anyone know what it is?

the change in kinetic energy of the car is None of these

perfectly inelastic collision - It is the physical process. Of one object which strike against another without conserving the kinetic energy of to object.

m₁ = 1,500 kg  (mass of car)

v₁ =  4.00m/s     (velocity of car)

m₂= 50000kg    (mass of truck)

v₂ = 1.80m/s       (velocity of truck)

for perfectly elastic collision

[tex]v_{f}[/tex]=?

[tex]m_{1} .v_{1,i} + m_{2} .v_{2,i} =( m_{1} + m_{2} ).v_{f}[/tex]

1500*4 + 50000*1.80

= (1500 + 50000) [tex]v_{f}[/tex]

[tex]v_{f}[/tex] = 1.864 m/s

Initial kinetic energy is

K.E (i) = [tex]\frac{1}{2} m_{1} .v_{1,i} ^{2} + \frac{1}{2} m_{2} .v_{2,i}^{2}[/tex]

[tex]\frac{1}{2}[/tex]× [(1500*4²) +(50000)*(1.8)²

= 93000J

Final kinetic energy is  

K.E = [tex]\frac{1}{2}(m_{1} +m_{2} )v_{f}^{2}[/tex]

[tex]\frac{1}{2}[/tex]× ((1500+50000)*(1.864 )²

= 89476 J

Change in kinetic Energy

K.E = K.E (f) -  K.E (i)  

 = 89476 -93000

= -3524 J

the change in kinetic energy of the car is None of these

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The content of matter in a body is called mass. Weight can be defined as the amount of force required to accelerate an object with mass, while volume is the space occupied by the mass.

On comparing with 1-kg block of solid iron, a 2-kg block of solid iron, mass is different for both the blocks.

So (a) mass cannot be the correct option.

According to this formula ,[tex]Density = \frac{Mass}{Volume}[/tex] , Density for both the blocks would be same, so volume will be different for both the blocks because of the different masses.

So (b) Volume cannot be the correct option.

Weight is mass dependent therefore

(c) weight cannot be the correct option

Since (a), (b), (c), are incorrect so (d) can not be the correct answer

Therefore on Comparing a 1-kg block of solid iron, with a 2-kg block of solid iron, (e) none of these would be same.

Correct option is (e)

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The correct options are B and C

Let us assume the amplitude of the two sinusoidal waves to be A1 and A2

Since, it is mentioned that the waves are identical, therefore let us assume A1=A2= a

The resultant wave of any two waves is given by the formula:

Resultant wave , R = [tex]\sqrt{A1^2+A2^2+2A1A2cos\alpha }[/tex]

where A1 and A2 are amplitudes of wave 1 and 2 and [tex]\alpha[/tex] is the angle between the waves

Now as per the formula, it is the the factor 2cos[tex]\alpha[/tex] that will decide that amplitude is greater than original or not

Hence we know that maximum value of cos [tex]\alpha[/tex] is when [tex]\alpha[/tex] is 0 degrees

Now it is known that value of cos[tex]\alpha[/tex] varies from 1 to 0 for o degrees to π/2

and the value varies from 0 to -1 for angle π/2 to π angle

Case 1: when [tex]\alpha[/tex]=0 degree

           Resultant wave R = [tex]\sqrt{A1^2+A2^2+2cos0A1A2}= \sqrt{2A^2+2A^2}=\sqrt{4A^2} =2A[/tex]

Case 2: When [tex]\alpha[/tex] is less than 90

let us consider [tex]\alpha = 45 degree[/tex]

 Resultant wave R =

[tex]\sqrt{A1^2+A2^2+2cos45A1A2}= \sqrt{2A^2+2A^2/2}=\sqrt{3A^2} =\sqrt{3} A[/tex]

case 3:

When [tex]\alpha[/tex] is less than 120

[tex]\sqrt{A1^2+A2^2+2cos120A1A2}= \sqrt{2A^2-2A^2/2}=A[/tex]

case 4: when [tex]\alpha[/tex] is less than 180 degrees

the value will be 0

Hence only option b and d are correct

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An astronaut could 6 times higher on the moon than on earth if his takeoff speed is the same in both locations.

