PLEASE HELP ME, A person has a 340 g can of organic frozen apple juice concentrate. In order to make the most dilute solution, which amount of water should he add? A. 709 mL B. 1064 mL C. 1419 mL D. 1774 mL

Answer:

The monkey will get hit by the dart. This is because when the dart is fired the force of gravity makes it fall from its original position. The monkey also has the same fall from the branches which means they will both be falling at the same gravitational rate.

This will eventually make the monkey to get hit by the dart.

Answer:

The current is  [tex]I = 0.4595 \ A[/tex]

Explanation:

From the question we are told that

    The length of the wire is  [tex]L = 64.7 \ cm = 0.647 \ m[/tex]

    The magnetic field is  [tex]B = 0.370 \ T[/tex]

      The magnetic force is  [tex]F = 0.110 \ N[/tex]

Generally the magnetic force exerted by the field is mathematically represented as

      [tex]F = IL B sin \theta[/tex]

Making I the subject we have

      [tex]I = \frac{F}{LB \ sin\theta}[/tex]

The angle here is  90° since the wire is perpendicular  to the direction of the magnetic field

      Substituting values

     [tex]I = \frac{0.110}{ (0.370) * 0.647 \ sin(90)}[/tex]

   [tex]I = 0.4595 \ A[/tex]

Answer:

I = 0.636*Imax

Explanation:

(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:

[tex]I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)[/tex]   (1)

I: intensity of light

Imax: maximum intensity of light

d: separation between slits = 0.200mm = 0.200 *10^-3 m

L: distance from the screen = 613cm = 0.613 m

y: distance to the central peak of the interference pattern

λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m

You replace the values of all variables in the equation (1):

[tex]I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}[/tex]

Hence, the fraction of the maximum intensity is I = 0.636*Imax

Answer:

259.52N

Explanation:

Force initiated on a spring that causes an extension is related by the expression below;

F= K×e

Where F is the Force

K is the spring constant

e is the extension caused by the spring.

By change of subject formula for K;

K = F/e

Now F is the same as weight and is given by mass× acceleration. In this case acceleration is g=9.8m/s2; this is because the child's mass is going to be under the influence of gravity as it swings up the harness}

Hence W =7.15×9.8=70.07N

Hence K = 70.07/0.27 =259.5185N/m

=259.52N/m to 2 decimal place.

Answer:

look at the data for the control group to see how much the fish grew without the new food.

Explanation:

i got it right

Answer:

Explanation:

The speed in meters per second is ...

  (90 km/h)(1000 m/km)(1 h/(3600 s)) = 25 m/s

The braking acceleration can be found from ...

  a = v²/(2d) = (25 m/s)²/(2×50 m) = 6.25 m/s²

Then the braking force is ...

  F = Ma = (1800 kg)(6.25 m/s²) = 11,250 N

The torque on the center of gravity is ...

  T = (11,250 N)(0.90 m) = 10,125 N·m

__

If we let x and y represent the normal forces on the front and rear wheels, respectively, then we have ...

  x + y = (10 m/s²)(1800 kg) = 18000 . . . . . newtons

  1.7x -1.3y = 10,125 . . . . . . . . . . . . . . . . . . . newton-meters

The latter equation balances the torque due to the wheel normal forces with the torque due to braking forces.

Multiplying the first equation by 1.3 and adding that to the second, we have ...

  3.0x = (1.3)(18,000) + 10,125

  x = 33,525/3 = 11,175 . . . . . . . . . . . newtons normal force on front tyres

  y = 18000 -11175 = 6,825 . . . . . . . .newtons normal force on back tyres

The least coefficient of friction is the ratio of horizontal to vertical acceleration, 6.25/10 = 0.625.

