Water enters a 180° elbow with an average velocity V1 = 3 m/s. The diameter at the inlet of the elbow is 6.0 cm and the exit diameter is 2.0 cm. The water exits to atmosphere. Determine the magnitude of the horizontal force required to hold the elbow in place. (Ignore the pressure component of the force.)

Answer:

Partial pressure of nitrogen gas,[tex]p^o_{N_2] =0.44 atm[/tex]

Explanation:

Pressure of the mixture of gases before adding nitrogen gas = 0.85 atm

Pressure of the mixture of gases after adding nitrogen gas = 988 mmHg

1 mmHg = 0.001315 atm

988 mmHg=[tex]988 mmHg\times 0.001315 atm = 1.29 atm[/tex]

Partial pressure of nitrogen gas,[tex]p^o_{N_2] = 1.29 atm - 0.85 atm = 0.44 atm[/tex]

According to Dalton's law of partial pressure , the total pressure of the mixture of gases is equal to sum of all the partial pressures of each gas present in the mixture.

[tex]P_{total}=\sum p^o_{i}[/tex]

So, on addition of nitrogen gas to the mixture the pressure of the mixture increases.

The distance from where the ball hits the ground is approximately 7.53 m.

To find the distance from where the ball hits the ground, we need to analyze the horizontal and vertical motion separately. First, we can find the time it takes for the ball to hit the ground using the vertical motion equation. The final position in the y-axis is 0, the initial position is 1 m, the initial vertical velocity is 8*sin(40°) m/s, and the acceleration is -9.8 m/s² (due to gravity). By solving the equation, we find that the time of flight is 1.15 s. Using this time, we can find the horizontal distance traveled by the ball using the horizontal motion equation. The initial horizontal velocity is 8*cos(40°) m/s, and multiplying it by the time of flight gives us the answer: approximately 7.53 m.

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Answer:

12500 J

Explanation:

m = 20 kg, u = 10 m/s, s = 2 m, v = 0

Use third equation of motion

v^2 = u^2 = 2 a s

0 = 100 - 2 a x 2

100 / 4 = - a

a = - 25 m/s^2

Force, F = m a = 20 x 25 = 500 N

Work done = F x s = 500 x 25 = 12500 J

Answer:

8.85 x 10⁴ Nm²/C

Explanation:

d = diameter of the conducting sphere = 0.10 m

r = radius of the conducting sphere = (0.5) d = (0.5) (0.10) = 0.05 m

Area of the sphere is given as

A = 4πr²

A = 4 (3.14) (0.05)²

A = 0.0314 m²

σ = Surface charge density = 150 x 10⁻⁶ C/m²

Q = total charge enclosed

Total charge enclosed is given as

Q = σA

Q = (150 x 10⁻⁶) (0.0314)

Q = 4.7 x 10⁻⁶ C

Electric flux through one of the side is given as

[tex]\phi = \frac{Q}{6\epsilon _{o}}[/tex]

[tex]\phi = \frac{4.7\times 10^{-6}}{6(8.85\times 10^{-12})}[/tex]

[tex]\phi [/tex] = 8.85 x 10⁴ Nm²/C

Answer:

[tex]\dot{m_{2}}=0.865 kg/s[/tex]

Explanation:

[tex]\dot{m_1}= 0.5kg/s[/tex]

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

[tex]\dot{m_{in}}=\dot{m_{out}}[/tex]

[tex]\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}[/tex]

according to energy balance equation

[tex]\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}[/tex]

[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]

[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s[/tex]

The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s

Given data :

m₁ = 0.5 kg/s

m₂ = ?

From steam tables

At 250 kPa

at T1 = 80℃ , h₁ = 335.02 kJ/kg

at T2 = 20℃, h₂ = 83.915 kJ/kg

at T3 = 42℃, h₃ = 175.90 kJ/kg

Given that :

Min = Mout    also m₁ + m₂ = m₃

applying the principle of energy balance

M₁h₁ + M₂ h₂  = ( m₁ + m₂ ) h₃ ---- ( 1 )

insert values into equation ( 1 )

m₂ = 0.865 kg/s

Hence In conclusion The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s

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Answer:

Angle of refraction.

Explanation:

Refractive index is given as the ratio of the angle of incidence to the angle of refraction.