From lunar circle, space explorers guided cameras through the window of their space apparatus toward catch photographs of the moon's surface. The nearest look we've had at the moon came from the sendoff of NASA's Apollo program during the 1960s. Somewhere in the range of 1967 and 1972, a progression of missions handled the primary men on the moon. At the point when the following space explorer to arrive at the moon strolls on the lunar surface in 2024, she'll confront radiation levels multiple times higher than on Earth. While Apollo mission space travelers. The prompt surface was extremely fine, as best we could see peering down from 15 feet. Off somewhere out there was an exceptionally clear skyline, perhaps with a rock.

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The speed of the wave in the rod is

[tex]The speed of the wave in the rod is $v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$[/tex]

5044m/s

A sound wave is the pattern of disruption brought on by the movement of energy moving through a medium as it propagates away from the source of the sound (such as air, water, or any other liquid or solid substance). Pressure waves are produced when an item vibrates, such as a ringing phone, and these waves are known as sound waves. The surrounding medium's particles are disturbed by the pressure wave, and those particles disrupt the particles next to them, and so on. Like ocean waves, the disturbance's pattern causes outward movement in all directions. Usually in all directions and with decreased intensity as it gets further away from the source, the wave transmits the sound energy through the medium.

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The electric field on the axis of the uniformly charged ring at the center of the ring at 5.00cm is 2.41×〖10〗^7 N/C

Where a positive charge (q) is given, always note that the charges point away from the electric field. The expression used to calculate the electric field away from the center of the ring is given by the equation below,

E=(K_e×b)/(r^3×Q)

Where

E is Electric field

Ke is electric field constant

a  is radius of the charge q from the center of the circle

Q is electric charge = 75.0μC

The radius of the charge is a = 10cm= 0.1m

Distance from the electric field = 5cm = 0.05m

Ke is a constant given as  8.99×〖10〗^(9  ) Nm^2/c^2

E=(8.99×〖10〗^(9  ) Nm^2/c^2×0.05)/(〖0.1〗^3×75×〖10〗^(-6) )

E=2.41×〖10〗^7N/C

Therefore, the electric field away from the center of the ring is 2.41×〖10〗^7N/C away from the center of the ring

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Ricardo came up with the three different paper airplane designs. Ricardo plans to fly each paper airplane five times to see which design flies the furthest. Ricardo should calculate the aircraft's flight distance after each flight and record it.

Two of the most popular methods in agricultural surveying involve taking measurements of angles and distances with simple equipment. even when the equipment is simple. A high level of precision can be reached with perseverance and practice.

The primary method for measuring distance is (1) pace. The mileage counter. (3) taping or chaining four phases and (6) methods for electronic distance measurement, optical range finders (EDM).

It will be helpful to first examine the common units of distance (displacement). Common English units are as follows:

A foot is made up of 12 inches (in)

Yd (yard) equals 3 ft (ft)

The length of one rod is 16.5 feet and 5.5 yards (yd)

5280 feet, 1760 yards, and 320 rods make up a mile.

The units cancellation method is very helpful when converting between different units of measurement.

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The Magnitude of force on the charged particle is -

F = (12.8 x 10⁻¹⁹C)i - (3.2 x 10⁻¹⁹C)j - (6.4 x 10⁻¹⁹C)k

We have a positively charged particle that moves through a region where both a uniform magnetic field and a uniform electric field exist.

We have to determine the total force on the moving particle.

The force exerted on a charged particle (q) moving with velocity (v) through an Electric field (E) and Magnetic field (B) is called Lorentz force. Mathematically -

|F| = qE + qv x B

According to the question, we have -

q = 3.20 × 10⁻¹⁹C

v = (2i + 3j - k) m/s

B = (2i + 4j + k) T

E = (4i - j - 2k) V

Now, using the formula to calculate Lorentz force -

F = q(4i - j - 2k) + q(2i + 3j - k) x (2i + 4j + k)

F = q(4i - j - 2k) + (2qi + 3qj - qk) x ((2i + 4j + k)

F = q(4i - j - 2k) + 0

F = q(4i - j - 2k)

F = (12.8 x 10⁻¹⁹C)i - (3.2 x 10⁻¹⁹C)j - (6.4 x 10⁻¹⁹C)k

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(ii) The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.

Now, since A = 0.1 m

Then, 2A = 2(0.1) = 0.2 m

As a result, the net displacement at the string's intersection of two pulses is 0.2 m.

