A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance, the Coulomb force between the two is to be maximized?

Answer:

The force will be zero

Explanation:

Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.

Answer:

volume of air breath = 495.75 mL per min

Explanation:

given data

oxygen required = 2.4 × 10² mL / Min = 240 mL

inhaled air = 20 %

exhaled air = 16 %

to find out

volume of air per breath

solution

we know here that breathes 12 time every minute

and  we consider here air breath = y per min

we know oxygen remain in body = 20 % - 16 % = 4%

and required 12 time air breath

so air required = 240 / 12 = 19.83 mL

so oxygen breath equation will be

y × 4% = 19.83

y = 19.83 / 0.04

y = 495.75

so volume of air breath = 495.75 mL per min

Answer:

it take for the proton to reach the negative plate is 0.1089 ns

Explanation:

given data

area = 21 cm² = 21 × [tex]10^{-4}[/tex] m²

separated d = 3 mm = 3 × [tex]10^{-3}[/tex] m

charge Q = 9.8 nC = 9.8 × [tex]10^{-9}[/tex] C

to find out

How long does it take for the proton to reach the negative plate

solution

we know that electric field between the plate is express as

electric field E = [tex]\frac{Q}{A*\epsilon}[/tex]   ..................1

here Q is charge and A is area and ∈ is permittivity of free space

put here value

electric field E = [tex]\frac{9.8 ×10^{-9}}{21×10^{-4}*8.85×10^{-12}}[/tex]

electric filed E = 527306.968 N/C

and

we know force on proton by electric field is

Force = electric filed × charge on proton

and force according to newton law

force = mass × acceleration

so

mass × acceleration = electric filed × charge on proton

here put all value and find acceleration

1.67 × [tex]10^{-27}[/tex] × acceleration = 527306.968 × 1.6 × [tex]10^{-19}[/tex]

acceleration = 5.05 × [tex]10^{13}[/tex] m/s²

and

final velocity of proton by equation of motion

v² - u² = 2as   ..........2

here u is zero and v is  final velocity and s is distance and a is acceleration

v² - 0 = 2(5.05 × [tex]10^{13}[/tex]) (3 × [tex]10^{-3}[/tex] )

velocity = 5.50 × [tex]10^{5}[/tex] m/s

so time is

time = [tex]\frac{velocity}{acceleration}[/tex]

time = [tex]\frac{5.50*10^{5}}{5.05*10^{13}}[/tex]

time = 0.1089 × [tex]10^{-9}[/tex] s

so  it take for the proton to reach the negative plate is 0.1089 ns

In a parallel plate, the capacitor plate is parallel while the plates are separated by distance. The time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.

What is a parallel plate capacitor?

It is a type of capacitor is in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.

A dielectric medium is a must in between these plates helps to stop the flow of electric current through it due to its non-conductive nature.

The following date given in the problem,

A id the area of plates  = 21 cm² = 21 × 10m²

d is the distance separated= 3 mm = 3 ×10⁻³m

q is the charge on capicitor = 9.8 nC = 9.8 ×10⁻⁹C

The formula of the electric field is as

[tex]\rm E = \frac{q}{A\varepsilon } \\\\ \rm E = \frac{9.8\times10^{-9}}{ 21\times10^{-4}\times8.85\times10^{-12}} \\\\\rm E = 527306\; N/C}[/tex]

The electric force is equal to the product of the charge and the electric field. Here the electric force is balanced by mechanical force.

[tex]\rm F= qE\\\\\rm{ F= ma}\\\\qE =ma\\\\ \rm{ a= \frac{qE}{m} }\\\\\rm{ a= \frac{1.6\times10^{-9}\times527306}{1.2\times10^{-10}} }\\\\ \rm a = 5.05\times10^{-31}\;m/sec^2[/tex]

From newton's third equation of motion

As the initial velocity is zero u=0

[tex]\rm{v^2= u^2-2as} \\\\ \rm{v^2= 2\times5.05\times10^-3\times10^{-3}} \\\\ \rm{v= \sqrt{2\times5.05\times10^-3\times10^{-3}}\\\\[/tex]

[tex]\rm v = 5.50\times10^5 \;m/sec.[/tex]

Time is defined as the ratio of velocity and acceleration.