If angle of incidence is i and the refractive index of the incident medium is n₁ , and if angle of refraction is r and refractive index of the refracting medium is n₂,

according to Snell's law, n₁ sin i = n₂ sin r,

relative refractive index = sin i / sin r

Answer:

The actual size is

69.33 nm.

Explanation:

It means that

3 cm is equivalent to 40 nm

So, 1 cm is equivalent to 40 / 3 nm

Thus, 5.2 cm is equivalent to

40 × 5.2 / 3 = 69.33 nm

Answer:

2*10^-5

Explanation:

B=IL

I=B/L

I=0.0020*10^-4/10

I=2*10^5

The magnitude of the current in a magnetic field is  2 x10⁵ Amperes.

The magnetic field is the region of space where a charged object experiences magnetic force when it is moving.

The formula for magnetic field is given as;

B=IL

I=B/L

A very long wire generates a magnetic field of 0.0020 x 10⁻⁴ T at a distance of 10 mm.

Plug the values, we get

I=0.0020 x 10⁻⁴/(10 x 10⁻³)

I=2 x10⁵

Thus, the magnitude of the current is 2 x10⁵ Amperes.

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Answer:

No, the car will not make it to the top of the hill.

Explanation:

Let ΔX be how long the slope of the hill is, Δx be how far the car will travel along the slope of the hill, Ф be the angle the slope of the hill makes with the horizontal(bottom of the hill), ki be the kinetic energy of the car at the bottom of the hill and vi be the velocity of the car at the bottom of the hill and kf be the kinetic energy of the car when it stop moving at vf.

Since Ф is the angle between the horizontal and the slope, the relationship between the angle and the slope and the height of the hill is given by

sinФ = 12/ΔX

Which gives you the slope as

ΔX = 12/sinФ

Therefore for the car to reach the top of the hill it will have to travel ΔX.

Ignoring friction the total work done is given by

W = ΔK

W = (kf - ki)

Since the car will come to a stop, kf = 0 J

W = -ki

m×g×sinФ×Δx = 1/2×m×vi^2

(9.8)×sinФ×Δx = 1/2×(10)^2

sinФΔx = 5.1

Δx = 5.1/sinФ

ΔX>>Δx Ф ∈ (0° , 90°)

(Note that the maximum angle Ф is 90° because the slope of a hill can never be greater ≥ 90° because that would then mean the car cannot travel uphill.)

Since the car can never travel the distance of the slope, it can never make it to the top of the hill.

Answer:

Part a)

Magnitude = 5.06 unit

Part b)

[tex]\theta = 36.2 ^0[/tex]

Explanation:

Part a)

Vector is given as

[tex]\vec A = 4.08 \hat x + 3.0 \hat y[/tex]

now from above we can say that

x component of the vector is 4.08

y component of the vector is given as 3.0

so the magnitude of the vector is given as

[tex]|A| = \sqrt{4.08^2 + 3^2}[/tex]

[tex]|A| = 5.06 unit[/tex]

Part b)

Now the angle made by the vector is given as

[tex]\theta = tan^{-1}(\frac{y}{x})[/tex]

[tex]\theta = tan^{-1}(\frac{3}{4.08})[/tex]

[tex]\theta = 36.3 degree[/tex]

Answer:

[tex]a_c = 120.4 m/s^2[/tex]

Explanation:

As we know that stone is revolving in horizontal circle

so here height of the stone is H = 1.9 m

now by kinematics we can find the time to reach it at the ground

[tex]H = \frac{1}[2}gt^2[/tex]

now we have

[tex]1.9 = \frac{1}{2}(9.81)t^2[/tex]

now we have

[tex]t = 0.622 s[/tex]

now the speed of the stone is given as

[tex]v = \frac{x}{t}[/tex]

[tex]v = \frac{8.9}{0.622} = 14.3 m/s[/tex]

now for finding centripetal acceleration we have

[tex]a_c = \frac{v^2}{R}[/tex]

we know that radius of circle is

R = 1.7 m

now we have

[tex]a_c = \frac{14.3^2}{1.7}[/tex]

[tex]a_c = 120.4 m/s^2[/tex]

Final answer:

The length of the string between Wanda's hand and the tennis ball is 4.5 cm.