The correct option is (c).

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The car's acceleration would be for a mass of 4 kg is 3 m/s²

Given the information below:

12 Newtons of force.

Increasing speed Equals

By using Newton's Second Law of Motion, we can determine the mass of the car:

According to Newton's second law of motion, force is equal to the rate at which momentum changes. Force is defined as mass times acceleration for a constant mass.

force= mass* acceleration

This formula provides Newton's Second Law of Motion mathematically;

12= mass*3

mass=12/ 3

mass=4 kilogram

By changing the parameters in the formula, we obtain;

= 4 kilos of mass.

As a result, the vehicle with a mass of 4 kg would accelerate at 3 m/s².

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The expansion of both will be the same. Then the correct option is C.

Thermal expansion, which often excludes phase transitions, is the propensity of matter to alter its form, area, volume, and concentration in response to a variation in heat.

The same metal is used to create two identically sized spheres, but only one is solid and the other is hollow. The same rate of warming is applied to the spheres.

The formula for the thermal expansion is given as,

[tex]\rm \alpha _L = \dfrac{1}{L} \cdot \dfrac{dL}{dT}[/tex]

[tex]\alpha_{L}[/tex] = thermal expansion

L = particular length measurement

dL = change in length

dT = change in temperature

If the material of the sphere is the same, then the coefficient of the expansion of the sphere is the same, then the expansion of both will be the same.

Both will expand at the same rate. Then, C is the right answer.

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Answer:

•So he stays healthy.

•He has energy and time to do it.

The Big Bang Theory is an example of the universe originated from the explosion and expansion of all matter and energy

The Big Bang Theory is the way that astronomers explain how the universe began as a tiny, dense, fireball that exploded 13.8 billion years ago. It includes Albert Einstein's general theory of relativity along with standard theories of fundamental particles.

The theory was born of the statement that other galaxies are moving away from our own at countless speed in all directions, as if they had all been forced by an ancient explosive force.

The solar system is the gravitationally bound system of the sun, the objects that orbit it or travel around the sun. It consists of an average star (sun), the planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto) and their moons, dwarf planets and countless asteroids, comets, and other small icy objects.

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The car’s velocity when it passed the checkpoint was of: 60 m/s

The formula and procedure we will use to solve this exercise is:

vi = [(2 *x)/ t] - vf

Where:

Information about the problem:

Applying the initial velocity formula we have:

vi = [(2 *x)/ t] - vf

vi = [(2 * 225m)/5 s ] - 30 m/s

vi = [(450 m)/5 s ] - 30 m/s

vi = 90 m/s - 30 m/s

vi = 60 m/s

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

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The work required to pump all the water over the top of the tank of radius 5 m and depth 2 m is 5.12 x 10^6 J.

The work required to pump the water over the top of the tank is equal to the gravitational potential energy of the water inside the tank.

The potential energy is given by the following equation,

U= mgh

Here, m is the mass of water present inside the tank.

We know that, Density( Mass/Volume

Hence, m= Density x volume

So, the equation becomes

U= ρ x V x g x h

Where V is the volume of hemispherical tank= 2/3π x r^3

Putting the given values in the above equation,

U= 1000 x 2/3 x 3.14 x 53 x 9.8 x 2 = 5.12 x 10^6 J

Hence, the work required to pump all the water over the top of a tank of radius 5 m and a depth of 2 m is 5.12 x 10^6 J.

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The acceleration of Mr. Henderson Porsche is 54545.45 mi/hr².

Velocity is the directional speed of an object, body, or particle in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

We know that acceleration of a body is Change in Velocity per unit time.

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

Here ,

v is the Final Velocity,

u is the Initial Velocity and;

t is the time taken.

In the given question ,

The Porsche accelerated from 0 to 60 miles per hour in four seconds.

u = 0 mi/hr

v = 60 mi/hr

t = 4 sec

[tex]t=\frac{4}{3600}{~}hr\\[/tex]

[tex]t=\frac{1}{900} {~}hr[/tex]

t  = 0.0011 hr

Then,

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

[tex]a=\frac{60-0}{0.0011}[/tex]

[tex]a=54545.45 {~}mi/hr^{2}[/tex]

Hence, the acceleration of the Porsche is 54545.45 mi/hr².

