Newton's fist law is just about what happens if no force is applied.

O True
O False

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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Answer:

i think a or b is the answer

Explanation:

i also don't know exactly

The weight of the car in kilogram is 4.04 kg.

The Mass of the substance in grams is 5,400 g.

The Distance of the marathoner in kilometers is 42.2 km.

The Volume of the Mountain dew in ounces is 33.8 oz.

The Capacity of the flash drive in centibytes is  8 x 10⁸ CB.

Unit conversion is used to convert the units of some measurement into other units.

The weight of the car in kilogram is calculated as follows;

1 pound = 0.453592 kg

8.9 pounds = ?

= (8.9 x 0.453592 kg)

= 4.04 kg

The mass of the substance in grams is calculated as follows;

1 kg = 1000 g

5.4 kg = ?

= (5.4 x 1000 g)

= 5,400 g

The distance of the marathoner in kilometers is calculated as follows;

1 mile = 1.609 km

26.2 miles = ?

= (26.2 x 1.609 km)

= 42.2 km

The volume of the Mountain dew in ounces is calculated as follows;

1 ml = 0.0338 oz

1000 ml = ?

= (1000 x 0.0338 oz)

= 33.8 oz

The capacity of the flash drive in centibytes is calculated as follows;

1 MB = 1 x 10⁸ CB

8 MB = ?

= (8 x 10⁸ CB)

= 8 x 10⁸ CB

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Answer:

The straight down arrow

Explanation:

Next time add a picture to also help those who would like to answer. :)

Answer: I don't know the right answer but it makes more sense for it to be balanced  

Explanation:

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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The car given in the question is not traveling at a constant speed as the time interval in which it to cover the equal distance is not equal.

Speed is the physical quantity that measures the rate at which the position of a body is changing with respect to its surrounding. Its S.I. unit is meters per second. And its dimensional equation is [[tex]LT^{-1}[/tex]]

Given in the Question

Time to travel first meter =  5 seconds

Time to travel second meter =  8 seconds

The body is said to be moving with a constant speed if it travels an equal distance in equal intervals of time.

But in the above-given condition, the car travels 1m in 5 seconds but travels the next 1m in 8 seconds. Here the distance traveled is equal but the time interval in which both distances are covered is not equal.

Therefore, the car is not traveling at a constant speed.

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The velocity of the horse as it travels through the track is 20 m/s.

Velocity can be defined as the ratio of distance or displacement to time.

To calculate the velocity of the the horse, we use the  formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the velocity of the horse as it travels through the track is 20 m/s.

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Answer:

A. 2s

Explanation:

time = distance ÷ speed

T = .2 ÷.1

T= 2

The time interval with respect to Ed is 4.88 ×10³ s.

A frame of reference is a set of reference points—geometric points whose positions are known mathematically and physically—that define the origin, orientation, and scale of an abstract coordinate system.

If a body does not continuously adjust its position in relation to its environment throughout the course of time, it is said to be at rest.

In the frame S', Dina and Owen are at rest.

The speed of the ball with respect to Owen, u =0.800c

The speed of the frame S' with respect to frame S,  v = 0.600c

Distance between Dina and Owen, L(p) = 1.8 × 10¹² m

Speed of light, c = 3 × 10⁸ m/s

Now, for time intervals using the Lorentz transformation equation:

Δt = 1 / [ √( 1 - (v²/c²)] ( Δt' + (vΔx'/c²))

Now, v = 0.6c as S' is moving with respect to frame S.

Hence, the time taken is:

Δt = 1 / [ √( 1 - ((0.6c)²/c²)] ( 7.5 ×10³ + [(-0.6c)(1.8 × 10¹²)/c²))

Δt = 1.25( 75000 - 3600)

Δt = 4.88 ×10³ s

The total time taken by the ball with respect to Ed is 4.88 ×10³ s.

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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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(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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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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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: 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 charge per unit length on the inner surface of the cylinder is 3λL.

The electric flux across any closed surface is proportional to the net electric charge q enclosed by the surface, according to Gauss's law of electricity, which reads as = q/0, where 0 is the electric permittivity of free space, which has a value of 8.854 10-12 square coulombs per newton per square meter.

Here, q stands for charge.

Volt-metres (V m), the SI base unit of electric flux, is also equivalent to newton-meters squared per coulomb (N m2 C-1). On top of that, the fundamental units of electric flux are kgm3s-3A-1.