Explanation:

To find the length of the string between Wanda's hand and the tennis ball, we need to use the formula for centripetal force:

F = (m * v^2) / r

where F is the tension force, m is the mass of the ball, v is the velocity, and r is the radius of the circular motion.

In this case, the tension force is 12 N, the mass of the ball is 60 g (0.06 kg), and the velocity is 9.0 m/s. We can rearrange the formula to solve for r:

r = (m * v^2) / F

Substituting the given values, we get:

r = (0.06 * 9.0^2) / 12

r = 0.045 m = 4.5 cm

Therefore, the length of the string between Wanda's hand and the tennis ball is 4.5 cm.

Final answer:

The argon temperature at the heater exit is determined to be 348.08 ℃, while the argon volume flow rate at the heater exit is calculated to be 3769.23 m³/s.

Explanation:

To determine the argon temperature and volume flow rate at the heater exit, we can use the principles of thermodynamics. The problem states that heat transfer in the rate of 150 kW is supplied to the argon. Since there is no information given about the heater efficiency, we can assume it to be 100%, meaning all the heat is transferred to the argon. Therefore, the heat transfer can be equated to the change in internal energy of the argon, which is given by:

Q = m * Cp * ΔT

Where Q is the heat transfer, m is the mass flow rate, Cp is the specific heat capacity of argon, and ΔT is the change in temperature.

Given that Q = 150 kW, m = 6.24 kg/s, and Cp = 0.52 kJ/kg·℃ (specific heat capacity of argon), we can rearrange the equation to solve for ΔT:

ΔT = Q / (m * Cp)

Substituting the values, we have ΔT = (150 * 10³ J/s) / (6.24 kg/s * 0.52 kJ/kg·℃), which equals 48.08 ℃.

Therefore, the argon temperature at the heater exit is 300 ℃ + 48.08 ℃, which equals 348.08 ℃.

To calculate the argon volume flow rate at the heater exit, we can use the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation to solve for V, we have:

V = (nRT) / P

Given that the initial conditions are 300 K and 100 kPa, and a mass flow rate of 6.24 kg/s, we can calculate the number of moles of argon:

n = m / M

Where M is the molar mass of argon. The molar mass of argon is 39.948 g/mol, so n = 6.24 / (39.948 * 10⁻³) = 156.172 mol.

Substituting the values into the ideal gas law equation, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100 kPa. Converting kPa to Pa, we have V = (156.172 mol * 8.314 J/mol·K * 300 K) / 100,000 Pa, which equals 3769.23 m³.

Therefore, the argon volume flow rate at the heater exit is 3769.23 m³/s.

Given:

heat generated by John's cooling system,  [tex]H = \rho A v^{3}[/tex]  = 45 W    (1)

If ρ, A, and v corresponds to John's cooling system then let [tex]\rho_{1}, A_{1}, v_{1}[/tex] be the variables for Mike's system then:

[tex]\rho  = 9.5\rho_{1}[/tex]

[tex]\rho_{1}  = \frac{\rho}{9.5}[/tex]

[tex]v_{1} =3.5 v[/tex]

Formula use:

Heat generated, [tex]H = \rho A v^{3}[/tex]

where,

[tex]\rho[/tex] = density

A = area

v = velocity

Solution:

for Mike's cooling system:

[tex]H_{2}[/tex] = [tex]v_{1}^{3}{1}A_{1}\rho_{1}[/tex]

⇒ [tex]H_{2}[/tex] = [tex](3.5v)^{3}[/tex] × A × [tex]\frac{\rho}{9.5}[/tex]

[tex]H_{2}[/tex] = 4.513[tex]v^{3}[/tex] A  [tex]\rho[/tex]

Using eqn (1) in the above eqn, we get:

[tex]H_{2}[/tex] = 4.513 × 45 = 203.09 W

We have that The Heat loss H2 is given as

From the question we are told

Generally the equation for the   is mathematically given as

H=PAV^3

Therefore

[tex]\frac{110}{h_2}=\frac{P_1}{p1/3.5}*(\frac{V_1}{3.5*v_1})^3[/tex]

H_2=1347.5w

Therefore

The Heat loss H2 is given as

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Answer:

Distance, x = 73.10 meters

Explanation:

It is given that,

Mass of the rock, m = 25 g = 0.025 kg

Acceleration due to gravity, a = 1.62 m/s²

We need to find the distance traveled by the rock when it falls from rest in 9.5 seconds if the only force acting on it was the gravitational force due to the Moon. It can be calculated using the second equation of motion i.e.