[tex]\rm t= \frac{v}{a} \\\\ \rm t= \frac{5.50\times10^5}{5.05\times10^{-3}} \\\\ \rm t= 0.1089\times 10^{-9}}\;sec[/tex]

Hence the time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.

To learn more about the parallel plate capacitor refer to the link;

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Answer:

The distance is 3.1 m

Explanation:

The position vector of the fly relative to the corner of the wall is

r = (3.1, 0.5).

The distance of the fly from the corner will be calculated as the magnitude of the vector "r"

magnitude of vector  [tex] r = \sqrt{(3.1 m)^{2} + (0.5 m)^{2}} = 3.1 m[/tex]

Since the numbers to be added have only one decimal place 3.1 and 0.5, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.

Answer:

(a): emf = [tex]\rm 2\pi f NBA\sin(2\pi ft).[/tex]

(b): Amplitude of alternating voltage = 20.942 Volts.

Explanation:

Given:

(a):

The magnetic flux through the coil is given by

[tex]\phi = \vec B \cdot \vec A=BA\cos\theta[/tex]

where,

[tex]\vec A[/tex] = area vector of the coil directed along the normal to the plane of the coil.

[tex]\theta[/tex] = angle between [tex]\vec B[/tex] and [tex]\vec A[/tex].

Assuming, the direction of magnetic field is along the normal to the plane of the coil initially.

At any time t, the angle which magnetic field makes with the normal to the plane of the coil is [tex]2\pi ft[/tex]

Therefore, the magnetic flux linked with the coil at any time t is given by

[tex]\phi(t) = NBA\cos(2\pi ft)[/tex]

According to Faraday's law of electromagnetic induction, the emf induced in the coil is given by

[tex]e=-\dfrac{d\phi}{dt}\\=-\dfrac{d(NBA\cos(2\pi ft))}{dt}\\=-NBA(-2\pi f\sin(2\pi ft))\\=2\pi f NBA\sin(2\pi ft).[/tex]

(b):

The amplitude of the alternating voltage is the maximum value of the emf and emf is maximum when [tex]\sin(2\pi ft)=1.[/tex]

Therefore, the amplitude of the alternating voltage is given by

[tex]e_o = 2\pi ft NBA.[/tex]

We have,

[tex]N=100\\A=10^{-2}\ m^2\\B=0.1\ T\\f=2000\ rev/ min = 2000\times \dfrac{1}{60}\ rev/sec=33.33\ rev/sec.[/tex]

Putting all these values,

[tex]e_o = 2\pi \times 33.33\times 100\times 0.1\times 10^{-2}=20.942\ Volts.[/tex]

Answer:

Ball A

Solution:

According to the question:

Two balls A and B are kicked at angle [tex]20^{\circ}[/tex] and [tex]75^{\circ}[/tex] respectively.

The component of the velocity that provides the ball with the forward movement is the horizontal component, i.e., cosine of the angle.

Since, cos[tex]20^{\circ}[/tex] is greater than cos[tex]75^{\circ}[/tex] and thus Ball A will have more speed than ball B at the top most point of the trajectory.

Final answer:

Ball A will have a bigger speed at the highest point of its trajectory in comparison to ball B, as it was kicked at a lower angle of 20°, resulting in a larger horizontal component of the initial velocity when compared to ball B kicked at 75°.

Explanation:

The question is asking about the speeds of two balls at the highest point of their trajectories when they are kicked with the same initial velocity but at different angles. To determine the speed at the highest point, we can use the principles of projectile motion. The speed of a projectile at its highest point consists only of the horizontal component of the initial velocity because the vertical component equals zero at that instant.

Since both balls A and B were kicked with the same initial speed, the horizontal component of that speed can be found using the cosine of their respective launch angles. For ball A, this component is cos(20°) multiplied by the initial speed, and for ball B, it is cos(75°) multiplied by the initial speed. Because the cosine of 20° is greater than the cosine of 75°, ball A will have a larger horizontal component of velocity and thus a higher speed at the peak of its trajectory.

Answer:

P.E.      =   -0.449 J

Explanation:

Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.