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The magnitude of the electric field is 0 N/C.

We choose a Gaussian sphere of radius r > R concentric with the solid sphere.

Apply Gauss's law from Equation (1):

[tex]\Phi_{E} = \oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

Since the electric field vector E is parallel to the area vector dA of the curved part the dot product can be the simple product EdA:

                                 [tex]\oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

since the electric field is constant we can put E out of the integral:

                                [tex]E \oint dA = \frac{q_{in}}{\in_{0} }[/tex]

                                    [tex]EA = \frac{q_{in}}{\in_{0} }[/tex]

Substitute for A from Equation (2) and for [tex]q_{in}[/tex] from Equation (3)

recognizing that the enclosed charge by the Gaussian sphere is less than Q:

                              [tex]E(4\pi r^{2}) = \frac{pVgaussian}{\in_{0} }[/tex]

Substitute for V gaussian (volume of the Gaussian surface) from Equation (4):

                                          [tex]E(4\pi r^{2}) = \frac{\rho (\frac{4}{3} \pie r^{2})} {\in_{0} }[/tex]

                                           [tex]E = \frac{\rho r}{3\in_{0} }[/tex]

Where p is the charge per unit volume of the solid sphere which is equal to the total charge Q of the sphere divided by the volume V of the sphere :

                             [tex]E = \frac{(\frac{Q}{V} )r}{3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{(\frac{4}{3} \pi R^{3})3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{4\pi \in_{0}R^{3} }[/tex]

Substitute numerical values:

                                                  [tex]E = \frac{(26\times10^{-6})(0)}{4\pi(8.8542 \times10^{-12})(0.4)^{3}}[/tex]

                                 [tex]E = 0 N/C[/tex]

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Answer:

8.33 m/s^2

Explanation:

The equation for force is Force = mass * acceleration. The force is 500 N, and the mass is 60 kg, so substituting those into the equation you get that 500 = 60 * acceleration. Divide 500 by 60 to get the acceleration, which is 8.33 m/s^2.

Hope this helps! Let me know if you have any confusion!

Answer: The diaphragm

Explanation:

The diaphragm is a thin dome-shaped muscle which separates the thoracic cavity (lungs and heart) from the abdominal cavity (intestines, stomach, liver, etc.)

The frictional force is 11 N and the block’s acceleration is 0.14 m/s².

It is defined as the numerical value that indicates the amount of friction present between the surfaces of two bodies. The lower the coefficient of friction, the lower the friction between the surfaces, and the higher coefficient of friction the higher the friction force between them.

For part (a):

Taking x component:

F(x) = F cosθ

Taking y component:

F(y) = -F sinθ

Applying Newton’s second law to the y-axis:

F(N) = 15 sin40 + (3.5)(9.8)

F(N) = 44 N

coefficient U = 0.25

f(k) = 11 N

For part (b):

Applying Newton’s second law to the x-axis:

a = ((15) cos40 - 11)/3.5

a = 0.14 m/s²

Thus, the frictional force is 11 N and the block’s acceleration is 0.14 m/s².

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Answer:

48.9 m/s

Explanation:

I am ignoring air resistance in this:

horizontal velocity stays constant, so we need to find vertical velocity

v_0y = 0 m/s, t = 1.23s, a = 9.81 m/s^2 <- (g)

v_y = v_0y + at = 0 + 9.81 * 1.23 = 12.066 m/s

the magnitude v = sqrt(v_x ^ 2 + v_y ^2) = sqrt(47.4^2 + 12.066^2) = 48.9m/s

Answer:

V = 49.2 m/s

Explanation:

Given:

Vₓ  = 47.4 m/s

t = 1.23 s

___________

V - ?

Vy =  Voy + g·t = 0 + 9.8·1.23  ≈ 12,1 m/s

V = √( (Vx)² + (Vy)² ) = √ ( 47,7² + 12,1²) = 49.2 m/s

Answer:

YOUR ANSWER IS HERE....

When an emission line of helium from a galaxy is shifted toward red, it means that the galaxy is moving away from the viewer and is denoted as option D.

This refers to the systems of stars and interstellar matter that make up the universe which are gravitationally bound together and are made up of four main components which are:

When an emission line of helium from a galaxy is shifted toward red, there is a displacement to a longer wavelengths and this alteration is referred to as a doppler shift.