[tex]x=ut+\dfrac{1}{2}at^2[/tex]

[tex]x=0+\dfrac{1}{2}\times 1.62\ m/s^2\times (9.5\ s)^2[/tex]

x = 73.1025 meters

or

x = 73.10 meters

So, the distance traveled by the rock is 73.10 meters. Hence, this is the required solution.

The correct answer is e) atom A becomes a positive ion and atom B becomes a negative ion. Ions are charged particles that result when a neutral atom gains or loses an electron.

Further Explanation:

In the problem, atom A loses an electron to atom B.

Suppose atom A has 11 protons and 11 electrons and atom B has 9 protons and 9 electrons.

BEFORE TRANSFER OF ELECTRONS

atom A:    11 protons, 11 electrons

atom B:     9 protons, 9 electrons

AFTER TRANSFER OF ELECTRONS

atom A:    11 protons, 10 electrons

atom B:      9 protons, 10 electrons

a) atom A becomes a negative ion and atom B becomes a positive ion FALSE because atom A has more protons than electrons turning it into a positive ion, and atom B now has more electrons making a negative ion.

b) atom A acquires more neutrons than atom B FALSE because only the number of electrons changed. The number of neutrons are unchanged in the formation of ions.

c) atom A becomes more negative than atom B FALSE because atom A becomes more positive since there is more protons (positively charged) than electrons (negatively charged) while atom B, with its extra electron becomes negatively charged

d) atom A acquires less neutrons than atom B FALSE because the number of neutrons does not change. Only an electron was transferred.

e) atom A becomes a positive ion and atom B becomes a negative ion TRUE because there are more protons than electrons in A after the electron transfer and there are more electrons in B than protons after receiving the electron.

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Keywords: ion, electron, atom

Answer:

d) None of the above

Explanation:

[tex]v_{o}[/tex] = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

[tex]v_{oy}[/tex] = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m  

t = time of travel

using the equation

[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.1 s

Using the kinematics equation

[tex]x =v_{ox} t+(0.5)a_{x} t^{2}[/tex]

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

[tex]v_{oy}[/tex] = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²

[tex]y_{o}[/tex] =initial vertical position at the time of launch = 20 m  

[tex]y[/tex] = vertical position at the maximum height = 20 m

[tex]v_{fy}[/tex] = final velocity along vertical direction at highest point = 0 m/s

using the equation

[tex]{v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})[/tex]

[tex]0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)[/tex]

[tex]y[/tex] = 20.02 m

h = height above the starting height

h = [tex]y[/tex] - [tex]y_{o}[/tex]

h = 20.02 - 20

h = 0.02 m

Answer:

Angle of refelction is 60°

Explanation:

Let two mirrors XY and ZY  meet at Y such that ∠XYZ = 30°

Let an incident ray PO incident on the 1st mirror XY and is parallel to the 2nd mirror YZ.

Let ON be the normal.

Now we know that , angle of incidence is equal to the angle of reflection

∴∠PON = ∠NOQ =60°

Therefore, the angle of reflection of the ray incident on the 1st mirror is 60 degree.

The angle of reflection of the light ray from Mirror 1 is 30°. This is in accordance with the Law of Reflection which states that the angle of incidence equals the angle of reflection. Examples of this principle are corner reflectors on bikes or cars and binoculars.

According to The Law of Reflection, the angle of incidence is equal to the angle of reflection. When a light ray approaches a mirror parallel to another, it will reflect off at the same angle it hit, but in the opposite direction. Therefore, if the light ray approaches Mirror 1 parallel to Mirror 2, and considering that the two mirrors meet at a 30° angle, the light should reflect off at a 30° angle to the normal (perpendicular line) of Mirror 1.

This behavior of light can be observed in corner reflectors used on bikes and cars, where light is reflected back exactly parallel to the direction from which it came. It's also a principle used in binoculars and periscopes, where light is made to reflect multiple times in the system at respective angles of incidence and reflection.

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Answer:

[tex]q = 598.9 \mu C[/tex]

Explanation:

As we know that when an uncharged capacitor is connected through a voltage source then the charge on the plates of capacitor will increase with time.

