An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.30 m before stopping. How far does the lighter fragment slide

Answers

Answer 1

Answer:

the distance d traveled by the lighter fragment is 58.1 m.

Explanation:

mass of the lighter fragment = m

the lighter fragment traveled a distance = ?

mass of the heavier fragment = 7m

the distance covered by the heavier fragment = 8.30 m

The two particles will be given the same amount of energy from the explosion. This energy is used to do work by the two fragments.

work done by heavier fragment w = mgd

where m is the mass

g is acceleration due to gravity

d is the distance traveled.

substituting, the work done by the heavier fragment is

w = 7m x g x 8.3 = 58.1mg

The same way, the lighter fragment does work of

w = mgd

equating the two work done since they are given the same amount of energy from the explosion, we have

58.1mg = mgd

mg cancels out, we have

the distance d traveled by the lighter fragment d = 58.1 m


Related Questions

Rank these significant figures numbers from the least to the most
a. 357
b. 0.006
c. 9520.00
d. 9256.0
e. 700.003
f. 6010

Answers

Answer:

0.006<357<700.003<6010<9256.0<9520.00

A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?

Answers

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

g Radiation of an unknown wavelength is used in a photoelectric effect experiment on a sodium surface. The maximum kinetic energy of the observed electrons is 0.7 eV. What is the wavelength of the light

Answers

Answer:

λ = 4.1638 10⁻⁷ m

Explanation:

The photoelectric effect was explained by Einstein assuming that the radiation acts like particles and the equation that describes the process is

           K = h f -Ф

where K is the kinetic energy of the emitted electrons, hf  the energy of the photons according to Planck's equation and Ф the work function of the material

In this case they give us the kinetic energy of the electrons

         K = 0.7 eV

The sodium work function is tabulated Ф = 2.28 eV

Let's find the frequency of the photons

            f = (K + Ф) / h

Planck's constant is

          h = 6.626 10⁻³⁴ J s (1 eV / 1.6 10⁻¹⁹ J) = 4.136 10⁻¹⁵ eV s

            f = (0.7 + 2.28) / 4.136 10⁻¹⁵

            f = 7.2050 10¹⁴ Hz

let's find the wavelength using the relationship between speed and frequency and wavelength

            c = λ f

            λ = c / f

            λ = 3 10⁸ / 7.205 10¹⁴

            λ = 4.1638 10⁻⁷ m

A metal sphere of radius 19 cm has a net charge of 2.4 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 370 V?

Answers

Answer:

1.29*10^6 N/C

1135.6 V

9.18 cm

Explanation:

Given that

radius of the metal, r = 19 cm

charge of the metal, q = 2.4*10^-8 C

coulomb's constant, k = 8.99*10^9

to find the electric field, we use the formula E = kq/r², where

E = electric field

k = coulomb constant

q = charge on the metal and

r = radius of the metal

E = (8.99*10^9 * 2.4*10^-8) / 0.19²

E = 215.76 / 0.0361

E = 1.29*10^6 N/C

to find the electric potential, we use this relation

V = kq/r

V = (8.99*10^9 * 2.4*10^-8) / 0.19

V = 215.76 / 0.19

V = 1135.6 V

V = kq/r,

r = kq/V

r = (8.99*10^9 * 2.4*10^-8)/ (1135.6 - 370

r = 215.76 / 765.6

r = 0.281 = 28.1 cm

distance from the sphere

28.18 - 19 = 9.18 cm

What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV. (in nm)

Answers

Answer: Wavelength of 424 nm  can produce photoelectrons from silver

Explanation:

[tex]\phi=h\times \nu_o=\frac{hc}{\lambda}[/tex]

[tex]\phi[/tex] = work function = energy of photon

h = Planck's constant =  [tex]6.63\times 10^{-34}Js[/tex]

[tex]\nu_0[/tex] = frequency of the metal

c = speed of light =  [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] =longest  wavelength of the radiation

Now put all the given values in the above formula, we get the wavelength of the photons.

[tex]\lambda=\frac{hc}{\phi}[/tex]

[tex]\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^8m/s}{2.93\times 1.6\times 10^{-19}J}[/tex]      ( as 1ev=[tex]1.6\times 10^{-19}J[/tex] )

[tex]\lambda=4.24\times 10^{-7}m[/tex]

[tex]1nm=10^{-9}m[/tex]

[tex]\lambda=424nm[/tex]

Here, wavelength of 424 nm  can produce photoelectrons from silver

A solenoid with a certain number of turns N and carrying a current of 2.000 A has a length of 34.00 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mT, what is the value of N?