Now Potential energy is defined by this formula,

P.E. = k q₁ q₂/ r

where P.E. is the potential energy.

k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)

q₁ = charge of one particle = +1.0μC

q₂ = charge of another particle = -5.0μC

r = distance = 0.1 m

Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m

          P.E.      =  -0.449 J

Answer:

50 km is equivalent to 50,000 metres

Answer:

[tex]I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}[/tex]

Explanation:

The intensity of the radiation emitted by a black body with a certain temperature T and frequency [tex]\nu[/tex], is given by Planck's law:

[tex]I(\nu,T)=\frac{2h\nu^3}{c^2}(\frac{1}{e^{\frac{h\nu}{kT}}-1})[/tex]

Considering the frequency range between [tex]\nu[/tex] and [tex]\nu + \delta \nu[/tex] and [tex]dI[/tex] the amount of energy emitted in this frequency range. Since an increase in frequency means a decrease in wavelength. Then:

[tex]I(a,T)da=-I(\nu,T)d\nu\\I(a,T)=-\frac{d\nu}{da}I(\nu,T)[/tex]

Now recall that [tex]\nu=\frac{c}{a}[/tex], differentiate both sides:

[tex]d\nu=-\frac{c}{a^2}da\\\frac{d\nu}{da}=-\frac{c}{a^2}[/tex]

Replacing this in previous equation:

[tex]I(a,T)=\frac{c}{a^2}I(\nu,T)\\I(a,T)=\frac{c}{a^2}(\frac{2h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{kT}}-1})[/tex]

Rewriting [tex]\nu^3[/tex] as [tex]\frac{c^3}{a^3}[/tex] and [tex]\nu[/tex] as [tex]\frac{c}{a}[/tex]

[tex]I(a,T)=\frac{c}{a^2}(\frac{2hc^3}{a^3c^2}\frac{1}{e^{\frac{hc}{akT}}-1})\\I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}[/tex]

Finally, we obtain Planck's radiation law in terms of wavelength

Answer:51.44 units

Explanation:

Given

x component of vector is [tex]-27.3\hat{i}[/tex]

y component of vector is [tex]43.6\hat{j}[/tex]

so position vector is

[tex]r=-27.3\hat{i}+43.6\hat{j}[/tex]

Magnitude of vector is

[tex]|r|=\sqrt{27.3^2+43.6^2}[/tex]

[tex]|r|=\sqrt{2646.25}[/tex]

|r|=51.44 units

Direction

[tex]tan\theta =\frac{43.6}{-27.3}=-1.597[/tex]

vector is in 2nd quadrant thus

[tex]180-\theta =57.94[/tex]

[tex]\theta =122.06^{\circ}[/tex]

Answer:

Explanation:

let the ball is thrown vertically downwards with velocity u.

So, initial velocity, = - u (downwards)

acceleration = - g (downwards)

let the velocity is v after time t.

use first equation of motion

v = u + at

- v = - u - gt

v = u + gt

So, it is a straight line having slope g and y intersept is u.

The graph I shows the velocity - time graph.

Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.

So, acceleration time graph is a starigh line parallel to time axis having slope zero.

the graph II shows the acceleration - time graph.

Use III equation of motion to find the final speed in terms height.

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

And the time is

v = u + gt

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

Final answer:

When the separation between two 1 kg masses is decreased by 2/3, the mutual gravitational force between them will increase. Conversely, when the separation is increased by a factor of 3, the mutual gravitational force will decrease.

Explanation:

When the separation between two 1 kg masses is decreased by 2/3, the mutual gravitational force between them will increase. As the separation decreases, the gravitational force between the masses becomes stronger. On the other hand, when the separation is increased by a factor of 3, the mutual gravitational force will decrease. As the separation increases, the gravitational force between the masses becomes weaker.

Answer:

d = 1.636 m

Explanation:

Initially charged sphere are far apart so their potential energy is zero . Kinetic energy will be of small sphere

K E of small sphere

= 1/2 m v²

= .5 x 3.1 x 10⁻³ x ( 5.6)²

=48.608 x 10⁻³ J

If d be the distance of closest approach between them , potential energy of

Charges

= k q₁ x q₂ / d²

[tex]\frac{9\times10^9\times2.6\times10^{-6}\times3.4\times10^{-6}}{d^2}[/tex]

= [tex]\frac{79.56\times10^{-3}}{d^2}[/tex]

Total kinetic energy at this point will be zero.

