The weights of bags filled by a machine are normally distributed with a standard deviation of 0.05 kilograms and a mean that can be set by the operator. At what level should the mean weight be set if it required that only 1% of the bags weigh less than 9.5 kilograms? Round the answer to 2 decimal places.

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, [tex]\theta_c=52^o[/tex]

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

[tex]n_1sin\theta_c=n_2sin\theta_2[/tex]

Here, [tex]\theta_2=90[/tex]

[tex]sin\theta_c=\dfrac{n_2}{n_1}[/tex]

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

[tex]n_1=\dfrac{n_2}{sin\theta_c}[/tex]

[tex]n_1=\dfrac{1}{sin(52)}[/tex]

n₁ = 1.269

or n₁ = 1.27

Hence, the refractive index of liquid is 1.27

Answer:

69°

Explanation:

Projectiles that land at the same elevation they're launched from will have the same range if the launch angles are complementary (add up to 90°).

The complement of 21° is 69°.

The second ball hit at a 69° angle.

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Explanation:

It is given that,

Number of turns in primary coil, [tex]N_p=60[/tex]

Number of turns in secondary coil, [tex]N_s=300[/tex]

Voltage in primary coil, [tex]V_p=120\ V[/tex]

Current in primary coil, [tex]I_p=2\ A[/tex]

We have to find the voltage and current in the secondary coli. Firstly calculating the voltage in secondary coil as :

[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}[/tex]

[tex]\dfrac{60}{300}=\dfrac{120}{V_s}[/tex]

[tex]V_s=720\ Volts[/tex]

Now, calculating the current present in the secondary coil as :

[tex]\dfrac{N_p}{N_s}=\dfrac{I_s}{I_p}[/tex]

[tex]\dfrac{60}{300}=\dfrac{I_s}{2\ A}[/tex]

[tex]I_s=0.4\ A[/tex]

Hence, this is the required solution.

Final answer:

The voltage present in the secondary is 600 V and the current is 0.4 A.

Explanation:

The question involves an ideal transformer where the primary coil has 60 turns and the secondary coil has 300 turns, with 120 V at 2.0 A applied to the primary. To find the voltage and current present in the secondary, we can use the transformer equations:

Vp/Vs = Np/Ns (where V is voltage and N is the number of turns in the primary (p) and secondary (s) coils)

According to the conservation of energy in an ideal transformer, Pp = Ps (where P is power), and since P = VI, we get:

VpIp = VsIs (where I is current)

Using these equations:

Is = (Np/Ns) \\times Ip = (60/300) \\times 2.0 A = 0.4 A

Thus, the voltage present in the secondary is 600 V and the current is 0.4 A.

Explanation:

To answer how far and how long the object travels, we need to know either the length of the table or the kinetic coefficient of friction between copper and glass.  We also need to know the kinetic coefficient of friction between the copper object and the steel incline.  This information wasn't provided, so for sake of illustration, I'll assume both coefficients are the same, and that μk = 0.35.

Let's divide the path into four sections to stay organized:

a) object moves up incline

b) object is in free fall

c) object slides across table

d) object falls onto books

a)

First, the object slides up the incline.  There are four forces acting on the object.  Weight pulling down, normal force perpendicular to the incline, tension up the incline, and friction down the incline.

Sum of the forces normal to the incline:

∑F = ma

N - W cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

T - F = ma

T - Nμ = ma

T - mgμ cos θ = ma

Now sum of the forces in the y direction on the hanging mass:

∑F = ma

T - W = M(-a)

T = Mg - Ma

Substituting:

Mg - Ma - mgμ cos θ = ma

Mg - mgμ cos θ = (m + M) a

a = g (M - mμ cos θ) / (m + M)

Given m = 33.1 g, M = 50.0 g, θ = 30°, μ = 0.35, and g = 9.8 m/s²:

a = 4.71 m/s²

So the velocity it reaches at the top of the incline is:

v² = v₀² + 2a(x - x₀)

v² = 0² + 2(4.71)(1.5)

v = 3.76 m/s

The time to reach the top of the incline is:

x = x₀ + v₀ t + ½ at²

1.5 = 0 + (0) t + ½ (4.71) t²

t = 0.798 s

The horizontal distance traveled is:

x = 1.5 cos 30°

x = 1.299 m

And the vertical distance traveled is:

y = 1.5 sin 30°

y = 0.750 m

b)