It is given by the equation

[tex]q = CV(1 - e^{-\frac{t}{RC}})[/tex]

here we know that

[tex]C = 13.1 \mu F[/tex]

[tex]R = 853 k ohm[/tex]

V = 64 volts

now we have

[tex]q = (13.1 \mu F)(64) ( 1 - e^{-\frac{14}{(853\times 10^3)(13.1 \times 10^{-6})}})[/tex]

[tex]q = 598.9 \mu C[/tex]

Answer:

A  1.4ly

Explanation:

speed v= 0.95c c= speed of light distance t₀=4.3ly

As per the time dilation priciple we know that

[tex] t=t_{0}\sqrt {\frac {c^2 -v^2}{c^2}} [/tex]

[tex]\Rightarrow t=4.3\sqrt {1-0.95^2}[/tex]

t=1.4 ly

therefore, it will take 1.4ly for the trip from the point of view of passenger on ship. ly here is light year.

Answer:

The acceleration of the car is 8.50 m/s²

Explanation:

Given that,

Initial velocity = 10 m/s

Final velocity = 61 m/s

Time = 6 s

We need to calculate the acceleration of the car

Using equation of motion

[tex]v=u+at[/tex].....(I)

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

Where, u = initial velocity

v = final velocity

t = time

Put the value in the equation (I)

[tex]a=\dfrac{61-10}{6}[/tex]

[tex]a=8.50\ m/s^2[/tex]

Hence, The acceleration of the car is 8.50 m/s²

Answer:

175  [tex]\frac{mi}{h}[/tex]

Explanation:

v = speedometer reading of a european sportscar = [tex]282 \frac{km}{h}[/tex]

we know that,

1 km = 0.621 miles

So the speedometer reading of a european sportscar in  mi/h is given as

v = [tex]282 \frac{km}{h}[/tex] = [tex]282\left ( \frac{km}{h} \right ) \left ( \frac{0.621 mi}{1 km} \right )[/tex]

v = (282 x 0.621) [tex]\frac{mi}{h}[/tex]

v = 175  [tex]\frac{mi}{h}[/tex]

Answer and Explanation:

Since black holes deal with dark matter and dark energy and are dark enough not to allow even light to escape them, so their detection is much more difficult.

There are two basic methods to detect a black hole in space and these are listed below:

Answer:

L = 30 m

Explanation:

As per mechanical energy conservation law we can say that total kinetic energy of the hoop is equal to the total gravitational potential energy at the top

So here we can say for initial total kinetic energy as

[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]

for pure rolling we will have

[tex]v = r\omega[/tex]

also for large hoop we will have

[tex]I = mR^2[/tex]

now we have

[tex]KE = \frac{1}{2}mR^2\omega^2 + \frac{1}{2}(mR^2)\omega^2[/tex]

[tex]KE = mR^2\omega^2[/tex]

[tex]KE = (3.8)(1.3)^2(6.7)^2[/tex]

[tex]KE = 288.3 J[/tex]

now for gravitational potential energy we can say

[tex]U = mg(Lsin\theta)[/tex]

now by energy conservation we have

[tex]288.3 = (3.8)(9.81)(L sin15)[/tex]

[tex]L = 30 m[/tex]

The horizontal distance traveled by the hoop is approximately 30 m.

The total kinetic energy of the ball is the sum of the translational and rotational kinetic of the ball.

K.E(total) = K.E(trans) + K.E(rotational)

[tex]K.E(t) = \frac{1}{2}mv^2 + \frac{1}{2} I \omega ^2\\\\K.E(t) = \frac{1}{2}mv^2 + \frac{1}{2} (mR^2) \omega ^2\\\\K.E(t) = \frac{1}{2}m(\omega R)^2 + \frac{1}{2} (mR^2) \omega ^2\\\\K.E(t) = m\omega ^2R^2\\\\K.E(t) = 3.8 \times (6.7)^2 \times (1.3)^2\\\\K.E(t) = 288.28 \ J[/tex]

The horizontal distance traveled by the hoop is calculated as follows;

K.E = P.E

288.28  = mg(Lsinθ)

288.28 = (3.8 x 9.8) x (L) x sin(15)

288.28 = 9.63 L

L = 30 m

Thus, the horizontal distance traveled by the hoop is approximately 30 m.