Applying the theory of conservation of mechanical energy

Potential energy at distance d = Kinetic energy at infinity

48.608 x 10⁻³ = [tex]\frac{79.56\times10^{-3}}{d^2}[/tex]

d = 1.636 m

Answer:

the buoyant force acting on the  ice cube is 0.119 N

Explanation:

given,

volume of ice cube = 15 cm³ = 15 × 10⁻⁶ m³

density of ice cube = 917 kg/m³

density of ethyl alcohol = 811 kg/m³

buoyant force = ?

The density of ice is more than ethyl alcohol hence it will sink.

buoyant  force acting on the ice cube = ρ V g

                                                                = 811 × 15 × 10⁻⁶ × 9.81

                                                                =0.119 N

so, the buoyant force acting on the  ice cube is 0.119 N

Answer:

D. pre schematic

Explanation:

During the _pre schematic____ developmental stage children are drawing their thoughts instead of merely observations.

Pre schematic stage in children is the age at which they drawing, symbols, spatial figures to express themselves. Their becomes more complex and full of lines as they age. They use single base lines, multiple base lines and fold up views.

Answer:

Discussed

Explanation:

Percentage error is the measure of how far is your measured value from the actual value. It give the magnitude of error in form of percentage.

it is calculated as ratio of  difference of calculated and the actual value to the actual value multiplied by 100.

If you are repeating the experiment the percent difference in your results will give the measure of how much variation there is in your own techniques, and therefore will give the precision of your experimental procedure.

Answer:

(a) 5.45 m

(b) no change

Explanation:

refractive index of water, n = 1.333

h = 2.4 m

Let the radius of circle is r.

Here we use the concept of total internal reflection.

Let i be the angle of incidence.

By using Snell's law

Refractive index of water with respect to air is the ratio of Sin of angle of incidence to the Sin of angle of refraction.

here refraction takes place from denser medium to rarer medium

So, [tex]\frac{1}{n}=\frac{Sin i}{Sin r}[/tex]

Here, angle r is 90 for total internal reflection

[tex]\frac{1}{1.33}=\frac{Sin i}{Sin 90}[/tex]

Sin i = 0.75

i = 48.6°

According to the triangle ΔOAB

[tex]tan i = \frac{AB}{OA}[/tex]

[tex]tan 48.6 = \frac{r}{2.4}[/tex]

r = 2.73 m

Diameter = 2 r = 2 x 2.73  = 5.45 m

Thus, the diameter of the circle is 5.45 m.

(b) As the value of r depends on the angle of incidence i and angle of incidence depends on the value of refractive index. As refractive index is constant, so the value of diameter remains same.

Due to refraction, a fish sees the world outside through a circle on the water's surface. The diameter of this circle, around 3.6 meters for a fish 2.4 m deep, decreases as the fish descends due to the narrowing range of incident light angles.

The question revolves around the concept of refraction, a phenomenon that occurs when light changes the medium, therefore, also changing direction. This comes into play when light travels from water to air (or vice versa), like how a fish observes the world outside the water body.

In response to the student's question part (a): When the fish is 2.4 m below the water's surface, it sees through a circle on the surface due to refraction. Knowing that the index of refraction of water is 1.333, the apparent depth of the fish to an observer on the surface would be the real depth divided by the refractive index, or around 1.8 meters. Therefore, the light from objects outside the lake would reach the fish within a circle having a diameter of about 3.6 meters (twice the apparent depth).

For part (b) of the question: If the fish were to dive deeper, the diameter of the circle through which it could observe the world outside would diminish. This is because as the actual depth increases, the range of incident angles for light rays entering the water from air, for which total internal reflection doesn’t occur, gets narrower. Therefore, the fish's field of view on the outside world decreases.

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Answer:

acceleration = 120 km/h²

so correct option is e 120 km/h²

Explanation:

given data

speed va = 30 km/h

speed vb = 60 km/h

time = 15 minutes

to find out

acceleration

solution

we know here car speed increases from 30 km/h to  60 km/h

sop change in speed is = vb - va

change in speed = 60 - 30 = 30  km/h

and

change in time = 15 minute = 0.25 hour

so acceleration is express as

acceleration = [tex]\frac{change in speed}{change in time}[/tex]  ........1

put here value

acceleration = [tex]\frac{30}{0.25}[/tex]

acceleration = 120 km/h²

so correct option is e 120 km/h²

Answer:

E = 2.88*10^6 N/C

v = 0.742 *10^6m/s

Explanation:

The two disks form a parallel-plate capacitor, which will cause an electric field equal to:

[tex]E = \frac{Q}{e_0 A}[/tex]

Where Q is the charge of the disks, A is the area of the disks and e_0 is the vacuum permisivity equal to 8.85*10^-12 C^2/Nm^2:

[tex]E = \frac{8*10^{-9} C}{(\frac{1}{4}\pi(0.02m)^2)*8.85*10^{-12}C^2/Nm^2}= 2.88*10^6 N/C[/tex]

Now, for the second part of the problem, we can use conservation of energy. The addition of potential and kinetic energy at launch point should be equal to the addition at the positive disk. Because the proton has positive charge, the potential energy of the proton will increase as its distance to the negative disk increases too. This is because the proton will be attracted towards the negative disk. The potential energy is given by:

[tex]E_p = V*q[/tex]

Where V is the difference in potential (voltage) between the disks. In a parallel-plate capacitor:

[tex]V = E*d[/tex], where d is the difference in position with the frame of reference. Our frame of reference will be the negative disk.

q is the charge of the proton.

The kinetic energy is given by:

[tex]E_k = \frac{1}{2}mv^2[/tex]

Then:

[tex]E_p_1+E_k_1 = E_p_2+E_k_2\\V_1*q+\frac{1}{2}mv_1^2= V_2*q+\frac{1}{2}mv_2^2 |v_2=0, E_p_1 = 0\\\frac{1}{2} mv_1^2 = E*d*q \\v = \sqrt{\frac{2Edq}{m}} = \sqrt{\frac{2*2.88*10^6N/m*0.001m*1.6*10^{-19}C}{1.67*10^{-27}}} = 0.742 *10^6m/s[/tex]

(a) The electric field strength between the disks is 2.88 N/C

(b) The launch speed of the proton to reach the positive disk is 7.43 x 10⁵ m/s.

The given parameters;

The electric field strength between the disks is calculated as follows;

[tex]E = \frac{Q}{\varepsilon _o A} \\\\E = \frac{Q}{\varepsilon _o \pi r^2} \\\\E = \frac{8 \times 10^{-9} }{8.85\times 10^{-12} \times \pi \times (0.01)^2} \\\\E = 2.88 \times 10^ 6 \ N/C[/tex]

The launch speed of the proton to reach the positive disk is calculated as follows;

[tex]K.E = W\\\\\frac{1}{2} mv^2 = Fd\\\\\frac{1}{2} mv^2 = Eqd\\\\mv^2 = 2Eqd\\\\v = \sqrt{\frac{2Eqd}{m} } \\\\v = \sqrt{\frac{2\times 2.88 \times 10^6 \times 1.6\times 10^{-19} \times 0.001 }{1.67 \times 10^{-27}} }\\\\v = 7.43 \times 10^5 \ m/s[/tex]

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Answer:

A) 594 Hz

B) 512 Hz

Explanation:

Howdy!

This is a typical Doppler effect problem, whose formula is:

[tex]f = \frac{v+v_{r} }{v+v_{s}} f_{0}[/tex]

Where :

[tex]v_{r}[/tex]

is the velocity of the receiver (in our case is equal to zero)

[tex]v_{s}[/tex]

is the velocity of the source (in our case is 92 km/h) which is positive when the source is moving away (second case) and negative otherwise.

[tex]v_{s}[/tex]

Is the velocity of the wave in the medium.

Before we start calculating we need to have the velocity of the source in the same units as the velocity of the waves:

92km/h = 92*(1000)/3600 m/s = 25.5 m/s

A)

When the source is moving towards the receiver the sign of the velocity  is negative, so:

[tex]f = \frac{343}{343-25.5} 550 Hz[/tex]

    f = 594.1 Hz

B) Now the velocity of the source must change sign:

[tex]f = \frac{343}{343+25.5} 550 Hz[/tex]

    f = 511.9 Hz

As a sanity check we know that when the source is moving towards the source the frequency is higher and when the source is moving away from the source the frequency is lower.

A) The frequency heard by a person standing beside the road in front of the car is 595.3 Hz. B) The frequency heard by a person standing beside the road behind the car is 514.5 Hz.