Answer:

a) [tex]a_{T}=8.50m/s^{2}[/tex], [tex]v_{T}=2.78 m/s[/tex] and [tex]N=279.83N[/tex]

b) [tex]a_{T}=0m/s^{2}[/tex], [tex]v_{T}=5.68 m/s[/tex] and [tex]N=795.2N[/tex]

Explanation:

a)

In order to solve this problem we need to start by drawing a diagram of what the problem looks like: See attached picture. Next, we can start by finding the initial height of the child which will happen at an angle of 20°. We can find this by subtracting the distance from the highest point and the initial point from the radius of the circle so we get:

[tex]h_{0}=2.5m-h_{1}[/tex]

so we get:

[tex]h_{1}=2.5m(sin (20^{o}))[/tex]

[tex]h_{1}=0.855m[/tex]

so

[tex]h_{0}=2.5m-h_{1}[/tex]

[tex]h_{0}=2.5m-0.855m[/tex]

[tex]h_{0}=1.645m[/tex]

once we got this value, we can find the final height the same way. This time the angle is 30° so we get:

[tex]h_{f}=2.5m-h_{2}[/tex]

[tex]h_{f}=2.5m-2.5m(sin 30^{o}))[/tex]

[tex]h_{f}=1.25m[/tex]

Once we have these heights, we can go ahead and use an energy balance equation to find the velocity at 30° so we get:

[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]

the initial kinetic energy is zero because its initial velocity is zero too, so the equation simplifies to:

[tex]U_{0}=U_{f}+K_{f}[/tex]

so now we substitute with the corresponding formulas:

[tex]mgh_{0}=mgh_{f}+\frac{1}{2}mv_{f}^{2}[/tex]

if we divided both sides of the equation by the mass, then the equation simplifies to:

[tex]gh_{0}=gh_{f}+\frac{1}{2}v_{f}^{2}[/tex]

and now we can solve for the final velocity so we get:

[tex]v_{f}=\sqrt{2g(h_{0}-h_{f}}[/tex]

and we can now substitute values:

[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m-1.25m)}[/tex]

which solves to:

[tex]v_{f}=2.78m/s[/tex]

which is our first answer. Once we got the velocity at 30° we can find the other data the problem is asking us for:

We can build a free body diagram (see attached  picture) and do a balance of forces so we get:

[tex]\sum{F_{x}}=ma_{T}[/tex]

in this case we only have one x-force which is the x-component of the weight, so we get that:

[tex]w_{x}=ma_{T}[/tex]

so we get:

[tex]mgcos\theta=ma_{T}[/tex]

and solve for the tangential acceleration so we get:

[tex]a_{T}=gcos\theta[/tex]

[tex]a_{T}=9.81m/s^{2}*cos(30^{o})[/tex]

[tex]a_{t}=8.50m/s^{2}[/tex]

Now we can find the centripetal acceleration by using the formula:

[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]

[tex]a_{c}=\frac{(2.78m/s)^{2}}{2.5m}[/tex]

so we get:

[tex]a_{c}=3.09m/s^{2}[/tex]

Next, we can find the normal force which is found by doing a sum of forces on y, so we get:

[tex]\sum{F_{y}}=ma_{c}[/tex]

so we get:

[tex]N-w_{y}=ma_{c}[/tex]

and we solve for the normal force so we get:

[tex]N=ma_{c}+w_{y}[/tex]

and substitute:

[tex]N=ma_{c}+mg sin\theta[/tex]

when factoring we get:

[tex]N=m(a_{c}+g sin\theta)[/tex]

and we substitute:

[tex]N=(35kg)(3.09m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]

which yields:

N=279.83N

b)

The procedure for part b is mostly the same with some differences due to the angle. First:

[tex]h_{0}=1.645m[/tex]

[tex]h_{f}=0m[/tex]

so

[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]

in this case the initial kinetic energy is zero because the initial velocity is zero and the final potential energy is zero because the final height is zero as well, so the equation simplifies to:

[tex]U_{0}=K_{f}[/tex]

so we get:

[tex]mgh_{0}=\frac{1}{2}mv_{f}^{2}[/tex]

so we solve for the final velocity so we get:

[tex]v_{f}=\sqrt{2gh_{0}}[/tex]

and we substitute:

[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m)}[/tex]

[tex]v_{f}=5.68m/s[/tex]

according to the free body diagram we get that:

[tex]a_{T}=gcos\theta[/tex]

[tex]a_{T}=9.81m/s^{2}(cos 90^{o})[/tex]

which yields:

[tex]a_{T}=0[/tex]

we can also find the centripetal acceleration, so we get:

[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]

[tex]a_{c}=\frac{(5.68m/s)^{2}}{2.5m}[/tex]

so we get:

[tex]a_{c}=12.91m/s^{2}[/tex]

and we can do a sum of forces on y to find the normal force:

[tex]\sum{F_{y}}=ma_{c}[/tex]

so we get:

[tex]N-w_{y}=ma_{c}[/tex]

and we solve for the normal force so we get:

[tex]N=ma_{c}+w_{y}[/tex]

and substitute:

[tex]N=ma_{c}+mg [/tex]

when factoring we get:

[tex]N=m(a_{c}+g)[/tex]

and we substitute:

[tex]N=(35kg)(12.91m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]

which yields:

N=795.2N

Following are the calculation to the angles:

For angle 30°:

consider the forces in tangential in direction:  

[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 30^{\circ} \\\\a_t = 8.496 \ \frac{m}{s^2}\\\\[/tex]

calculate the speed of the child:

[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]  

Integrating the equation:

[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{30} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 30-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 30-\sin 20) = \frac{V^2}{2}\\\\v=2.79 \ \frac{m}{s}\\\\[/tex]

consider the forces in the normal declaration:  

[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 30) = \frac{35 \times 2.78^2}{2.5}\\\\N= 279.87\ N[/tex]

For angle 90°:

consider the forces in tangential in direction:  

[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 90^{\circ} \\\\a_t = 0\ \frac{m}{s^2}\\\\[/tex]

calculate the speed of the child:  

[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]  

Integrating the equation:

[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{90} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 90-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 90-\sin 20) = \frac{V^2}{2}\\\\V=\sqrt{\frac{9.81 \times 2.5 (\sin 90-\sin 20) }{2}}[/tex]

   [tex]=\sqrt{\frac{9.81 \times 2.5 (1 -0.342)}{2}}\\\\=\sqrt{8.06}\\\\=2.83[/tex]

consider the forces in the normal declaration:  

[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 90) = \frac{35 \times 2.78^2}{2.5}\\\\N= \frac{35 \times 2.78^2}{2.5 \times 35\times 9.81\times 1}\\[/tex]

    [tex]= \frac{270.494}{858.375}\\\\=0.315[/tex]

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Answer:

Protons contribute the most to the size of the atom

Answer:

Protons

Explanation:

Protons contribute more to both the mass and size of an atom. Protons contribute more to an atom's mass while electrons contribute more to its size.

Complete Question

The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ . The average rate of electric energy consumption in one house is 1.0 kW .

If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.

Answer:

The area is  [tex]A = 1.26 *10^{7} m^2[/tex]

Explanation:

From the question we are told that

      The efficiency is  [tex]\eta =[/tex]12%

      The number of houses is  [tex]N = 350000[/tex]

       The light energy per day is [tex]E = 20 \ MJ[/tex]

       The average rating of electric energy for a house is  [tex]E_h = 1.0 \ k W = 1000W[/tex]

Generally the electric energy which the solar cells covering [tex]1 \ m^2[/tex] produces in a day is

       [tex]E_s = \eta * E[/tex]

           [tex]E_s = 0.12 * 20*10^{6}[/tex]

          [tex]E_s = 2.4 MJ m^{-2}[/tex]

Energy for required by one house for one day is  

        [tex]E_H = E_h * 1 \ day[/tex]  

       [tex]E_H = 1000 * 24 * 3600[/tex]  

        [tex]E_H = 86.4 MJ[/tex]

Energy needed for 350000 house is

      [tex]E_z = 86.4 *10^{6} * 350000[/tex]  

     [tex]E_z = 3.02 *10^{7} MJ[/tex]

The area covered is mathematically represented as

            [tex]A = \frac{3.02*10^{7} \ MJ}{2.4 \ MJ m^{-2}}[/tex]

           [tex]A = 1.26 *10^{7} m^2[/tex]

Answer:

B) The lighter gases boiled off of the protoplanets closest to the sun, leaving dust and metals behind to form the inner planets.