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Answer:

0.217 m/s

Explanation:

The protons in the beam passes undeflected when the electric force is equal to the magnetic force:

qE = qvB

where

q is the proton's charge

E is the magnitude of the electric field

v is the speed of the protons

B is the magnitude of the magnetic field

Re-arranging the equation,

[tex]v=\frac{E}{B}[/tex]

And by substituting

E = 0.5 N/C

B = 2.3 T

We find

[tex]v=\frac{0.5}{2.3}=0.217 m/s[/tex]

The speed of the protons that pass through the parallel plates undeflected is determined by equalizing the electric and magnetic forces. By applying the formula v = E/B with E = 0.5 N/C and B = 2.3 T, we find that the speed is approximately 2.17 x 10^8 m/s.

To determine the speed of the protons that pass through a pair of parallel plates undeflected, we need to understand that the electric force and the magnetic force must be equal and opposite for the protons to travel in a straight line. The electric force (FE) is given by FE = qE, where q is the charge of the proton (q = 1.602 x 10-19 C) and E is the electric field strength.

The magnetic force (FB) on a moving charge is given by FB = qvB sin(θ), where v is the velocity of the proton, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field direction. Since the magnetic field is perpendicular to the velocity of the protons, sin(θ) = 1.

For the proton beam to pass through undeflected, FE must be equal to FB, hence qE = qvB. By canceling out the charge q from both sides and rearranging the equation, we get the velocity v = E/B. Substituting the given values, E = 0.5 N/C and B = 2.3 T, the speed of the protons is v = 0.5 N/C / 2.3 T which computes to approximately 2.17 x 108 m/s.

Answer:

The meteoroid's speed is 18.5 km/s

Explanation:

Given that,

Speed = 14.8 km/s

Distance [tex]d= 3.84\times10^{8}[/tex]

We need to calculate the meteoroid's speed

The total initial energy

[tex]E_{i}=K_{i}+U_{i}[/tex]

[tex]E_{i}=\dfrac{1}{2}mv_{i}^2-\dfrac{GM_{e}m}{r}[/tex]

Where, m = mass of  meteoroid

G = gravitational constant

[tex]M_{e}[/tex]=mass of earth

r = distance from earth center

Now, The meteoroid hits the earth then the distance of meteoroid from the earth 's center will be equal to the radius of earth

The total final energy

[tex]E_{f}=K_{f}+U_{f}[/tex]

[tex]E_{f}=\dfrac{1}{2}mv_{f}^2-\dfrac{GM_{e}m}{r_{e}}[/tex]

Where,

[tex]r_{e}[/tex]=radius of earth

Using conservation of energy

[tex]E_{i}=E_{j}[/tex]

Put the value of initial and final energy

[tex]\dfrac{1}{2}mv_{i}^2-\dfrac{GM_{e}m}{r}=\dfrac{1}{2}mv_{f}^2-\dfrac{GM_{e}m}{r_{e}}[/tex]

[tex]v_{f}^2=v_{i}^2+2GM_{e}(\dfrac{1}{r_{e}}-\dfrac{1}{r})[/tex]

Put the value in the equation

[tex]v_{f}^2=(14.8\times10^{3})^2+2\times6.67\times10^{-11}\times5.97\times10^{24}(\dfrac{1}{6.37\times10^{6}}-\dfrac{1}{3.84\times10^{8}})[/tex]

[tex]v_{f}=\sqrt{(14.8\times10^{3})^2+2\times6.67\times10^{-11}\times5.97\times10^{24}(\dfrac{1}{6.37\times10^{6}}-\dfrac{1}{3.84\times10^{8}})}[/tex]

[tex]v_{f}=18492.95\ m/s[/tex]

[tex]v_{f}=18.5\ km/s[/tex]

Hence, The meteoroid's speed is 18.5 km/s

To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. The final velocity of the meteoroid is approximately 13.4 km/s.

To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. Since there is no air friction, the mechanical energy of the meteoroid is conserved as it falls towards Earth. The initial kinetic energy of the meteoroid is equal to the final kinetic energy plus the gravitational potential energy.

