A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x axis at 37 m/s in the positive direction. The second, with mass 22 g , moves along the y axis at 34 m/s in the positive direction. Find the velocity of third piece.

Answers

Answer 1

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

applying conservation law of momentum

p + .444 i + .748 j  = 0

p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .


Related Questions

If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the

Answers

Answer:

The wave is travelling in the ±z-axis direction.

Explanation:

An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the  electric and magnetic field.

In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.

When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?

Answers

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.70 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface.
A) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.
B) Find the height of the image formed by paraxial rays incident on the convex surface.
C) Is the image erect or inverted?

Answers

Answer:

A) 0.1477

B) 0.65388 mm

C) object is inverted

Explanation:

The formula for object - image relationships for spherical reflecting surface is given as;

n1/s + n2/s' = = (n2 - n1)/R

Where;

n1 & n2 are the Refractive index of both surfaces

s is the object distance from the vertex of the spherical surface

s' is the image distance from the vertex of the spherical surface

R is the radius of the spherical surface

We are given;

index of refraction of glass; n2 = 1.60

s = 24 cm = 0.24 m

R = 4 cm = 0.04 m

index of refraction of air has a standard value of 1. Thus; n1 = 1

a) So, making s' the subject from the initial equation, we have;

s' = n2/[((n2 - n1)/R) - n1/s]

Plugging in the relevant values, we have;

s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]

s' = 0.1477

b) The formula for lateral magnification of spherical reflecting surfaces is;

m = -(n1 × s')/(n2 × s) = y'/y

Where;

m is the magnification

n1, n2, s & s' remain as earlier explained

y is the height of the object

y' is the height of the image

Making y' the subject, we have;

y' = -(n1 × s' × y)/(n2 × s)

We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.

Thus;

y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)

y' = - 0.00065388021 m = -0.65388 mm

C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.

Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.

Thus, the object is inverted since m is negative.

Question 5 of 10
Heat is being transferred through currents within a liquid. When will this heat
transfer mostly end?
O A. When the substance changes state and becomes a gas
O B. When the entire liquid is a single temperature
O C. When the substance is very hot on top and cold beneath
O D. When the particles stop bumping into each other
SUBMIT

Answers

Answer:

When the entire liquid is a single temperature

Explanation:

When a liquid is heated, a convection current is set up. Convection is the movement of

fluid particles in response to a temperature gradient.

When you start heating a liquid, the particles near the base of the heating vessel increase in temperature, become less dense and rise upwards while the denser particles move downwards. This convection current will continue until an equilibrium temperature is obtained throughout the liquid.

What is the smallest value of n for which the wavelength of a Balmer series line is less than 400 nm

Answers

Answer:

The smallest value is n= 2

Explanation:

The balmer equation is given below

1/λ = R(1/4 - 1/n₂²).

R= 1.0973731568508 × 10^7 m^-1

λ= 400*10^-9 m

(400*10^-9)= 1.0973731568508 × 10^7 (1/4-1/n²)

(400*10^-9)/1.0973731568508 × 10^7

= 1/4 - 1/n²

364.51 *10^-16= 1/4 - 1/n²

1/n²= 1/4 -364.51 *10^-16

1/n² = 0.25-3.6451*10^-14

1/0.25= n²

4= n²

√4= n

2= n

The smallest value is N= 2

A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy

Answers

Answer: her rotational kinetic energy increases

An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Answers

Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

pls give brainliest

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

                      [tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]

Substitute,  [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]

                   [tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]

Hence, The new length of aluminum rod is 17.435 cm.

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An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The electron reaches a closest distance of 15 cm before being repelled.

What is the plate's surface charge density?

Answers

Answer:

2.29e-9C/m²

Explanation:

Using E = σ/ε₀ means the force on the electron is F = eE = eσ/ε₀.

The work done on the electron is W = Fd = deσ/ε₀. This equals the kinetic energy lost, ½mv².

½mv² = deσ/ε₀

d = 75cm – 15cm = 60cm = 0.6m

σ = mv²ε₀/(2de)

. .= 9.11e-31 * (7.4e6)² * 8.85e-12 / (2 * 0.6 * 1.6e-19)

. .= 2.29e-9 C/m² (i.e. 2.29x10^-9 C/m²)

A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.