Explanation:

Inner planets are smaller and rockier than outer gas planets,outer planets are larger,because of their lower gravity they don't attract extreme amounts of gas in their planets.The four outer planets were so far from the Sun that its winds could not blow away their ice and gases

Answer:

A.  Crest

Explanation:

Longitudinal wave is a type of wave that is characterized by the particles of the medium's movement in a parallel direction in comparison to the direction in which wave travels, such that, in compression of longitudinal wave, the density of the wave medium is at its highest due to its closeness together than natural state, while in rarefaction, the density is at its lowest due to wave medium spread apart than normal.

Similarly, in Transverse wave, the crest of a wave implies the medium has reached the highest point while the trough of the wave depicts the lowest point the wave medium has reached.

Therefore, longitudinal wave's compression and rarefaction equates accordingly to the crest and trough of a transverse wave.

Hence, a compression of a longitudinal wave is like a CREST of a transverse wave.

Answer:

Electron cloud model describes the region in an atom which negatively charged and which has probability to find an electron. according to this model, an electron can move closer or away from the nucleus that is it can be inside the nucleus but according to Bohr's model, an electron is always at a fixed distance from the nucleus. Thus, it is the limitation of the electron cloud model but still it is a widely accepted model.

Answer:

(a)   ΔФ = -0.109W

(b)  emf = 28.43V

(c)   Iin = emf/R

Explanation:

(a) In order to calculate the magnetic flux you use the following formula:

[tex]\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)[/tex]   (1)

B: magnitude of the magnetic field = 1.40T

A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2

Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.

Then, you obtain:

[tex]\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W[/tex]

The change in the magnetic flux is -0.109 W

(b) During the rotation of the coil the emf induced is given by:

[tex]emf=-N\frac{\Delta \Phi}{\Delta t}[/tex]         (2)

N: turns of the coil = 60

ΔФ: change in the magnetic flux = 0.109W

Δt: lapse time of the rotation = 0.230s

You replace the values of the parameters in the equation (2):

[tex]emf=-(60)(\frac{-0.109W}{0.230s})=28.43V[/tex]

The induced emf is 28.43V

(c) The induced current in the coil is given by:

[tex]I_{in}=\frac{emf}{R}[/tex]      (3)

R: resistance of the coil     (it is necessary to have this value)

emf :induced emf  = 28.43V

Answer:

F/L =  8*10^-4 N/m

Explanation:

To calculate the magnitude of the force per meter in the central wire, you take into account the contribution to the force of the others two wires:

[tex]F_N=F_{1,2}+F_{2,3}[/tex]   (1)

F1,2 : force between first and second wire

F2,3 : force between second and third wire

The force per meter between two wires of the same length is given by:

[tex]\frac{F}{L}=\frac{\mu_oI_1I_2}{2\pi r}[/tex]

μo: magnetic permeability of vacuum =  4pi*10^-7 T/A

r: distance between wires

Then, you have in the equation (1):

[tex]\frac{F_N}{L}=\frac{\mu_oI_1I_2}{2\pi r}+\frac{\mu_oI_2I_3}{2\pi r}\\\\\frac{F_N}{L}=\frac{\mu_oI_1}{2\pi r}[I_2+I_3][/tex]

But

I1 = I2 = I3 = 10A

r = 10cm = 0.1m

You replace the values of the currents and the distance r and you obtain:

[tex]\frac{F_N}{L}=\frac{\mu_oI^2}{\pi r}\\\\\frac{F_N}{L}=\frac{(4\pi*10^{-7}T/A)(20A)^2}{2\pi (0.1m)}=8*10^{-4}\frac{N}{m}[/tex]

hence, the net force per meter is 8*10^-4 N/m

Answer:

B

Explanation:

Resolve the 75N force into 2 components; horizontal and vertical. And remember that there is no acceleration in the downward direction, so apply Newton's second law and equate it to 0.