To solve this problem, we will use the Doppler effect formula, which relates the observed frequency to the actual frequency of the source, the speed of sound, and the relative velocity between the source and the observer. The formula for the observed frequency [tex]\( f' \)[/tex] when there is relative motion between the source and the observer is given by:

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

where:

f is the frequency of the source,

v is the speed of sound in the medium,

[tex]v_o[/tex] is the speed of the observer relative to the medium (positive if moving towards the source, negative if moving away),

[tex]v_s[/tex] is the speed of the source relative to the medium (positive if moving away from the observer, negative if moving towards).

A) For the person standing beside the road in front of the car:

The observer is stationary, so [tex]\( v_o = 0 \)[/tex].

The car is moving towards the observer, so [tex]\( v_s = -92 \text{ km/h} \)[/tex].

We need to convert the speed of the car from km/h to m/s to match the units of the speed of sound:

[tex]\[ 92 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 25.56 \text{ m/s} \][/tex]

Now we can plug the values into the Doppler effect formula:

[tex]\[ f' = \left( \frac{343 \text{ m/s} + 0}{343 \text{ m/s} - 25.56 \text{ m/s}} \right) \times 550 \text{ Hz} \][/tex]

[tex]\[ f' = \left( \frac{343}{317.44} \right) \times 550 \][/tex]

[tex]\[ f' \approx 595.3 \text{ Hz} \][/tex]

B) For the person standing beside the road behind the car:

The observer is again stationary, so [tex]\( v_o = 0 \)[/tex].

The car is moving away from the observer, so [tex]\( v_s = 92 \text{ km/h} \)[/tex] or [tex]\( 25.56 \text{ m/s} \)[/tex].

Using the Doppler effect formula:

[tex]\[ f' = \left( \frac{343 \text{ m/s} + 0}{343 \text{ m/s} + 25.56 \text{ m/s}} \right) \times 550 \text{ Hz} \][/tex]

[tex]\[ f' = \left( \frac{343}{368.56} \right) \times 550 \][/tex]

[tex]\[ f' \approx 514.5 \text{ Hz} \][/tex]

Answer:

a) yield strength

   [tex]\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa[/tex]

b) modulus of elasticity

strain calculation

[tex]\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046[/tex]

strain for offset yield point

[tex]\varepsilon_{new} = \varepsilon_0 -0.002[/tex]

                              =0.0046-0.002 = 0.0026

now, modulus of elasticity

 [tex]E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}[/tex]

    = 184615.28 MPa = 184.615 GPa

c) tensile strength

 [tex]\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa[/tex]

d) percentage elongation

[tex]\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%[/tex]

e) percentage of area reduction

[tex]\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%[/tex]                            

The mechanical properties of the tensile test specimen, such as yield strength, modulus of elasticity, tensile strength, percent elongation, and percent reduction in area, are determined by performing calculations based on the given dimensions, yield force, maximum load, and deformations observed during the test.

To determine the mechanical properties of a tensile test specimen, the following calculations are performed:

Answer:

Metal is more dense than water.

Explanation:

As we know, the molecules of the metal are tightly closed as compared to that of water and the density of a material is defined as the mass of the material per unit volume.

In a certain volume of metal, there are more numbers of molecules than that of water in the same amount of volume, therefore the density of metal is greater than that of water.

Also, according to Archimedes' principle, if there is an object in a fluid, then the buoyant force on that object is equal to the weight of the fluid that it displaces.

When the density of the object is larger than that of the fluid then it overcomes that buoyant force and sinks.

Thus, an object sinks in a fluid if its density is larger than that of the fluid and floats otherwise.

Since, the metal sink in water, it means Metal is more dense than water.

Answer:

[tex]|A|=\sqrt{1^2+(-2)^2+(3)^2}=3.74[/tex]

Explanation:

Given that,

Vector [tex]A=i-2j+3k[/tex]............(1)

We need to find the magnitude of vector A. Let us suppose vector in the form of, R = xi + yj + zk. Its magnitude is given by :

[tex]|R|=\sqrt{x^2+y^2+z^2}[/tex]

In equation (1),

x = 1

y = -2

z = 3

So, the magnitude of vector A is given by :

[tex]|A|=\sqrt{1^2+(-2)^2+(3)^2}[/tex]

|A| = 3.74

So, the magnitude of given vector is 3.74. Hence, this is the required solution.