In the second stage, the object is in free fall.  v₀ = 3.76 m/s and θ = 30°.  The object lands on the table at the point where its speed is a minimum.  This is at the highest point of the trajectory, when the vertical velocity is 0.

v = at + v₀

0 = (-9.8) t + (3.76 sin 30°)

t = 0.192 s

The horizontal distance traveled is:

x = x₀ + v₀ t + ½ at²

x = 0 + (3.76 cos 30°) (0.192) + ½ (0) (0.192)²

x = 0.625 m

The height it reaches is:

v² = v₀² + 2a(y - y₀)

(0)² = (3.76 sin 30°)² + 2(-9.8)(y - 0)

y = 0.180 m

c)

The object is now on the glass table.  As it slides, there are three forces acting on it.  Normal force up, gravity down, and friction to the left.

In the y direction:

∑F = ma

N - W = 0

N = mg

In the x direction:

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

The time it takes to reach the end of the table is:

v = at + v₀

0 = (-9.8×0.35) t + (3.76 cos 30°)

t = 0.949 s

And the distance it travels is:

v² = v₀² + 2a(x - x₀)

(0)² = (3.76 cos 30°)² + 2(-9.8×0.35) (x - 0)

x = 1.546 m

d)

Finally, the object falls onto the books.  There are 9 books, each 5 cm thick, so the height is 45 cm.  As the book falls, there are two forces acting on it: drag up and gravity down.

∑F = ma

D - W = ma

0.05 - (0.0331)(9.8) = 0.0331 a

a = -8.29 m/s²

Adding the vertical distances from parts a and b, we know the height of the table is 0.930 m.  So the time it takes the object to land is:

y = y₀ + v₀ t + ½ at²

0.45 = 0.93 + (0) t + ½ (-9.8) t²

t = 0.313 s

Adding up the times from all four parts, the total time is:

t = 0.798 + 0.192 + 0.949 + 0.313

t = 2.25 s

Adding up the horizontal distances, the total distance traveled is:

x = 1.299 + 0.625 + 1.546

x = 3.47 m

Answer:

The linear density of the wire is 0.314 g/m.

Explanation:

It is given that,

Acceleration, [tex]a=5.52\ m/s^2[/tex]

Mass of the ball, m = 112 gm

Speed of the transverse wave, v = 44.4 m/s

The speed of the transverse wave is given by :

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Where

T = tension in the wire

[tex]\mu[/tex] = mass per unit length

[tex]\mu=\dfrac{T}{v^2}[/tex]

[tex]\mu=\dfrac{ma}{v^2}[/tex]

[tex]\mu=\dfrac{112\ g\times 5.52\ m/s^2}{(44.4\ m/s)^2}[/tex]

[tex]\mu=0.3136\ g/m[/tex]

or

[tex]\mu=0.314\ g/m[/tex]

So, the linear density of the wire is 0.314 g/m. Hence, this is the required solution.

Answer:

The frequency is 0.34 Hz and the wave speed is 1.05 m/s.

Explanation:

Given that,

The distance between two successive minima of a transverse wave is wavelength.

Wavelength = 3.10 m

Time = 14.7 s

(I). We need to calculate the frequency

[tex]f=\dfrac{number\ of\ wave\ propagated\ per\ second }{times}[/tex]

[tex]f=\dfrac{5}{14.7}[/tex]

[tex]f=0.34\ Hz[/tex]

(II). We need to calculate the wave speed

Formula of wave speed

[tex]v= \lambda\times f[/tex]

Where,

[tex]\lambda = wavelength[/tex]

[tex]f = frequency[/tex]

Put the value into the formula

[tex]v=3.10\times0.34[/tex]

[tex]v=1.05\ m/s[/tex]

Hence, The frequency is 0.34 Hz and the wave speed is 1.05 m/s.

The frequency of the wave is 0.34 Hz and the wave speed is 1.054 m/s.

