Hey guys,I hope u r gonna answer this question fast,SI system is extended from of MKS system.Why? I will be waiting for the answer. Good luck thank u​

ANSWER; KE=5mv^2 so it is proportional to v^2.

Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.

Answer:

Weight, acceleration when it falls vertically, are not same as that of earth.

Explanation:

Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.

So, the weight of object is not same as that on earth.

The mass is defined as the amount of matter contained in the object.

So, the mass of the object is same as that of earth.  

The volume of the object is defined as the space occupied by the object.

So, the volume of the object is same as that of earth.  

The density is defined as the ratio of mass of the object to its volume.

So, the density of the object is same as that of earth.  

The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.

So, the acceleration is not same as that of earth.

The color of the object is its characteristic.

It is same as that of earth.

"Longitudinal wave" is the wave where the difference between the coils increases as well as decreases.

Thus the above answer is correct.

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Answer:

A denser object will usually have a high index of

refraction.

A denser object will usually have a high index of refraction.

The refractive index is the ratio of the speed of light in vacuum and speed of light in any medium.

n = c/v

The density is greater for the denser medium (water, oil, mercury, etc) then the rarer medium (any gas or air).

When a light ray travels through denser medium, its velocity is reduced and the refracted ray bends towards normal.

As, the index of refraction is inversely proportional to the velocity of light in the medium, index of refraction will be high for denser object.

Thus, denser object have high index of refraction.

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When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

Assumptions:

∴

For a motion under constant acceleration, we can apply the kinematic equation:

[tex]\mathsf{v^2 = u^2 + 2as}[/tex]

where;

From the above equation, making acceleration (a) the subject of the formula:

[tex]\mathsf{v^2 - u^2 =2as }[/tex]

[tex]\mathsf{a = \dfrac{v^2 - u^2 }{2s}}[/tex]

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

[tex]\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}[/tex]

[tex]\mathsf{a = \dfrac{- 771.7284 }{2}}[/tex]

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration, and velocity is;

v = u + at

where;

v = 0

-u = at

[tex]\mathsf{t = \dfrac{-u}{a}}[/tex]

[tex]\mathsf{t = \dfrac{-27.78}{-385.86}}[/tex]

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that in order to estimate how fast the airbag must inflate to effectively protect the driver, the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

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Answer:

The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".

Explanation:

As the outer spherical shell is conducting, so there is no electric field in side from

⇒ [tex]r_b_1 < r < r_b_2[/tex].

So the electric potential at all points inside the conducting shell that from

⇒ [tex]r_b_1<r<r_b_2[/tex]  

and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:

⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]

Thus the above is the right solution.

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L)   ==>   n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n ≈ 14.0 g/mol

Answer:

Force = mass × acceleration

[tex]F =(100 \times 1000) \times 10 \\ = 1 \times {10}^{6} \: newtons[/tex]

Answer:

The total number of maxima that can be seen is 11

Explanation:

Given the data in the question

wavelength λ = 500 nm = 5 × 10⁻⁷ m

if the third order maximum is 32

i.e m = 3 and θ = 32°

Now, we know that condition for diffraction maximum is as follows;

d × sinθ = m ×  λ

so we substitute in our given values

d × sin( 32° ) = 3 ×  5 × 10⁻⁷ m

d × sin( 32° ) = 1.5  × 10⁻⁶ m

d = [ 1.5  × 10⁻⁶ m ] / sin( 32° )

d = 2.83 × 10⁻⁶ m

Now, maxima n when θ = 90° will be;

sin( 90° ) = nλ / d

1 =  nλ / d

d =  nλ

n = d / λ

we substitute

n = [ 2.83 × 10⁻⁶ m ]  / [ 5 × 10⁻⁷ m ]

n = 5.66

so 5 is the max value

hence, total maxima value is;

⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11

Therefore, total number of maxima that can be seen is 11

Answer:

a)  [tex]d_g=1.95m[/tex]

b) [tex]d_g'=2.3m[/tex]

Explanation:

From the question we are told that:

Depth [tex]d=2.60[/tex]

a)

Generally the equation for distance to ground is mathematically given by

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

Where

[tex]n=\frac{4}{3}[/tex]

Therefore

[tex]d_g=\frac{d}{n}[/tex]

[tex]d_g=\frac{2.6}{4/3}[/tex]

[tex]d_g=1.95m[/tex]

b)

For when The pool is filled halfway with water

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

[tex]d_g'=\frac{1.3}{4/3}[/tex]

[tex]d_g'=0.98m[/tex]

Therefore

[tex]d_g'=(1.3+0.98)[/tex]

[tex]d_g'=2.3m[/tex]

We have that the angle is

[tex]\theta=32.53[/tex]

From the Question we are told that

E⃗ =5.00×105ı^N/C

Generally the equation for Tension   is mathematically given

[tex]W=Tcos\theta[/tex]

Where

[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]

[tex]\theta=32.53[/tex]

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Answer:

[tex]\rho_{avg}=4000kg/m^3[/tex]

Explanation:

From the question we are told that:

Density of Material 1 [tex]\rho_1=2000kg/m^3[/tex]

Density of Material 2 [tex]\rho_2=6000kg/m^3[/tex]

Generally the equation for Average density is mathematically given by

[tex]\rho_{avg}=frac{\rho _1+rho _2}{2}[/tex]

[tex]\rho_{avg}=\frac{2000+6000}{2}[/tex]

[tex]\rho_{avg}=4000kg/m^3[/tex]

Answer:

The electric potential energy is 6.72 x 10^-11 J.