Answer:

The electric flux through the car's bottom is [tex]1.75\times 10^{4} Wb[/tex]

Solution:

As per the question:

Magnitude of vertical Electric field, [tex]E_{v} = 1.85\times 10^{4} N/C[/tex]

The area of the rectangular surface of the car, [tex]A_{bottom} = 6\times 3 = 18 m^{2}[/tex]

Downward slope at an angle, [tex]\angle\theta = 19.0^{\circ}[/tex]

Now, the electric flux, [tex]\phi_{E}[/tex] is given by:

[tex]\phi_{E} = \vec{E_{v}}.\vec{A_{bottom}} = E_{v}A_{bottom}cos\theta[/tex]

Now, substituting the appropriate values in the above formula:

[tex]\phi_{E} = 1.85\times 10^{4}\times 18cos19.0^{\circ}[/tex]

[tex]\phi_{E} = 1.75\times 10^{4} Wb[/tex]

Answer:

The total time to reach ground is 24.89 seconds

Explanation:

Since the height of the helicopter is given by

[tex]h(t)=3.30t^{3}[/tex] thus at time t = 2.30 seconds the height of the helicopter is

[tex]h(2.30)=3.30\times (2.30)^{3}=40.151m[/tex]

The velocity of helicopter upwards at time t = 2.30 is given by

[tex]v=\frac{dh(t)}{dt}\\\\v=\frac{d}{dt}(3.30t^{3})\\\\v(t)=9.90t^{2}\\\\\therefore v(2.30)=9.90\times (2.30)^2=120.45m/s[/tex]

Now the time after which it becomes zero can be obtained using the equations of kinematics as

1) Time taken by the mailbag to reach highest point equals

[tex]v=u+gt\\\\0=120.45-9.81\times t\\\\\therefore t_{1}=\frac{120.45}{9.81}=12.28s[/tex]

2) Time taken by the mailbag to reach ground from a height of 40.151 meters equals

[tex]s=ut+\frac{1}{2}gt^{2}\\\\40.151=120.45t+4.9t^{2}[/tex]

Solving for t we get[tex]t_{2}=0.3289secs[/tex]

Now the total time of the journey is

[tex]\\\\2\times t_{1}+t_{2}\\\\=2\times 12.28+0.3289=24.89secs[/tex]

Answer:

Explained

Explanation:

Dynamically continuous innovation:

- Falls in between continuous and discontinuous innovation.

-Changes in customer habits are not as large as in discontinuous innovation and not as negligeble as in continuous innovation.

best example can as simple as transformation in  Television. New HD TVs have flat panels, wide screens and improved performance The Added features are considered dynamically improved.

Discontinuous innovation:

- discontinuous innovation comprise of new to world product only so they are discontinuous to every customer segment.

- these product are so fundamentally different from the the product that already exist that they reshape market and competition.

For example- the mobile and the internet technology are reshaping the market through regular innovation and change.

Dynamically continuous innovation involves moderate changes to an existent product, seen, for example, in the evolution from traditional cell phones to smartphones. Discontinuous innovation involves the creation of entirely new products, directly impacting consumer behavior, such as the invention of the internet.

Dynamically continuous innovation and discontinuous innovation are two concepts centered around improvements to existing products or introduction of radically new products. A dynamically continuous innovation signifies moderate changes to existing products. An example could be the transition from a traditional cell phone to a smartphone. The base product- phone remained the same, only the features were enhanced. This innovation still requires consumers to adapt to new functionalities, but to a lesser degree than discontinuous innovations.

On the other hand, discontinuous innovation involves creating products that did not exist before, fundamentally changing consumer behavior. An example would be the creation of the internet. Before its invention, nothing similar existed and it greatly impacted how people live and work.