First, we find the initial kinetic energy of the meteoroid using the formula KE = (1/2)mv^2, where m is the mass of the meteoroid and v is its initial velocity relative to the center of the Earth. Since the mass is not given, we can assume it cancels out in the equation.

Next, we calculate the gravitational potential energy of the meteoroid using the formula PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height from which the meteoroid fell. The height can be calculated by subtracting the radius of the Earth from the distance from the center of the Earth to the moon's orbit (h = 3.84 × 10^8 m - 6.37 x 10^6 m).

Solving for the final velocity, we equate the initial kinetic energy and the sum of the final kinetic energy and gravitational potential energy. Rearranging the equation, we find that the final velocity is the square root of (initial velocity squared minus 2 times g times h).

Plugging in the given values, the final velocity of the meteoroid as it hits the Earth is approximately 13.4 km/s.

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Answer:

1.7 × 10^8 N/C

Explanation:

Q = 7.6 nC, r = 2 cm = 0.02 m

E = KQ/r^2

E =

8.99 × 10^9 × 7.6 × 10^(-6)/(0.02 × 0.02)

E = 1.7 × 10^8 N/C

Answer:

Heat energy needed = 1243.45 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 0.500 kg water from 50°C to vapor at 110°C

First the water changes to 100°C from 50°C , then it changes to steam and then its temperature increases from 100°C to 110°C.

Mass of water = 500 g

Heat energy required to change water temperature from 50°C to 100°C

                   [tex]H_1=mC\Delta T=500\times 4.18\times (100-50)=104.5kJ[/tex]

Heat energy required to change water from 100°C to steam at 100°C

                   [tex]H_2=mL=500\times 2257=1128.5kJ[/tex]

Heat energy required to change steam temperature from 100°C to 110°C

                   [tex]H_3=mC\Delta T=500\times 2.09\times (110-100)=10.45kJ[/tex]

Total heat energy required

                   [tex]H=H_1+H_2+H_3=104.5+1128.5+10.45=1243.45kJ[/tex]

Heat energy needed = 1243.45 kJ

To find the electric field induced at a radius of 1.00 cm from the axis of the solenoid, we can use the formula for the magnetic field inside a solenoid:

B = μ₀nI,

where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.

First, let's find the magnetic field at a radius of 2.00 cm from the axis of the solenoid using the given values:

B = μ₀nI,

B = (4π×10⁻⁷ T·m/A)(1.65×10³ turns/m)(6.00 sin 90πt A),

B = (4π×10⁻⁷)(1.65×10³)(6.00 sin 90πt) T.

Now, we can use Ampere's law to find the electric field at a radius of 1.00 cm from the axis of the solenoid. Ampere's law states that the line integral of the magnetic field around a closed loop is equal to the permeability times the current enclosed by the loop.

∮B·dl = μ₀I_enclosed.

For a solenoid, the magnetic field is constant along any circular loop inside the solenoid. Therefore, the left side of the equation simplifies to B multiplied by the circumference of the loop, 2πr.

B(2πr) = μ₀I_enclosed.

Since the electric field is induced by a changing magnetic field, we can use Faraday's law of electromagnetic induction to relate the electric field to the time derivative of the magnetic field:

E = -d(B·A)/dt,

where E is the electric field, B is the magnetic field, A is the cross-sectional area of the loop, and dt is the change in time.

To find the electric field at a radius of 1.00 cm, we need to differentiate the magnetic field with respect to time and multiply by the cross-sectional area of the loop. Since the cross-sectional area of the loop is proportional to the square of the radius, A = πr², we have:

E = -d(B·A)/dt,

E = -(d(B·πr²)/dt,

E = -πr²(dB/dt).

Taking the time derivative of the magnetic field B, we get:

dB/dt = (4π×10⁻⁷)(1.65×10³)(6.00 cos 90πt) T/s.

Substituting this expression back into the equation for the electric field, we have:

E = -πr²(dB/dt),

E = -π(1.00×10⁻² m)²[(4π×10⁻⁷)(1.65×10³)(6.00 cos 90πt)] T/s,

E = -6.27×10⁻⁸π cos 90πt mV/m.

So, the electric field induced at a radius of 1.00 cm from the axis of the solenoid is -6.27×10⁻⁸π cos 90πt mV/m.

