following  acceleration:

a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj

you know that force F1 and F2 are:

F1 = (30.0N)ˆi + (16.0N)ˆj

F2 = −(12.0N)ˆi + (8.00N)ˆj

and the force F3 is unknown:

F3 = F3x ˆi + F3y ˆj

The second Newton law is given by the following equation:

[tex]\vec{F}=m\vec{a}[/tex]

F: the total force = F1 +F2 + F3

m: mass of the object = 2 kg

By the properties of vectors you have:

[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]

Both x and y component must be equal in the previous equality, then you have:

[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]

Hence, the vector F3 is:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

Answer:

The curve should be banked at an angle of 13 degrees.

Explanation:

We have,

Radius of a highway curve is 274 m

Speed of car on this curve is 25 m/s

Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.

[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]

g = 10 m/s²

Plugging all the values in above formula,

[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]

So, the curve should be banked at an angle of 13 degrees.

Answer:

0.173 N.

Explanation:

We will calculate the mass and then use the following calculations on the surface of planet X that is :

                           [tex]W=mg[/tex]

We would use the following equation to get the value of g for planet X that is :

                   [tex]y_f-y_i=v_{yi}t+\frac{1}{2}gt^2[/tex]

Then, put the values in the above equation.

                          [tex]16=0+\frac{1}{2}\times g\times(2.90)^2[/tex]

                           [tex]\bf\mathit{g=3.80\;m/s^2}[/tex]

Now, we will measure the ball weight on planet X's surface:

                          [tex]m=\frac{100}{1000} \;\;\;\;\;\;\;\;\;\;[1kg=1000g][/tex]

Then, we have to put the value in the above equation.

                        [tex]W=0.1\times 1.73=0.173\:N[/tex]

Answer:

The number of molecules displaced in a vibration makes the amplitude of a sound.

Answer:

a = -0.8 m/s²

Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.

Explanation:

First we find the angular acceleration of the ball from the following formula:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = 7 rad/s

ωi = initial angular velocity = 13 rad/s

t = Time taken = 15 s

Therefore,

α = (7 rad/s - 13 rad/s)/15 s

α = - 0.4 rad/s

negative sign shows that acceleration is in opposite direction to the direction of motion.

Now, for the linear acceleration, we use the formula:

a = rα

where,

a = linear acceleration = ?

r = radius of circular path = length of rope = 2 m

therefore,

a = (2 m)(- 0.4 rad/s²)

a = -0.8 m/s²

Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.

Speed = (distance) / (time)

Speed = (

Velocity = speed, and its direction

The velocity of the plane is 10.2 miles per second East.

(about 48 times the speed of sound)

Answer:

a. SO2

Explanation:

Answer:

The distance from the cornea vertex to the retina is 2.36 cm

Explanation:

The question is incomplete.

The complete question is as follows;

A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.

If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?

Solution.

We use image-object reaction to calculate the distance from the cornea vertex to the retina.

Mathematically;

n1/s + n2/s’ = n2-n1/R

From the question, we identify the following;

n1 ; Refractive index of air = 1

n2 ; Refractive index of lens = 1.4

S ; Object Distance = 36 cm

S’ = ?

R ; Radius of curvature of the cornea = 0.65

Substituting these values into the equation above;

1/36 + 1.4/S’ = (1.4-1)/0.65

{S’+ 36(1.4)}/36S’ = 0.4/0.65

{S’ + 50.4}/36S’ = 0.62

S’ + 50.4 = 22.32S’

50.4 = 22.32S’ -S’

21.32S’ = 50.4

S’ = 50.4/21.32

S’ = 2.36 cm

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

Answer:

Loudness: decreases

Amplitude: decreases

Pitch: stays the same

Frequency: stays the same

Explanation:

1.

An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.

2.

Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.

3 and 4.

The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.

Hope this helps!

Answer:

Explanation:

Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...

Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...