Moving away from the view means that the galaxy will be viewed in the red spectrum of the visible light which is a part of the electromagnetic spectrum.

This is therefore the reason why option D was chosen as the most appropriate choice.

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Answer:

[tex]{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance = \frac{50}{ {4.5}^{2} } }} \\ \\ { \rm{resistance = 2.5 \: ohms}}[/tex]

(i) The largest mass will be of the object B.

(ii) The smallest mass will be of the object C.

(iii) The ratio of the mass of object A and b will be 3:5.

According to the Newton’s 2 law for the same force applied acceleration and mass will be in the inverse ratio .it means the more the acceleration the less will be the mass of the object for the same force applied. As the acceleration of the object B is less so its mass will be more and vice versa. And the ratio of the masses will be the inverse ratio of the acceleration of object.

Hence, the ratio will become inverse ratio of acceleration of both the object that is 3:5.

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The number of molecules in the given volume is 5.33 x 10⁻¹² moles.

The number of molecules of the gas is calculated as follows;

PV = nRT

n = PV/RT

where;

1 torr = 133 Pa

1 x 10⁻¹⁰ torr =  ?

= 1.33 x 10⁻⁸ Pa

n = (1.33 x 10⁻⁸ x 1) / (8.314 x 300)

n = 5.33 x 10⁻¹² moles

Thus, the number of molecules in the given volume is 5.33 x 10⁻¹² moles.

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The distance at which the sound level is 100 dB is 30m.

Distance is a measurement of how far apart two objects or locations are. Distance in physics or common language can refer to a physical length or an estimate based on other factors (e.g. "two counties over"). |AB| is a symbol that can be used to represent the distance between two points. "Distance from A to B" and "Distance from B to A" are frequently used interchangeably. A distance function or metric is a technique to describe what it means for elements of some space to be "close to" or "far away" from each other in mathematics. It is a generalization of the idea of physical distance. Distance is a non-numerical unit of measurement in psychology and other social sciences.

Explanation:

We anticipate the radial distance to be 10 times greater, or 30 m, when the intensity is reduced by 20 dB, or by a factor of 102. A further 90 dB reduction might translate to an additional factor of 104.5 in terms of distance, or roughly 3030000m or 1000 miles. We measure sound intensity using the inverse square law and the decibel scale definition. According to the sound level definition,

β=10log[tex](\frac{1}{10^{-12}W/m^{2} } )[/tex]

We can compute the intensities corresponding to each of the levels mentioned as

[tex]I=[10^{\frac{\beta }{10} } ]10^{-12} W/m^{2}[/tex]

a. The power passing through any sphere around the source is

P=4π[tex]r^{2}[/tex]l

If we ignore absoption of sound by the medium,

conservation of energy requires that

[tex]r_{120} ^{2} I_{120} = r_{100}^{2} I_{100} =r_{10} ^{2} I_{10}[/tex]

then,[tex]r_{100} =r_{120} \sqrt{x} \frac{I_{120} }{I_{100} } =(3.00m)\sqrt{\frac{1\frac{W}{m^{2} } }{10^{-2}\frac{W}{m^{2} } } } =30 m[/tex]

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Answer:

i think a or b is the answer

Explanation:

i also don't know exactly

The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

It is very hard when pulling  water and if a person need to consistently pull air out of the tube before water makes it to the pump, it is very  difficult and some pumps may or cannot do it.

In regards to pushing, when a person is pushing above the Center of Gravity, one is pushing the said object into the ground and thus is increasing the rate of friction.

If a person is pulling you are said to be pulling above the Center of Gravity and lowering friction.

So, The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

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Option(c) is Kilometer is correct. Kilometer unit for length is greater than a meter.

Worldwide adoption of the International System of Units (SI) as a system of measurement units is widespread. For convenience, this updated version of the metric system is based on the number 10. The metric prefixes or the SI prefixes are a collection of prefixes that have been established. According on the prefixes, the unit is either a multiple or a fraction of base ten. It enables the zeros of very small or very high numbers to be reduced, for example, from 7,500,000 Joules to 1 nanometer and from 0.000000001 meter to 7.5 Megajoules, respectively. Additionally, some SI prefixes have a group of symbols before the unit symbol.

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