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

Answer:a

Explanation: they are all positive

The equivalent vectors of both the vectors lie in the first quadrant. So, the right choice is a. first quadrant.

The resultant of the vectors can be calculated using the Parallelogram theorem of vector addition. In this theorem, a parallelogram is formed using the vectors, having the vectors as the adjacent side and the diagonal of this parallelogram is the resultant of both the vectors.

The vectors A and B are drawn at the given angle, angles of 20° and 80°  respectively. Using the vectors, A and B, the dotted parallelogram is drawn, and the vector C is the resultant.

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Answer:

r = 4.23 m

Explanation:

To find the radius of the circular path of the electron you use the following formula:

[tex]r=\frac{mv}{qB}[/tex]  (1)

This formula can be used because the motion of the electron is perpendicular to the direction of the magnetic field vector.

m: mass of the electron = 9.1*10^-31 kg

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 1.02*10^-5 T

v: velocity of the electron = 7.6*10^6 m/s

You replace the values of m, v, q and B in the equation (1):

[tex]r=\frac{(9.1*10^{-31}kg)(7.6*10^6 m/s)}{(1.6*10^{-19}C)(1.02*10^{-5}T)}\\\\r=4.23m[/tex]

hence, the raiuds of the orbit of the electron is 4.23m

your answer is b to be presided

Answer:

Explanation:

The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression

ω²R where ω is angular velocity and R is radius of the circular path .

angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.

ω = 2π / T

= 2π / 60

= .1047

angular acceleration =  .1047² x .22

= 2.41 x 10⁻³ rad / s² .

So, the required radial acceleration is [tex]a=0.21 cm/s[/tex]

The rate of change of velocity with respect to time.

Acceleration is a vector quantity.

The formula for the acceleration is,

[tex]a=\frac{V^2}{R}[/tex]

It is given that the radius is 22 cm

Now, substituting the given values into the above formula we get,

[tex]a=\frac{V^2}{22}[/tex]

Here, 1 second is equal to [tex]\frac{2\pi}{60}[/tex] then,

[tex]\\w=\frac{2\pi}{60}\\v=wR\\v=\frac{2\pi}{60}\times 19\\a=(\frac{2\pi}{60}\times 19)^2\times 19\\a=0.21 cm/s[/tex]

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Answer:

A

Explanation:

A)

INCORRECT

No, Gauss's Law can not be derived from Coulomb's Law alone. Since, Coulomb's Law provides electric field produced by a single point charge. Therefore, it is important to assume principle of super position along with Coulomb's law to get the field due to all charges. So, Gauss's Law can be proved by the help of both Super Position Principle and Coulomb's law.

B)

CORRECT

Yes, Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface.

C)

CORRECT

Yes, Coulomb's law can be derived from Gauss' law and symmetry.

D)

CORRECT

Yes, Gauss's Law applies to a closed surface of any shape.

E)

CORRECT

Yes, if a closed surface encloses no charge, the electric field is zero everywhere on the surface.

Answer:

608.4m/s

Explanation:

We are given that

Mass of Sleigh,M=1200 kg

Speed of Sleigh,u=322 m/s

Speed of jet,u'=680 m/s

Mass of jet,m=4800 kg

Total mass=M+m=1200+4800=6000 kg

We have to find the final velocity of the two objects after the collision.

The collision is inelastic .