The distance between two successive minima in a transverse wave is the same as the wavelength (λ). Given that the question provided this distance as 3.10 m, we can use this as the wavelength. On the other hand, the frequency (f) of the wave is the number of wave cycles that pass a point per unit time. The information from the question gives us five crests (complete cycles) over 14.7 seconds. Thus, we can calculate the frequency by dividing the total cycles by the total time, i.e., f = 5 cycles / 14.7 seconds = 0.34 Hz.

The speed of the wave (v) can be obtained from the equation v = λf. Substituting the values we have, v = (3.10 m) * (0.34 Hz) = 1.054 m/s. Therefore, the frequency of the wave is 0.34 Hz, and the wave speed is 1.054 m/s.

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Answer:

[tex]F = 1.5 \times 10^{-16} N[/tex]

this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

[tex]KE = 1 keV[/tex]

[tex]KE = 1 \times 10^3 (1.6 \times 10^{-19}) J[/tex]

[tex]KE = 1.6 \times 10^{-16} J[/tex]

now the speed of electron is given as

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

now we have

[tex]v = \sqrt{\frac{2 KE}{m}}[/tex]

[tex]v = 1.87 \times 10^7 m/s[/tex]

now the maximum force due to magnetic field is given as

[tex]F = qvB[/tex]

[tex]F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})[/tex]

[tex]F = 1.5 \times 10^{-16} N[/tex]

Now if this force is compared by the gravitational force on the electron then it is

[tex]\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}[/tex]

[tex]\frac{F}{F_g} = 1.68 \times 10^{13}[/tex]

so this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force

Final answer:

The maximum magnetic force on an electron in Earth's magnetic field is calculated using the charge of the electron, its velocity derived from kinetic energy, and the field strength. This force is much larger than the gravitational force acting on the electron due to its minuscule mass.

Explanation:

The maximum possible magnetic force on an electron moving through Earth's magnetic field can be found using the formula F = qvB sin(\theta), where q is the charge of the electron, v is its velocity, B is the magnetic field strength, and \theta is the angle between the velocity and the magnetic field. Since the maximum force occurs when the angle is 90 degrees (sin(\theta) = 1), we first need to calculate the velocity from its kinetic energy, 1 keV. Given the kinetic energy E = 1 keV = 1 × 103 eV and using the relation E = \frac{1}{2}mv2, we can find v. The electron charge is q = -1.6 × 10-19 C, and the magnetic field is B = 0.5 G = 5 × 10-5 T.

For the gravitational force, we use F = mg, where m is the mass of the electron and g is the acceleration due to gravity, approximately 9.8 N/kg on the surface of Earth. The mass of the electron is about 9.11 × 10-31 kg. Comparing these forces, the magnetic force is several orders of magnitude greater than the gravitational force acting on an electron due to its negligible mass.

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

[tex]F = ma[/tex]

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

[tex]-10000 =2000\times a[/tex]

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

[tex]a = -\dfrac{10000}{2000}[/tex]

[tex]a=-5\ m/s^2[/tex]

For the distance,

Using third equation of motion

[tex]v^2-u^2=2as[/tex]

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

[tex]0-30^2=2\times(-5)\times s[/tex]

[tex]s = 90\ m[/tex]

Hence, The car traveled the distance before stopping is 90 m.

Try this option (shortly):

According to the 'law of consevation of pulse' the passenger in the smaller car receives greater energy then the one in the larger car.

Almost all the received energy is spent to injure the passenger in the smaller car. :(

Final answer:

Passengers in a smaller mass car are at greater risk in a head-on collision due to higher acceleration and force resulting from the collision. Cars with similar masses experience a more even distribution of momentum changes, potentially reducing impact severity. The duration of impact also affects the decelerating force felt by occupants, illustrating the importance of seatbelts and crumple zones.

Explanation:

A passenger in the smaller mass car is more likely to be injured in a head-on collision with a larger mass car because the smaller car will experience a larger acceleration due to the force of impact, according to Newton's second law (force equals mass times acceleration). This greater acceleration translates to a stronger force exerted on the occupants of the smaller car. Additionally, according to the principle of conservation of momentum, if both cars are moving at the same speed and collide, the car with less mass will undergo a more significant change in velocity.