Q 1 = L as a result of the wire's induction of the cylinder's inner surface.

"q cylinder" is equal to "q in + q out" (cylinder).

L = L + q out

⇒q out =λL+(2λ)

L = q out =3λL

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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 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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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:

480 meters

Explanation:

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 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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The currents in the wires are in opposite directions.

If this flowing through the wire are in same direction then they the wire attract each other and if this flowing through the wire are opposite in direction then they the wire repel each other .

As it is given that a pair of wires repel each {other} which suggests that current inside the wires area unit in other means .

Hence current flowing through wires area unit opposite in direction.

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At a depth of 10 m, the manatee's neutral buoyancy keeps it at the same level in water as if there were no air inside its lungs. However, because more air is now contained in the manatee's lungs than before it dove, its weight increases by 9.81 kg and therefore it accelerates downwards (due to momentum). This downward acceleration continues until equilibrium is reached and then ceases due to gravity pulling on both objects equally.

As a result, the manatee now becomes negatively buoyant and must rise to compensate.

When something has neutral buoyancy, it is not affected by the weight of anything above or below it.

This means that objects with neutral buoyancy will stay in place regardless of how high or low they are located in a liquid medium.

Examples of things with neutral buoyancy include air balloons and swimming caps.

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Answer:

YOUR ANSWER IS HERE....

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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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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The man moves across the ice with a velocity of -1 m/s.

Formula:

To calculate the velocity we would be using the formula,

M₁U₁ + M₂U₂ = M₁V₁+ M₂V₂

Given:

After applying the law of linear momentum, we get the following value before the throw;

M₁ = Man's initial mass= 80kg,

U₁ = Initial velocity of Man = 0,

M₂= Initial mass of Book  = 4kg,

U₂= Initial velocity of Book = 0.

After the throw, the book's velocity will be

M₁ = Man final mass = 80kg,

V₁ = Man's final velocity = V.

M₂ = Final mass of Book = 4kg,

V₂ = Final velocity of Book = 20 m/s

After substituting the values in the formula, we get,

M₁U₁ + M₂U₂ = M₁V₁+ M₂V₂

Or, (80 x 0) + (4 x 0) = (80V) + (4 x 20)

Or, 0 = 80V + 80

Or, 80V = -80

Or, V = -80/80

V= -1m/s

[Note: The -ve sign shows the opposite direction]

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The electric field at a midpoint is 1.425 x 10⁴ N/C.

We need to know about the electric fields to solve this problem. The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation

E = k.Q/r²

where E is the electric field, Q is the charge and r is the radius.

From the question above, we know that

k = 9 x 10⁹ Nm²/C²

Q1 =e = 1.6 x 10¯¹⁹ C

Q2 = e = 1.6 x 10¯¹⁹ C

x = 897 nm = 8.97 x 10¯⁷ m

from midpoint (r = 4.495 x 10¯⁷ m)

The electric field radially outward from proton to electron so that the total electric field between the charges will be added up.

Etotal = E1 + E2

Etotal = k.e/r² + k.e/r²

Etotal = 2. k.e/r²

Etotal = 2 x 9 x 10⁹ x 1.6 x 10¯¹⁹/(4.495 x 10¯⁷)²

Etotal = 1.425 x 10⁴ N/C

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[tex]Given data : f01 = 560 hz f02 = 480 hz vs = 340 m/s ( velocity of sound ) vx = velocity of source ( police car )Condition 1 when source approach observer : f01 =[ vs /( vs - vx )].fs ...........(1)Condition 2 when source passes observer : f02 = [vs /( vs + vx) ].fs ............(2)eq(1) : eq (2) we get : f01 / f02 = ( vs + vx )/( vs - vx ) 7/6 = ( 340 + vx ) / ( 340 - vx )[/tex][tex]solve this equation get vx = 26.15 m/s[/tex]

The frequency of a repeated event is its number of instances per unit of time. In certain cases, it is also referred to as temporal frequency or ordinary frequency to underline differences with spatial and angular frequencies, respectively. One (event) per second is equal to one hertz (Hz), which is how frequency is stated. The period is the reciprocal of the frequency since it represents the length of time for one cycle in a recurring occurrence. For instance, the period, T—the space between beats—of a heart beating at a frequency of 120 beats per minute (2 hertz), is equal to 0.5 seconds (60 seconds divided by 120 beats).

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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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