Answers

Answer:

The number of turns of the solenoid is 1217 turns

Explanation:

Given;

current in the solenoid, I = 2 A

length of the solenoid, L = 34 cm = 0.34 m

magnitude of the magnetic field, B = 9 mT = 0.009 T

Number of turns of the solenoid = N

The magnitude of magnetic field at the center of the solenoid is given by;

B = μnI

Where;

μ is permeability of free space = 4π x 10⁻⁷ m/A

I is the current in the solenoid

n is the number of turns per length

n = B/μI

n = (0.009) / (4π x 10⁻⁷)(2)

n = 3580.52 turns/m

N = nL

N =(3580.52 turns/m) x (0.34 m)

N = 1217 turns

Therefore, the number of turns of the solenoid is 1217 turns

A part of a circuit contains an inductor. The current through the inductor is changing uniformly from 2.40A to 0.30A over the course of 1.75s. If the EMF ( voltage change ) from one side of the inductor to the other is 5.70 Volts, what is the value of the inductance

Answers

Answer:

The value of the inductance is 4.75 H

Explanation:

Given;

initial current through the inductor, I₁ = 2.4 A

final current through the inductor, I₂ = 0.3 A

duration of change of current, dt = 1.75 s

voltage change of the inductor, V = 5.7 Volts

The voltage change of the inductor is given by;

[tex]V_L = -L\frac{di}{dt}\\\\ V_L = -L(\frac{I_2-I_1}{dt} )\\\\V_L = L(\frac{I_1-I_2}{dt} )\\\\[/tex]

Where;

L is the inductance of the coil;

[tex]5.7 = L(\frac{2.4-0.3}{1.75} )\\\\5.7 = 1.2 L\\\\L = \frac{5.7}{1.2}\\\\ L = 4.75 \ H[/tex]

Therefore, the value of the inductance is 4.75 H

The half-life of element X is 20 years. If there are 48 g initially a) How much is there after 80 years

Answers

Answer:

After 80 years there will be 3 g of element X remaining

Explanation:

Given;

the half life of element X = 20 year

initial mass of element X = 48 g

a) How much is there after 80 years

0 year --------------------------> 48 g

20 years -----------------------> (48g / 2) = 24 g

40 years ------------------------> 12 g

60 years ------------------------> 6 g

80 years --------------------------> 3 g

Therefore, after 80 years there will be 3 g of element X remaining.

Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal

Answers

Answer:

1.41eV

Explanation:

Kinetic Energy of photoelectron(K. Emax)

E = Wo + K.Emax

E = hc/ λ

h = planck's constant = 6.63 * 10^-34

c = speed of light = 3×10^8 m/s

λ = 400nm

Work function (Wo) = 1.70eV

1 eV = 1×10^-19

E = [(6.63×10^-34) * (3×10^8)] / 400×10^-7

E = (19.89 × 10^(-26))/400×10^-7

E = 0.049725×10^-19

K.Emax = E - Wo

K.Emax = (0.049725×10^-19) - (1.7×10^-38)

0.049725×10^-19 interms of eV = (0.049725/1.6)×10^-19 =

K.Emax = 3.11eV - 1.70eV

K.Emax = 1.41eV

Suppose a big chunk of gold is submerged in water and its volume is found to be 12.5 cm?
Compute the mass of the chunk of gold in grams if you know the density is 19.3 g/cm2. Round
appropriately.

Answers

Answer:

The mass of the chunk of gold is 241.25 g

Explanation:

Since density is defined as mass divided by volume, we can solve for the mass (m) via its equation:

[tex]density=\frac{mass}{volume} \\19.3=\frac{m}{12.5} \\m=19.3\,(12.5)\, grams\\m = 241.25 \,\,g[/tex]

If you are pushing on a crate on a frictionless surface in one direction, and your friend is pushing on the crate in the opposite direction with an equal amount of force. Which of the following statements is the most accurate? a. The crate will not move as the forces cancel each other out b. Because the surface is frictionless, the crate will always move regardless of who is pushing c. The crate can continue to move, but it will move at a constant velocity d. The net force is towards the direction that you are pushing, as you started the crate's motion

Answers

Answer:

Its not A..

Explanation:

I chose A - was incorrect

A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%. What is her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started? Show all your work.

Answers

Her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s

What is mechanical energy?

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.