When cars with more similar masses collide, the change in momentum is distributed more evenly between the two, potentially reducing the severity of the impact on both vehicles. This is due to the momentum of the two-car system remaining constant if external forces are negligible (as stated by the law of conservation of momentum). The impulse, or change in momentum, is also important to consider. If cars crumple together instead of rebounding, there's a smaller change in momentum, leading to a smaller force on the occupants, which is why crumple zones are critical safety features in cars.

The scenario with a seatbelt demonstrates the importance of the time over which a force acts. A seatbelt helps extend the time of a collision's impact, reducing the average decelerating force experienced by the occupant, compared to a sudden stop against the dashboard.

Answer:

approximately 3.68

Explanation:

Explained in the attached picture.

Answer:

Minimum stopping distance = 164.69 ft

Explanation:

Speed of car = 35.6 mi/hr = 15.82 m/s

Stopping distance = 40.8 ft = 12.44 m

We have equation of motion

             v² = u² + 2as

             0²=15.82²+ 2 x a x 12.44

              a = -10.06 m/s²

Now wee need to find minimum stopping distance, in ft, for the same car moving at 71.5 mi/h.

          Speed of car = 71.5 mi/h = 31.78 m/s

We have

           v² = u² + 2as

          0² = 31.78² -  2 x 10.06 x s

           s = 50.2 m = 164.69 ft

Minimum stopping distance = 164.69 ft

Answer:

2 m/s

Explanation:

Momentum is conserved:

mv = MV

where m is mass of the boxcar, v is its initial velocity, M is the mass of all six box cars, and V is the final velocity.

v = 12 m/s, and M = 6m, so:

m (12 m/s) = 6m V

12 m/s = 6V

V = 2 m/s

In the case of resistors in series and in parallel to a battery, the resistances can be determined by applying Ohm's law and using equations for total resistance in both series and parallel configurations. The calculations yield the values R1 = 1.53 ohms and R2 = 5.48 ohms.

The problem involves applying Ohm's law and equations for series and parallel resistors. When the resistors are in series, we add the resistances so R1 + R2 = 12.7 V / 1.81 A = 7.01 ohms. In a parallel connection, the total resistance R is given by 1/R = 1/R1 + 1/R2 or 12.7 V / 9.06 A = 1.40 ohms. Solving this pair of equations we get R1 = 1.53 ohms and R2 = 5.48 ohms, giving the pair of resistances asked for.

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Answer:

18816π ≈ 59112 Joules

Explanation:

Draw a picture (like the one attached).  The cone has a radius R and height H.  The spout has height a.  The tank is filled with water to height h.

Cut a thin slice from the volume of the water.  This slice is a cylindrical disc with a radius r, a thickness dy, and a height y.

Using similar triangles, we can say:

r / y = R / H

The work required to lift this slice up to the spout is:

dW = dm g (H + a − y)

where dm is the mass of the slice and g is the acceleration due to gravity.

Mass is density times volume, so:

dW = dV ρ g (H + a − y)

Substituting the volume of the cylindrical disc:

dW = dy π r² ρ g (H + a − y)

From our similar triangles equation, we know that r = R/H y, so:

dW = dy π (R/H y)² ρ g (H + a − y)

Rearranging:

dW = π (R/H)² ρ g y² (H + a − y) dy

dW = π (R/H)² ρ g ((H + a)y² − y³) dy

The work to lift all the slices between y=0 and y=h is:

W = ∫ dW

W = π (R/H)² ρ g ∫₀ʰ ((H + a)y² − y³) dy

Integrating:

W = π (R/H)² ρ g (⅓(H + a)y³ − ¼y⁴) |₀ʰ

W = π (R/H)² ρ g (⅓(H + a)h³ − ¼h⁴)

Given:

R = 2 m

H = 5 m

a = 1 m

h = 2 m

g = 9.8 m/s²

ρ = 1000 kg/m³

W = π (2/5)² (1000) (9.8) (⅓(5 + 1)(2)³ − ¼(2)⁴)

W = 18816π

W ≈ 59112 Joules

It takes approximately 59.1 kJ of work.

Final answer:

To calculate the work needed to pump water from a conical tank, integrate the volume of water being moved over the distance it must be lifted, considering the tank's dimensions, the spout's height, and the effects of gravity.