Explanation:

Potential difference, V = 4.2 x 10^8 V

charge of electron, q = - 1.6 x 10^-19 C

Let the potential energy is U.

U = q V

U = 1.6 x 10^-19 x 4.2 x 10^8

U = 6.72 x 10^-11 J

Answer:

No

Explanation:

You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.

Answer:

The car will make the turn perfectly

Explanation:

Given that the centripetal force= mv^2/r

M= mass of the car

v = speed of the car

r= radius

Hence;

F = 1000 × (14)^2/50

F= 3920 N

The frictional force = μmg

μ = coefficient of static friction

m= mass

g = acceleration due to gravity

Frictional force= 0.6 × 1000× 10

Frictional force = 6000 N

The car will not skid off the curve because the frictional force is greater than the centripetal force.

Answer:

120 km/s

Explanation:

Given data :

Radius of the star is r = 19 km

Rotational time period of the star is T = 1 s

Therefore, we know that the velocity of the star is given by :

[tex]$V=\frac{2\pi r}{T}$[/tex]

[tex]$V=\frac{2 \times 3.14 \times 19\times 10^3}{1}$[/tex]

V = 119380.52 m/s

Therefore, the velocity of the point on the equator of the star is = 120 km/s

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The answer is...

C. Work and time.

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Answer:

Final kinetic energies of both astronauts will be the same.

Explanation:

If we ignore all the friction present between the ramp and the person. Then essentially there is no loss of energy in the system. Hence the initial potential energies of the astronauts must be equal to their final kinetic energies.

Now the potential energy depends upon mass, height and acceleration due to gravity. All these parameters for both the astronauts. Therefore,both astronauts have same initial potential energies.

Similarly, the final kinetic energies of astronauts will also be the same.

Answer:

The energy from food and then from plants and then from sun.

As sun is the ultimate source of energy.

Explanation:

Distance = 100 m, 1000m, marathon

As the distance is covered by the person, so the muscular energy is used and thus the energy comes form out food.

As we know that the energy can neither be created nor be destroyed it can transform from one form to another.

So, the energy form the food which we consume is converted into the kinetic energy as we run.

A body of mass m feels a gravitational force due to the planet of

F = GmM/R ² = ma

where

• G = 6.67 × 10⁻¹¹ N•m²/kg² is the universal gravitational constant

• M is the mass of the planet

• R is the distance between the body and the planet's center

• a is the acceleration due to gravity

Solving for a gives

a = GM/R ²

Notice that 28,000 km is twice 14,000 km. The equation says that the acceleration varies inversely with the square of the distance. So if R is changed to 2R, we have a new acceleration of

GM/(2R)² = 1/4 × GM/R ² = a/4

so the acceleration of the body at 28,000 km from the planet's center would be (32 m/s²)/4 = 8 m/s².

Answer:

(a) the velocity of the shirt is 2.14 m/s

(b) the velocity of the shirt is 5.3 m/s

Explanation:

Given;

initial velocity of the shirt, u = 5.3 m/s

height of the platform above the ground, h = 4.00 m

(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]

(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = [tex]\frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150} }{2}[/tex]

       t = [tex]\frac{12.5 \ \pm 27.5}{2}[/tex]

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

Answer:

4.926 L Y 7.389 L

Explanation:

first you calculate the tank volume

V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]

then you convert to liters

[tex]12315 cm^{3}[/tex] = 12.315 l

then you calculate the liters of water

2/5(12.35 l) = 4.926 l

finally we calculate the amount without water

12.315 l - 4.926 l = 7.389 l

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Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]

Therefore, the resulting temperature is 23.37 ⁰C

Answer:

do this Q yourself because i havent read the chapter

The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.

This is calculated using the coefficient of static friction between the penny and block, which is 0.495.

The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.

This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.

The angular velocity of the disk is then calculated from this force and the radius of the disk.

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Answer:

24 meters

Explanation:

Find the final velocity. 12m/s

d=[final-initial]/2×time

D=(6m/s)×4=24 m/s

Answer:

Explanation:

From the question we are told that:

Number of turns [tex]N=10[/tex]

Area [tex]a=0.23m^2[/tex]

Magnetic field [tex]B=0.947T[/tex]

Generally the equation for maximum flux is mathematically given by

 [tex]\phi=NBa[/tex]

 [tex]\phi=10*0.047*0.23[/tex]

 [tex]\phi=0.1081wbi[/tex]

Therefore induced emf

 [tex]e= \frac{d\phi}{dt}[/tex]

Since

 [tex]t=0[/tex]

Therefore

 [tex]e=0[/tex]

Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]

           v₂ = 15.24 m / s