As given in the problem A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg in total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%

The potential energy is getting converted into kinetic energy with an efficiency of 52 %

1/2mv² = 0.52 (mgh)

v²= 1.04gh

v = √(1.04gh)

v= √(1.04×9.81×10)

v = 10.1 m/s

Thus, her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be  10.1 m/s

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A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stretched to a tension of 122 N. During what time interval will a transverse wave travel the entire length of the two wires

Answers

Answer:

The time taken is  [tex]t = 0.356 \ s[/tex]

Explanation:

From the question we are told that

  The length of steel the wire is  [tex]l_1 = 31.0 \ m[/tex]

   The  length of the  copper wire is  [tex]l_2 = 17.0 \ m[/tex]

    The  diameter of the wire is  [tex]d = 1.00 \ m = 1.0 *10^{-3} \ m[/tex]

     The  tension is  [tex]T = 122 \ N[/tex]

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              [tex]t = t_s + t_c[/tex]

Where  [tex]t_s[/tex] is the time taken to transverse the steel wire which is mathematically represented as

         [tex]t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ][/tex]

here  [tex]\rho_s[/tex] is the density of steel with a value  [tex]\rho_s = 8920 \ kg/m^3[/tex]

   So

      [tex]t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]

      [tex]t_s = 0.235 \ s[/tex]

 And

        [tex]t_c[/tex] is the time taken to transverse the copper wire which is mathematically represented as

      [tex]t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ][/tex]

here  [tex]\rho_c[/tex] is the density of steel with a value  [tex]\rho_s = 7860 \ kg/m^3[/tex]

 So

      [tex]t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]

      [tex]t_c =0.121[/tex]

So  

   [tex]t = t_c + t_s[/tex]

    [tex]t = 0.121 + 0.235[/tex]

    [tex]t = 0.356 \ s[/tex]

a. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
b. What are the electron's speed and energy in this state?

Answers

Answer:

a

  [tex]n = 23[/tex]

b

  [tex]v = 87377.95 \ m/s[/tex]

Explanation:

From the question we are told that

   The diameter is [tex]d = 61\ nm = 61 *10^{-9} \ m[/tex]

   

Generally the radius electron orbit  is mathematically represented as

      [tex]r = \frac{61 *10^{-9}}{2}[/tex]

=>   [tex]r = 3.05*10^{-8} \ m[/tex]

This radius can also be represented mathematically  as

      [tex]r = n^2 * a_o[/tex]

Here n is the quantum number and [tex]a_o[/tex] is  the Bohr radius with a value

    [tex]a_o = 0.0529 *10^{-9} \ m[/tex]

So

   [tex]n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }[/tex]

=>   [tex]n = 23[/tex]

Generally the angular momentum of the electron is mathematically represented as

          [tex]L = m * v * r = \frac{n * h }{2 \pi}[/tex]

Here  h is the Planck constant and the value is  [tex]h = 6.626*10^{-34} J \cdot s[/tex]

          m is the mass of the electron with values [tex]m = 9.1*10^{-31} \ kg[/tex]

         So

               [tex]v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }[/tex]

                [tex]v = 87377.95 \ m/s[/tex]

       

Determine the inductance L of a 0.65-m-long air-filled solenoid 3.2 cm in diameter containing 8400 loops.

Answers

Answer:

The inductance is  [tex]L = 0.1097 \ H[/tex]

Explanation:

From the question we are told that

  The  length is  [tex]l = 0.65 \ m[/tex]

   The  diameter is  [tex]d = 3.2 cm = 0.032 \ m[/tex]

    The  number of loops is  [tex]N = 8400[/tex]

Generally the radius is evaluated as

       [tex]r = \frac{ 0.032 }{2} = 0.016 \ m[/tex]

The  inductance is mathematically represented as

      [tex]L = \frac{ \mu_o * N^2 * A }{ l }[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

A is the cross-sectional area which is mathematically evaluated as

             [tex]A = \pi r^2[/tex]

=>        [tex]A = 3.142 * (0.016)^2[/tex]

=>        [tex]A = 0.000804 \ m^2[/tex]

 =>  [tex]L = \frac{ 4\pi * 10^{-7} * 8400^2 * 0.000804 }{ 0.65 }[/tex]

=>    [tex]L = 0.1097 \ H[/tex]

A narrow beam of white light is incident on a sheet of quartz. Thebeam disperses in the quartz, with red light (λË700nm)traveling at an angle of 26.3o with respect to thenormal and violet light (λË400nm) traveling at25.7o. The index of refraction of quartz for red lightis 1.45. What is the index of refraction of quartz for violetlight?

Answers

Answer:

The index of refraction of quartz for violet light is 1.47.

Explanation:

It is given that, a narrow beam of white light is incident on a sheet of quartz.