Answers

Answer:

82.2 mL

Explanation:

The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;

C1V1=C2V2

Where;

C1= concentration of stock solution

V1= volume of stock solution

C2= concentration of dilute solution

V2= volume of dilute solution

V2= C1V1/C2

V2= 1.48 × 55.6/ 1.0

V2= 82.2 mL

A converging lens 7.50 cm in diameter has a focal length of 330 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of resolving power of the human eye. Part A If the resolution is diffraction limited, how far away can an object be if points on it transversely 4.10 mm apart are to be resolved (according to Rayleigh's criterion) by means of light of wavelength 600 nm

Answers

Answer:

D Is 430m

Explanation:

See attached file

A person looks horizontally at the edge of a swimming pool. If its length is 5 m, and the pool is filled to the surface, to what depth (in m) could the observer see

Answers

Answer:

The observer could see to a depth of 4.38 m

Explanation:

Please check attachment for diagram.

Mathematically, from Snell law;

n1sin theta = n2 sin theta

1 * sin 90 = n2 * sin θR

where n2 = 1.33

1/1.33 = sin θR

Sin θR = 0.7519

θR = arc sin 0.7519

θR = 48.76

Now to get the height, we use the triangle

Using trigonometric ratio;

Tan( 90- θR) = H/5

H = 5 Tan( 90 - θR)

H = 5 Tan( 90-48.76)

H = 5 Tan41.24

H = 4.38 m

What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light in a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s

Answers

Answer:

The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

Explanation:

To find the speed of light in air and in polystyrene we need to use the following equation:

[tex] c_{m} = \frac{c}{n} [/tex]

Where:

[tex]c_{m}[/tex]: is the speed of light in the medium

n: is the refractive index of the medium

In air:

[tex]c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s[/tex]

In polystyrene:

[tex]c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s[/tex]  

Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

I hope it helps you!

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1.0 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Required:
a. Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

b. Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm^2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

Answers

Answer:

1. 6.99x 10^-6V/m

2. 18m

Explanation:

See attached file

Mars Rover When the Mars rover was deployed on the surface of Mars in July 1997, radio signals took about 12 minmin to travel from Earth to the rover.
How far was Mars from Earth at that time?

Answers

Answer:

s = 2.16 x 10¹¹ m

Explanation:

Since, the waves travelling from Earth to the Mars rover are electromagnetic. Therefore, there speed must be equal to the speed of light. So, from the equation given below:

s = vt

where,

s = the distance between Earth and Mars = ?

v = speed of the wave = speed of light = 3 x 10⁸ m/s

t = time taken by the radio signals to reach the rover from Earth

t = (12 min)(60 s/1 min) = 720 s

Therefore,

s = (3 x 10⁸ m/s)(720 s)

s = 2.16 x 10¹¹ m

if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?

Answers

Answer:

0.35 m³/s

Explanation:

When the pipe's depth is 0.4 m, the area of the circular segment is:

A = ½ R² (θ − sin θ)

The depth of the water is:

h = R (1 − cos(θ/2))

Solving for θ:

0.4 = 0.5 (1 − cos(θ/2))

0.8 = 1 − cos(θ/2)

cos(θ/2) = 0.2

θ/2 = acos(0.2)

θ = 2 acos(0.2)

θ ≈ 2.74 rad

The area is therefore:

A = ½ (0.5 m)² (2.74 − sin 2.74)

A = 0.338 m²

The cross-sectional area when the pipe is full is:

A = π (0.5 m)²

A = 0.785 m²

The flow velocity is constant:

v = v

Q / A = Q / A

(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)

Q = 0.35 m³/s

Calculate the range of wavelengths (in m) for AM radio given its frequency range is 540 to 1,600 kHz. smaller value m larger value m (b) Do the same for the ultraviolet frequency range of 760 to 30,000 THz. smaller value m larger value m

Answers

Answer:

a)  λ = 555.5 m,  λ = 187.5 nm

 

Explanation:

The velocity of a wave is given by the relation

           c = λ f

           λ = c /f

a) length of the radii AM

            λ = 3 10⁸/540 10⁻⁷

            λ= 5.555 102 m

            λ = 555.5 m

             f = 1600 kHz

            λ = 3 108/1600 103

             λ = 1.875 102 m

            lam  = 187.5 nm

b) light = 760 Thz = 760 10-12 Hz

               

         λ=  c/f

            λ = 3 108/760 10-12

            λ = 3.947 10-9

           λ= 30000 Thz

           λ= c/f

      λ = 3 10⁸/ 30000 10-12

       λ = 1  m

Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?

Answers

Answer:

L = 0.8 m

Explanation:

Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:

Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m

but,

Δx = λL/d

λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m

d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m

L = Distance between slits and screen = ?