Answer:

a = 5.34 m/s²

T = 37.86 N

Explanation:

This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:

a = g(m₂ - m₁)/(m₂ + m₁)

where,

a = acceleration of both masses = ?

g = 9.8 m/s²

m₂ = heavier mass = 8.5 kg

m₁ = lighter mass = 2.5 kg

Therefore,

a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)

a = (9.8 m/s²)(6 kg)/(11 kg)

a = 5.34 m/s²

The formula for tension in cable is derived to be:

T = 2m₁m₂g/(m₁ + m₂)

T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)

T = 37.86 N

Before the bus starts moving, the bus and the skater are both standing still.

When the bus starts moving and pulls away from the bus-stop, the skater stays right where she is.  

The people outside on the sidewalk see her standing still, and they see the bus moving out from under her.  

The other passengers on the bus see her rolling backwards down the aisle, toward the back of the bus.

Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1]  = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

Answer:

14.68m/s

Explanation:

As per the question, the data provided is as follows

Mass = M = 0.148 kg

Height = h = 11 m

Initial velocity = U = 0 m/s

Final velocity = V

Gravitational force = F

Mass = M

Based on the above information, the speed that hit to the ground is

As we know that

Work to be done = Change in kinetic energy

[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]

[tex]V^2 - U^2 = 2gh[/tex]

[tex]V^2 - 0 = 2gh[/tex]

[tex]V = \sqrt{2 g h}[/tex]

[tex]= \sqrt{2\times9.8\times11}[/tex]

= 14.68m/s

Answer:

To things that are positive charged

Answer:

1.25kg

Explanation:

Simply multiply volume and density together

Answer:

C = 74.53°

Explanation:

Let the magnitudes of 5.00 and 9.00 be vectors A and B respectively, hence the dot product of this vector is defined as

A.B = |A||B|cosC; let C be the angle between the vectors

12 = 5×9 cos C

Hence cos C = 12/45

C = cos^-1(12/45)

C = 74.53°

Answer:

A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)

Explanation:

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

Answer:

[tex]T_{1}[/tex] = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about [tex]T_{1}[/tex] we have:

(M + m) g * 0.5L - [tex]T_{2}[/tex](L - d) = 0

⇒ [tex]T_{2}[/tex] = [(M + m) g * 0.5L] ÷ (L - d)

[tex]T_{2}[/tex] = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

[tex]T_{2}[/tex]= 59.535 ÷ 2.4

[tex]T_{2}[/tex] = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

[tex]T_{1}[/tex] + [tex]T_{2}[/tex] = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

[tex]T_{1}[/tex] + 24.81 = 26.46 + 13.23

[tex]T_{1}[/tex] = 14.88 N

Answer:

(a)  emf = 1.18 mV

(b) counter-clockwise sense

Explanation:

(a) The induced emf is given by the following formula:

[tex]emf=-\frac{d\Phi_B}{dt}[/tex]     (1)

where:

ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)

A = πr^2

B = 0.800 T

You replace the expression for the magnetic flux in the equation (1):

[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]

A1: initial area

A2: final area

t2-t1: time interval  = 9.0s

Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m

[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]

You calculate the initial area A1:

[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]

After 9.0 second the circumference will be:

[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]

the new radius and the final area is:

[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]

[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]

Finally, you replace in the equation (1):

[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]

The induced emf in the circular loop is 1.18mV

(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.

Answer:

(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.

Answer:

173.75 km/hr in the NW direction.

Explanation:

Velocity is the time rate of change in displacement of a body. Mathematically:

v = d / t

where d = displacement

t = time

Therefore, the velocity of the car is:

v = 556 / 3.2 = 173.75 km/hr

The velocity of the car is 173.75 km/hr in the NW direction.

The velocity of a car will be "173.75 km/hr".

The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.

Displacement, d = 556 km

Time, t = 3.2 hours

We know the relation,

→ Velocity = [tex]\frac{Displacement}{Time}[/tex]

or,

→ V = [tex]\frac{d}{t}[/tex]

By substituting the values, we get

      = [tex]\frac{556}{3.2}[/tex]

      = [tex]173.75[/tex] km/hr

Thus the response above is right.

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