By using law of conservation of momentum

[tex]Mu+mu'=(m+M)v[/tex]

Using the formula

[tex]1200\times 322+4800\times 680=6000v[/tex]

[tex]6000v=3650400[/tex]

[tex]v=\frac{3650400}{6000}[/tex]

[tex]v=608.4m/s[/tex]

Hence, the final  velocity of two objects after the collision=608.4m/s

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

ΔS = 49.13 ft

Answer:

(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ  (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.

Note: find an attached copy of a part of the solution to the given question below.

Explanation:

Solution

Recall that:

d = 6.6 um

λ₀ =d/10

λ₀ = 6.6 um

Now,

(a) We find the wavelength λ of the light in water.

Thus,

λ water = (λ₀ )/n

= 0.66/1.33

So,

λ water = λ = 0.496 um

(b) We find the phase difference between the waves from slit 1 and 2

Now,

if a <<d  and a<<λ

Then the path difference between the rays will be

Δ S₂N = Δ d sinθ

Thus, the phase difference becomes,

S = 2π Δ/λ is S= 2π Δ d sinθ/ λ

(c) The next step is to derive an expression for the intensity  I as function of O and other relevant parameters.

Now,

Let p be the point where these two rays interfere with each other.

Thus,

The electric field vector coming out from slot and and slot 2 is

E₁= E₀₁ cos (ks₁ p - wt) i

E₂ = E₀₂ cos (ks₂ p - wt) i

Note: Kindly find an attached copy of a part of the solution to the given question below.

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

[tex]f'=f(\frac{v-v_o}{v+v_s})[/tex]    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

[tex]v=\lambda f\\\\[/tex]

v: speed of sound

λ: wavelength = 0.700 m

[tex]f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz[/tex]

Next, you replace the values of all parameters in the equation (1):

[tex]f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz[/tex]

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

[tex]I=\frac{P}{4\pi r^2}[/tex]

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

[tex]I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}[/tex]

hence, the intensity is 8.12*10^⁻4 W/m^2

Answer:

The new volume is  [tex]11.6L[/tex]

Explanation:

[tex]pressure \alpha \frac{1}{volume}[/tex]

Given data

initial pressure p1= 100atm

initial volume v1= 3.7L

final pressure p2= 32 atm

final volume v2= ?

Apply Boyle's law we have

[tex]p1v1= p2v2\\[/tex]

[tex]v2= \frac{p1v1}{p2}[/tex]

Substituting our data we have

[tex]v2= \frac{100*3.7}{32} \\v2=\frac{370}{32} \\v2= 11.6 L[/tex]

The new volume is [tex]11.6L[/tex]

Answer:

protons

Explanation:

Answer:

The angle of deflection will be "1.07 × 10⁻⁷°".

Explanation:

The given values are:

Mass of a cow,

m = 1100 kg

Mass of bob,

mb = 1 kg

The total distance between a cow and bob will be,

d = 2 m

Let,

The tension be "t".

The angle with the verticles be "[tex]\theta[/tex]".

Now,

Vertically equating forces

⇒  [tex]T\times Cos \theta =mb\times g[/tex] ...(equation 1)

Horizontally equating forces

⇒  [tex]T\times Sin \theta = G\times M\times \frac{mb}{d^2}[/tex] ...(equation 2)

From equation 1 and equation 2, we get

⇒  [tex]tan \thata=\frac{G\times M}{g\times d^2}[/tex]

O putting the estimated values, will be

⇒  [tex]\theta = tan(\frac{6.674\times 10^{-11}\times 1100}{9.8\times 2^2} )[/tex]

⇒  [tex]\theta = 1.07\times 10^{-7}^{\circ}[/tex]

Answer:

A. mass

Explanation:

Mass determines the quantity of inertia for an object. Mass is the quantity that depends upon the inertia of an object. The inertia that an object has is directly proportional to the mass of the object.

An object that has more mass has a greater tendency as compared to the object that has less mass to resist changes in its state of motion.