Therefore, using the values, we get:

2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)

L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)

L = 0.8 m

within which type of system is the total mass conserved but not the total energy

Answers

In a closed system the mass is conserved, but the energy is not conserved.

To find the answer, we have to study about different systems in thermodynamics.

What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.

Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.

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Sunlight reflected from a smooth ice surface is completely polarized. Determine the angle of incidence. (nice = 1.31.)

Answers

Answer:

Explanation:

We shall apply Brewster's law to solve the problem . Let i be the angle of reflection required. According to this law ,

tan i = μ where i is angle of incidence , μ is refractive index of the medium .

Here i = ? , μ = 1.31

tani = 1.31

i = 53° approx .

At the angle of 53° incidence , the reflected ray will be completely polarised .

"How many wavelengths wide must a single slit be if, at a point 8o from the central maximum, there is a 72 rad phase difference between the top and bottom rays?"

Answers

Answer

82.3 wavelengths

The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter and 1.2 m long. The starter motor runs for 0.760 s until the car engine starts.Required:a. How much charge passes through the starter motor? b. How far does an electron travel along the wire while the starter motor is on?(mm)

Answers

Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

Q= required charge

Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

diameter of the conductor is 4.20 mm

But Radius= diameter/2= 4.20/2=

The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then

radius of the conductor is 0.0021m.

We can now calculate the area of the conductor which is

A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

We can proceed to calculate the current density below

J = 140/13.85*10^-6 = 10108303A/m

According to the listed reference:

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s

Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

(a) The charge passes through the starter motor is 106.4C.

(b) An electron travel along the wire while the starter motor is on 0.5676mm.

Electron

Answer (a)

I= current

t= time taken

Q= required charge

Q= I*t

Q= 140*0.760

Q= 106.C

Answer (b)

The n electron travel along the wire while the starter motor is on:

Diameter of the conductor is 4.20 mm

Radius= diameter/2= 4.20/2

Radius =2.1mm

Radius of the conductor is 0.0021m.

A = π*r^2

A= π*(0.0021)^2

A= 13.85*10^-6 m^2

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )

Vd  =0.0007468m/s

Vd =0 .7468 mm/s

The distance traveled is:

x = v*t

x= 0.7468 * 0.760

x = 0.5676mm

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A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=5.0rad/s. A second cylinder with rotational inertia I2=1.0kg·m2 rotates counterclockwise about the same axis with angular speed ω2=8.0rad/s. If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?

Answers

Answer:

a) 0.67 rad/sec in the clockwise direction.

b) 98.8% of the kinetic energy is lost.

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia [tex]I[/tex] = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia [tex]I[/tex] = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = [tex]I[/tex]ω

where [tex]I[/tex] is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = [tex]I[/tex]ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = [tex]I[/tex]ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s

When they are coupled coupled together, their total rotational inertia [tex]I_{t}[/tex] = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = [tex]I_{t}[/tex][tex]w_{f}[/tex]

where [tex]I_{t}[/tex] is their total rotational inertia

[tex]w_{f}[/tex] = their final angular speed together

Final angular momentum = 3 x [tex]w_{f}[/tex] = 3[tex]w_{f}[/tex]

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 =  3[tex]w_{f}[/tex]

[tex]w_{f}[/tex] = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s

From the first statement, the direction is clockwise

b) Rotational kinetic energy = [tex]\frac{1}{2} Iw^{2}[/tex]

where [tex]I[/tex] is the rotational inertia

[tex]w[/tex] is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = [tex]\frac{1}{2} Iw^{2}[/tex] = [tex]\frac{1}{2}*2*5^{2}[/tex] = 25 J

for second cylinder = [tex]\frac{1}{2}*1*8^{2}[/tex] = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = [tex]\frac{1}{2}I_{t}w^{2} _{f}[/tex]

where

where [tex]I_{t}[/tex] is the total rotational inertia of the cylinders

[tex]w_{f}[/tex] is final total angular speed of the coupled cylinders

Final kinetic energy =  [tex]\frac{1}{2}*3*0.67^{2}[/tex] = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%

A) The angular speed of the combination of the two cylinders is; ω₃ = 0.67 rad/s

B) The percentage of the original kinetic energy lost to friction is;

percentage energy lost = 98.82%

We are given;

Rotational Inertia for first cylinder; I₁ = 2 kg.m²

Angular speed of first cylinder; ω₁ = 5 rad/s

Angular speed of second cylinder; ω₂ = 8 rad/s

Rotational Inertia for second cylinder; I₂ = 1 kg.m²

From conservation of angular momentum, we know that;