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Answer:

Volume of Tank 2, V' = [tex]2.17 m^{3}[/tex]

Equilibrium Pressure, [tex]P_{eq} = 278.82 kPa[/tex]

Given:

Volume of Tank 1, V = 1 [tex]m^{3}[/tex]

Temperature of Tank 1, T = [tex]25^{\circ}C[/tex] = 298 K

Pressure of Tank 1, P = 500 kPa

Mass of air in Tank 2, m = 5 kg

Temperature of tank 2, T' = [tex]35^{\circ}C[/tex] = 303 K

Pressure of Tank 2, P' = 200 kPa

Equilibrium temperature, [tex]20^{\circ}C[/tex] = 293 K

Solution:

For Tank 1, mass of air in tank can be calculated by:

PV = m'RT

[tex]m' = \frac{PV}{RT}[/tex]

[tex]m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg[/tex]

Also, from the eqn:

PV' = mRT

V' = volume of Tank 2

Thus

V' = [tex]\frac{mRT}{P}[/tex]

V' = [tex]\frac{5\times 0.287\times 303}{200} = 2.17 m^{3}[/tex]

Now,

Total Volume, V'' = V + V' = 1 + 2.17 = 3.17[tex]m^{3}[/tex]

Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg

Final equilibrium pressure, P'' is given by:

[tex]P_{eq}V'' = m''RT_{eq}[/tex]

[tex]P_{eq} = \frac{m''RT_{eq}}{V''}[/tex]

[tex]P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa[/tex]

[tex]P_{eq} = 287.82 kPa[/tex]

Answer:

a. a = [tex]6.41 m/s^2[/tex]

b. T = -0.81 N

Explanation:

Given,

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

[tex]w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]

From the f.b.d. of the heavier block,

[tex]w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)[/tex]

From eqn (1) and (2), we get,

[tex]w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\[/tex]

[tex]\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\[/tex]

[tex]\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.[/tex]

part (b)

From the eqn (2), we get,[tex]T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N[/tex]

Answer:

(a) 2.793 m/s^2

(b) 0.335 N

Explanation:

Let a be the acceleration in the system and T be the tension in the string.

W1 = 3 N

W2 = 7 N

m1 = 3 /1 0 = 0.3 kg

m2 = 7 / 10 = 0.7 kg

θ = 30°

μ1 = 0.13, μ2 = 0.31

Let f1 be the friction force acting on block 1 and f2 be the friction force acting on block 2.

By the laws of friction

f1 = μ1 x N1

Where, N1 be the normal reaction acting on block 1.

So, f1 = 0.13 x W1 Cosθ = 0.13 x 3 x cos 30 = 0.337 N

By the laws of friction

f2 = μ2 x N2

Where, N2 be the normal reaction acting on block 2.

So, f2 = 0.31 x W2 Cosθ = 0.31 x 7 x cos 30 = 1.88 N

Apply Newton's second law for both the blocks

W1 Sin30 - T - f1 = m1 a

3 Sin 30 - T - 0.337 = 0.3 x a

1.163 - T = 0.3 a ..... (1)

W2 Sin30 + T - f2 = m2 a

7 Sin30 + T - 1.88 = 0.7 x a

1.62 + T = 0.7 a ..... (2)

By solving equation (1) and (2) we get

a = 2.793 m/s^2

(b) Put the value of a in equation (2), we get

1.62 + T = 0.7 x 2.793

T = 0.335 N

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

[tex]v^2=u^2+2as[/tex]

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

[tex]0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s[/tex]

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

[tex]v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s[/tex]

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

[tex]v=u+at[/tex]

where symbols have the usual meaning

Applying the given values we get

[tex]t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds[/tex]

Final answer:

To find the distance the cat strikes from the edge of the piano, we can break the motion into horizontal and vertical components, and solve for the time of flight and horizontal distance traveled. By substituting the given values into relevant equations and solving, we can determine that the cat will strike the floor approximately 1.71 meters from the edge of the piano.

Explanation:

To solve this problem, we can break the motion into horizontal and vertical components. The horizontal component of the initial velocity will not change, so the cat will travel a horizontal distance of 3m/s *cos(37°) * t, where t is the time of flight. To find t, we can use the equation h = v0y * t + (1/2) * g * [tex]t^2[/tex], where h is the vertical displacement, v0y is the vertical component of the initial velocity, and g is the acceleration due to gravity.

Substituting the given values into the equation, we get 1.3m = 3m/s * sin(37°) * t - (1/2) * 9.8m/[tex]s^2[/tex] * [tex]t^2[/tex]. Solving this quadratic equation, we find two possible values for t: approximately 0.364s and 0.203s. Since the cat jumps off the piano, we only consider the positive value of t. So the cat will strike the floor approximately 3m/s * cos(37°) * 0.364s = 1.71 meters from the edge of the piano.