The beam disperses in the quartz, with red light at an angle, [tex]\theta_r=26.3^{\circ}[/tex] wrt to the normal and violet light traveling at an angle of [tex]\theta_v=25.7^{\circ}[/tex]

The index of refraction of quartz for red light is 1.45.

We need to find the index of refraction of quartz for violet light.

Using Snell's law of red light as follows :

[tex]\mu_a\sin\theta_i=\mu_r\sin\theta_r[/tex]

Here,

[tex]\mu_a[/tex] is the refractive index of air

[tex]\theta_i[/tex] is the angle of incidence

We can find the value of angle of incidence as follows :

[tex]\sin\theta_i=\dfrac{\mu_r \sin\theta_r}{\mu_a}\\\\\sin\theta_i=\dfrac{1.45\times \sin(26.3)}{1}\\\\\theta_i=\sin^{-1}(0.642)\\\\\theta_i=39.79^{\circ}[/tex]

Now again using Snell's law for violet light as follows :

[tex]\mu_a\sin\theta_i=\mu_v\sin\theta_v\\\\\mu_v=\dfrac{\mu_a\sin\theta_i}{\sin\theta_v}\\\\\mu_v=\dfrac{1\times \sin(39.79)}{\sin(25.7)}\\\\\mu_v=1.47[/tex]

So, the index of refraction of quartz for violet light is 1.47.

"Find the change in gravitational potential energy that the box undergoes as it rises to its final height."

Answers

Explanation:

Given that,

Mass of a box is 2.6 kg

(1) We need to find the work done on the box by this force as it is pushed up the 5.00-m ramp to a height of 3.00 m.

It means that the position of the object is 3 m i.e. h = 3 m

Work done = Fd

= mgh

So,

[tex]W=2.6\times 9.8\times 3\\\\W=76.44\ J[/tex]

(2) Now the gravitational potential energy that the box undergoes as it rises to its final height is equal to the work done by the box. So,

Change in potential energy = 76.44 J

As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:___________.
A) approaches zero.
B) approaches infinity.
C) approaches unity.
D) none of the given answers

Answers

Answer:

B) approaches infinity

Explanation:

The capacitive reactance of an AC capacitor is given by;

[tex]X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}[/tex]

Where;

C is the capacitance

f is the frequency of the ac voltage

[tex]X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity[/tex]

Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.

The correct option is (B) approaches infinity

A scientist claims that a certain chemical will make fabric waterproof. Which
option describes a controlled experiment that will produce evidence that will
support or refute her claim?

Answers

Answer: She adds different amounts of the chemical to the material and then puts them in the water

Explanation:

Answer: One group of fabric is treated with the chemical, and the other group is not. Then each group is exposed to water.

Explanation:

While taking the stairs it takes you 10 seconds to reach the top. The next time you take the same stairs, it takes you 5 seconds to reach the top stair. During which of these trips up the stairs did you use more power to climb? Explain your answer in complete sentences with proper spelling, grammar, and other language mechanics.

Answers

Answer:

  P₂ = 2 P₁

we conclude that in the second time the power used is double that in the first rise

Explanation:

In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.

Now we can analyze the required power,

         P = W / t

From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,

therefore the first time the power is

           P₁ = E / 10

           P₁ = 0.1 E

for the second time the power is

          P₂ = E / 5

          P₂ = 0.2 E

we see that the power in the second case is

         P₂ = 2 P₁

Therefore, we conclude that in the second time the power used is double that in the first rise.

What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?

Answers

Answer:

4.003" (inches )

Explanation:

The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .

Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is

An airplane is in level flight over Antarctica, where the magnetic field of the earth is mostly directed upward away from the ground. As viewed by a passenger facing toward the front of the plane, is the left or the right wingtip at higher potential? Does your answer depend on the direction the plane is flying?

Answers

Answer:

It does not depend on direction of plane and the left windtips more potential

Explanation:

Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction

A 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds. During the 10 seconds, the truck has traveled a distance of

Answers

Answer:

250 m

Explanation:

1. First, we have to calculate the acceleration:

[tex]\boxed{\mathsf{v=v_o+a\cdot t}}[/tex]

where:

v = present velocity

vo = initial velocity

a = acceleration

t = time

2. Let us use given information and substitute in the expression above. Hence:

[tex]\mathsf{50=0+a\cdot10}\\\\\mathsf{50=10\cdot a}\\\\\mathsf{a=\dfrac{50}{10}}\\\\\therefore \boxed{\mathsf{a=5\,m/s^2}}}[/tex]

3. Now we can calculate traveled distance with Torricelli's equation:

[tex]\boxed{\mathsf{v^2=v_o^2+2\cdot a \cdot d}}[/tex]

where:

v = present velocity

vo = initial velocity

a = accelerarion

d = distance

4. So, we get:

[tex]\mathsf{50^2=0^2+2\cdot 5 \cdot d}\\\\\mathsf{2500=10\cdot d}\\\\\mathsf{d=\dfrac{2500}{10}}\\\\\therefore \boxed{\mathsf{d=250\,m}}[/tex]

Conclusion: during 10 seconds the truck has traveled a distance of 250 m.