Initial angular Momentum([tex]L_{i}[/tex]) = Final angular Momentum([tex]L_{f}[/tex])

Thus;

I₁ω₁ + I₂ω₂ = I₃ω₃

Where;

ω₃ is the angular speed when the two cylinders are combined

I₃ = I₁ + I₂

I₃ = 2 + 1

I₃ = 3 kg.m²

Since the second cylinder rotates in an anticlockwise direction, then its' angular speed will be negative. Thus;

(2 * 5) + (1 * -8) = 3ω₃

10 - 8 = 3ω₃

3ω₃ = 2

ω₃ = 2/3

ω₃ = 0.67 rad/s

B) Let us find initial kinetic energy;

E_i = ¹/₂I₁ω₁² + ¹/₂I₂ω₂²

E_i = ¹/₂((2 * 5²) + (1 * 8²)

E_i = 57 J

Final kinetic energy is;

E_f = ¹/₂I₃ω₃²

E_f = ¹/₂ * 3 * 0.67²

E_f = 0.67335 J

Energy lost = 57 - 0.67335 = 56.32665 J

percentage energy lost = (56.32665/57) * 100%

percentage energy lost = 98.82%

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Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.

Answers

Answer:

se the principle of induction.

place the two metallic spheres together,  now we bring the positively charged bar closer to the first sphere.

The charge that was induced in the sphere is distributed as infirm as possible,

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive

Explanation:

For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.

Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.

Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.

The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.

At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.

Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.

Answers

Complete Question

An emf is induced in a conducting loop of wire 1.12m long as its shape is.

changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.

Answer:

The induced emf is  [tex]\epsilon = 0.0863 \ V[/tex]

Explanation:

From the question we are told that

      The  time taken is  [tex]\Delta t = 0.125 \ s[/tex]

       The magnitude of the magnetic field is  B =  0.504 T

        The length of the loop wire is  [tex]l = 1.12 \ m[/tex]

Generally the circumference of the wire when in circular form is  

          [tex]C = 2 \pi r[/tex]

=>        [tex]l = 2 \pi r[/tex]

=>         [tex]r =[/tex][tex]\frac{l}{2 \pi}[/tex]

=>          [tex]r =[/tex][tex]\frac{1.12}{2 * 3.142}[/tex]

=>        [tex]r =[/tex][tex]0.1782 \ m[/tex]

Now the area of the wire as a circle is

           [tex]A = \pi r^2[/tex]

    =>     [tex]A = 3.142 * (0.1782)^2[/tex]      

     =>    [tex]A = 0.0998 \ m^2[/tex]

The  length of one side of the square is

         [tex]b = \frac{l}{4}[/tex]

         [tex]b = \frac{1.12}{4}[/tex]

         [tex]b = 0.28 \ m[/tex]

Now the area of the wire as a square is

          [tex]A_s = b^2[/tex]

=>          [tex]A_s =(0.28 )^2[/tex]

             [tex]A_s = 0.0784 \ m^2[/tex]

Generally the induced emf is mathematically represented as

        [tex]\epsilon = \frac{B * [A - A_s ]}{\Delta t }[/tex]

=>      [tex]\epsilon = \frac{0.504 * [0.0998 - 0.0784 ]}{0.125 }[/tex]

=>      [tex]\epsilon = 0.0863 \ V[/tex]

   

An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?

Answers

Answer:

3.9E-8

Explanation:

We know that

Mv²/r = Bqv

So

r= mv/Bq

But E is 1/2mv² which is 5.53eV

m²v² =2m x 5.53eV

mv = √( 2 x 9.1E-31)

So

r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19

= 3.9x10-8cm

Explanation:

You have a horizontal grindstone (a disk) that is 95 kg, has a 0.38 m radius, is turning at 87 rpm (in the positive direction), and you press a steel axe against the edge with a force of 16 N in the radial direction.
(a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone.
(b) How many turns will the stone make before coming to rest?

Answers

Answer:

Explanation:

The moment of inertia of the disk I  = 1/2 m R² where R is radius of the disc and m is its mass .

putting the values

I = .5 x 95 x .38²

= 6.86 kg m²

n = 87 rpm = 87 / 60 rps

n = 1.45 rps

angular velocity ω = 2π n , n is frequency of rotation .