The induced electric field at ( r = 1.00 ) cm from the solenoid axis is approximately [tex]\( -12\pi^2 \cos(90\pi t) \) mV/m.[/tex]  

Electric field induced:  [tex]\( 0.111 \cos(90\pi t) \) mV/m at \( r = 1.00[/tex] \) cm from solenoid axis.

To find the electric field induced at a radius of ( r = 1.00 ) cm from the axis of the solenoid, we can use Faraday's law of electromagnetic induction.

The induced electric field  [tex](\( E \))[/tex] is given by:

[tex]\[ E = -\frac{d\Phi}{dt} \][/tex]

Where  [tex]\( \Phi \)[/tex]  is the magnetic flux through the solenoid's cross-sectional area.

For a solenoid, the magnetic flux  [tex]\( \Phi \)[/tex]  is given by:

[tex]\[ \Phi = BA \][/tex]

Where:

- B  is the magnetic field inside the solenoid,

-  A  is the cross-sectional area of the solenoid.

Given the current  [tex]\( I(t) = 6.00 \sin(90\pi t) \)[/tex]  A, and using Ampere's law for the magnetic field inside a solenoid:

[tex]\[ B = \mu_0 n I \][/tex]

Where:

[tex]- \( \mu_0 \)[/tex] is the permeability of free space [tex](\( 4\pi \times 10^{-7} \) T m/A),[/tex]

-  n  is the number of turns per meter.

Substitute  [tex]\( I(t) \)[/tex]  into the equation for ( B ) and then calculate  [tex]\( \frac{d\Phi}{dt} \) to find \( E \).[/tex]

Given:

- [tex]\( n = 1.65 \times 10^3 \)[/tex]  turns/m

- [tex]\( I(t) = 6.00 \sin(90\pi t) \)[/tex] A

- [tex]\( A = \pi r^2 \)[/tex]  (for a circular cross-section)

We calculate the magnetic field [tex]\( B \):[/tex]

[tex]\[ B = \mu_0 n I(t) = 4\pi \times 10^{-7} \times 1.65 \times 10^3 \times 6.00 \sin(90\pi t) \][/tex]

Next, we calculate the magnetic flux  [tex]\( \Phi \):[/tex]

[tex]\[ \Phi = BA = 4\pi \times 10^{-7} \times 1.65 \times 10^3 \times 6.00 \sin(90\pi t) \times \pi \times (0.01)^2 \][/tex]

Now, we take the time derivative of [tex]\( \Phi \) to find \( \frac{d\Phi}{dt} \):[/tex]

[tex]\[ \frac{d\Phi}{dt} = 4\pi \times 10^{-7} \times 1.65 \times 10^3 \times 6.00 \times 90\pi \cos(90\pi t) \times \pi \times (0.01)^2 \][/tex]

Finally, we plug in the values and calculate:

[tex]\[ \frac{d\Phi}{dt} = 12\pi^2 \cos(90\pi t) \times 4.95 \times 10^{-10} \][/tex]

Now, we have [tex]\( \frac{d\Phi}{dt} \),[/tex]  the induced electric field [tex]\( E \) is given by \( E = -\frac{d\Phi}{dt} \).[/tex]

[tex]\[ E = -12\pi^2 \cos(90\pi t) \times 4.95 \times 10^{-10} \][/tex]

Therefore, the induced electric field at ( r = 1.00 ) cm from the solenoid axis is approximately [tex]\( -12\pi^2 \cos(90\pi t) \) mV/m.[/tex]

The external resistance R is approximately 10.63 Ohms while the internal resistance of the battery is approximately 2.43 Ohms.

This problem pertains to the principles of electric circuits, namely Ohm's Law (V = IR) and power (P = VI). To solve this, we'll first compute the resistance R of the external load and then calculate the internal resistance (r) of the battery.

(a) We begin by using the power formula P = VI, rearranging to find resistance R: P/V = I. So, the current I flowing through the circuit is 14.0 W/12.2 V which equals to about 1.148 A. Then, we apply Ohm's Law to get R = V/I, which results in: R = 12.2 V /1.148 A = 10.63 Ω.

(b) The difference between the emf (ε) and the terminal voltage (Vt) gives the voltage drop across the internal resistance (i.e., ε - Vt = Ir). Thus, r = (ε - Vt) / I = (15V - 12.2V) / 1.148 A = 2.43 Ω.

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