Have a nice day! : )

If a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds. During the 10 seconds, the truck traveled a distance of 250 meters.

What are the three equations of motion?

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

By using the first equation of motion,

v = u + at

50 = 0 + 10 a

a = 50/10

a= 5 meters/second²

As given in the problem a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds.

The total distance traveled by truck,

S = ut + 0.5at²

S = 0 + 0.5 ×5 ×10²

S = 250 meters

Thus, the total distance traveled by truck would be 250 meters.

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For the microscope to be in focus, how far should the objective lens be from the specimen?

Answers

Answer:

p ≈ f_ objective     Therefore for the object to be in focus it must be close to the focal length

Explanation:

A microscope is an optical instrument that uses two lenses, or a long focal length objective lens that forms a real image of the object and an eyepiece that forms a virtual image of the object. Therefore for the object to be in focus it must be close to the focal length

           p ≈ f_ objective

p distance objetive

A resistor is connected across an oscillating emf. The peak current through the resistor is 2.0 A. What is the peak current if:

a. The resistance R is doubled?
b. The peak emf εo is doubled?
c. The frequency ω is doubled?

Answers

Answer:

(a) When the resistance R is doubled, I = 1 A

(b) When the peak emf εo is doubled, I = 4 A

(c)  When the frequency ω is doubled, I = 2 A

Explanation:

Given;

peak current through the resistor, I = 2.0 A

According to ohms law the peak current through the circuit is given by;

[tex]I = \frac{V}{R}[/tex]

(a) When the resistance R is doubled;

[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]

(b)When the peak emf εo is doubled

[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]

(c) When the frequency ω is doubled

Peak current through resistor is independent of frequency

I₂ = 2.0 A

what single property was the most important in jesseca's material

Answers

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation: The Graphite was the material in the passage that had reflected most of the light.

Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
What was his average speed in mph over the last 400 m?

Answers

Answer: His average speed in mph over the last 400 m is 7.7 m/s.

Explanation:

Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.

We know that , speed = [tex]\dfrac{distance}{time}[/tex]

Here , distance = 400m and time = 51.9 s

Then, speed =  [tex]\dfrac{400}{51.9}\approx7.7\ m/s[/tex]

Hence, his average speed in mph over the last 400 m is 7.7 m/s.

Answer: 17.2 MPH

Explanation:

Average speed = distance/time

400m÷ 51.9s= 7.7 m/s

Now convert m/s →MPH

m→km→mi and s→min→hr

[tex]\frac{7.7m}{s} x[/tex] [tex]\frac{1 km}{1000 m} x\frac{0.6214 mi}{1 km} x\frac{60 s}{1 min} x\frac{60 mins}{1 hr} = 17.2 mph[/tex]

A 50g marble is moving at 2m/s when it strikes a 20g marble at rest. Immediately after the collision, the 50g ball is moving at 1m/s. Is this an elastic collision

Answers

Answer:

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Explanation:

Given;

mass of first marble, m₁ = 50g = 0.05 kg

initial velocity of the first marble, u₁ = 2 m/s

mass of second marble, m₂ = 20 g = 0.02 kg

initial velocity of the second marble, u₂ = 0

final velocity of the first marble, v₁ = 1 m/s

Let the final velocity of the second marble, = v₂

Determine the final velocity of the second marble by applying principle of momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂

0.1 = 0.05 + 0.02v₂

0.02v₂ = 0.1 - 0.05

0.02v₂ = 0.05

v₂ = 0.05 / 0.02

v₂ = 2.5 m/s

During inelastic collision both objects will move at the same velocity after collision.

During elastic collision both objects will move at different velocities after collision.

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delivered by the engine? (1 hp 746 W)

Answers

Answer:

90 hp

Explanation:

Power = work / time

P = ½ (1500 kg) (25 m/s)² / 7.0 s

P = 67,000 W

P = 90 hp

Define fluid flow. Write five difference between uniform and non uniform flow.​

Answers

Answer:

Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.

Difference:

Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.

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