= 2 x 3.14 x 1.45

= 9.106 radian /s

frictional force = 16 x .2

= 3.2 N

torque created by frictional force = 3.2 x .38

= 1.216 N.m

angular acceleration = torque / moment of inertia

= - 3.2 / 6.86

α  = - 0.4665 rad /s²

b ) ω² = ω₀² +  2 α θ , where α is angular acceleration

0 = 9.106² - 2 x .4665 θ

θ = 88.87 radian

no of turns = 88.87 / 2π

= 14.15  turns

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m ​ 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.

Answers

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.

For this "type" of motion, displacement (Δx) can be determined by:

[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]

[tex]v_{i}[/tex] is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]

[tex]\Delta x = 30 - 2.36[/tex]

[tex]\Delta x[/tex] = - 42

Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.

A scuba diver fills her lungs to capacity (6.0 L) when 10.0 m below the surface of the water and begins to ascend to the surface. Assume the density of the water in which she is swimming is 1000 kg/m3 and use g = 10 m/s2A. Assuming the temperature of the air in her lungs is constant, to what volume must her lungs expand when she reaches the surface of the water?B. What effect would the warming of the air in her lungs have on the volume needed when she surfaces?C. Assuming the temperature of the air in her lungs is constant, what effect does her ascent have on the vrms of the air molecules in her lungs?

Answers

Answer:

Explanation:

As temperature is constant , we shall apply Boyle's law

P₁V₁ = P₂V₂

P₁ = pressure at depth of 10 m

= P + hdg , h = 10 , d = 10³ , g = 10

P is atmospheric pressure which is 10⁵ Pa

P₁ = 10⁵ + 10 x 10³ x 10

= 2 x 10⁵

applying the formula

2 x 10⁵ x 6 = 10⁵ x v

v = 2 x 6 = 12 L

volume will be doubled at the surface .

B )

warming of air at the surface will increase the volume of air in her lungs so so she will need more lung capacity .

C )

The rms value of a gas depends upon the temperature of the gas . As temperature of the gas is constant , the rms value of the gas particles will remain constant when she goes to the surface .

The lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.

What is Boyle's law?

According to the law, the pressure of the gas is inversely proportional to the volume of the gas. In other words when the pressure of the gas increases the volume of the gas decreases.

We know the pressure at the 10 meters depth:

[tex]\rm P_1 = P+h\times \rho\times g[/tex]

Where P = Atmospheric pressure

            h = Depth

            ρ =Density of the water

We have: [tex]\rm P = 10^5 \ Pa[/tex], h = 10 meters, and [tex]\rm \rho = 1000 \ kg/m^3[/tex], and [tex]\rm g = 10 \ m/s^2[/tex]

Putting the values in the above equation, we get:

[tex]\rm P_1 = 10^5+ 10\times 1000\times 10[/tex]

[tex]\rm P_1 = 2\times 10^5[/tex]

From the Boyle's law:

[tex]\rm P_1\times V_1 = P_2\times V_2[/tex]

[tex]\rm 2\times10^5\times 6 = 10^5\times V_2[/tex]

[tex]\rm V_2 = 12 \ L[/tex]

We know that as the air at the surface warms, the volume of air in her lungs expands, requiring more lung capacity.

The temperature of the gas is constant and [tex]\rm V_{rms}[/tex] values for gas depend on the temperature of the gas, but here the temperature of the gas is constant thus, the  [tex]\rm V_{rms}[/tex] will remains constant.

Thus, the lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.

Learn more about the Boyle's law here:

https://brainly.com/question/23715689

Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.

Requried:
What is the frequency the horns emit?

Answers

Answer: f ≈ 8.5Hz

Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.

For a source moving and a stationary observer, to determine the frequency:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

where:

[tex]f_{0}[/tex] is frequency of observer;

[tex]f_{s}[/tex] is frequency of source;

c is the constant speed of sound c = 340m/s;

[tex]v_{s}[/tex] is velocity of source;

Rearraging for frequency of source:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]

Replacing and calculating:

[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]

[tex]f_{s} = 9.(0.9412)[/tex]

[tex]f_{s} =[/tex] 8.5

Frequency the horns emit is 8.5Hz.

When a ray of light passes from glass to water it is?

Answers

Answer:

[tex]\huge\boxed{Refracted}[/tex]

Explanation:

When a ray of light passes from glass to water, it

1) is Slightly refracted (bending of light)

2) moves away from the normal.

Whenever a light ray travels from a denser medium to a rarer medium, it bends away from the normal.

Answer:

refraction

Explanation:

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