Explanation:

The question involves finding the work needed to pump water from a conical tank with a spout at the top. To do this, we need to consider the volume of water to be moved, the height it must be lifted, and the forces involved, including gravity. The water's density (1000 kg/m3) and gravitational acceleration (9.8 m/s2) are key to calculating this work.

The work needed to lift water involves the product of the force required to lift the water, the distance it must be moved, and the volume of water. Given the tank's dimensions and the additional height from the spout, one would typically integrate along the height of the water to compute the total work, considering the changing cross-sectional area of the water column as it is being pumped out.

To find this, one could use the formula for the volume of a cone to model the water's volume at any height and then integrate this volume over the distance it must be moved, accounting for the force of gravity. This problem combines principles of physics and calculus to find a solution.

Answer:

6300 J = 1.51 Calories

Explanation:

We have 1 Calorie = 1000 cal

1 cal = 4.184 J

That is 1 Calorie = 1000 cal = 1000 x 4.184 = 4184 J

Here we have 6300 J

That is

         [tex]1Calorie=4184J\\\\1J=\frac{1}{4184}Calorie\\\\6300J=\frac{6300}{4184}Calorie=1.51Calorie[/tex]

    6300 J = 1.51 Calories

Answer:

197.2 s

Explanation:

First of all, let's convert the mass of the ocean liner into kilograms:

[tex]m=40,000 Mg = 40,000,000 kg = 4\cdot 10^7 kg[/tex]

and the initial velocity into m/s:

[tex]u=4 km/h =1.11 m/s[/tex]

The force applied is

[tex]F=-225 kN = -2.25\cdot 10^5 N[/tex]

So we can find first the deceleration of the liner:

[tex]a=\frac{F}{m}=\frac{-2.25\cdot 10^5 N}{4\cdot 10^7 kg}=-5.63\cdot 10^{-3} m/s^2[/tex]

And now we can use the following equation:

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

with v = 0 being the final velocity, to find t, the time it takes to bring the liner to rest:

[tex]t=\frac{v-u}{a}=\frac{0-1.11 m/s}{-5.63\cdot 10^{-3} m/s^2}=197.2 s[/tex]

Answer:

Escape velocity, v = 12.6 km/s

Explanation:

It is given that,

Mass of earth like planet, [tex]m=6.5\times 10^{24}\ kg[/tex]

Radius of planet, [tex]r=55\times 10^5\ m[/tex]

The escape velocity of a planet is given by the following formula as:

[tex]v=\sqrt{\dfrac{2Gm}{r}}[/tex]

G = universal gravitational constant

[tex]v=\sqrt{\dfrac{2\times 6.67\times 10^{-11}\ m^3kg^{-1}s^{-2}\times 6.5\times 10^{24}\ kg}{55\times 10^5\ m}[/tex]

v = 12556.05 m/s

or

v = 12.6 km/s

Hence, this is the required solution.

Answer:

a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m

Explanation:

This can be answered with either force analysis and kinematics, or work and energy.

a) Using force analysis, we can draw a free body diagram for the snowboarder.  There are two forces: normal force perpendicular to the slope and gravity down.

Sum of the forces parallel to the slope:

∑F = ma

mg sin θ = ma

a = g sin θ

Therefore, the velocity at the bottom is:

v² = v₀² + 2a(x - x₀)

v² = (0)² + 2(9.8 sin 60°) (45.7 / sin 60° - 0)

v = 29.9 m/s

Alternatively, using energy:

PE = KE

mgh = 1/2 mv²

v = √(2gh)

v = √(2×9.8×45.7)

v = 29.9 m/s

b) Drawing a free body diagram, there are three forces on the snowboarder.  Normal force up, gravity down, and friction to the left.

Sum of the forces in the y direction:

∑F = ma

N - mg = 0

N = mg

Sum of the forces in the x direction:

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Therefore, the snowboarder's final speed is:

v² = v₀² + 2a(x - x₀)

v² = (29.9)² + 2(-9.8×.102) (10 - 0)

v = 29.6 m/s

Using energy instead:

KE = KE + W

1/2 mv² = 1/2 mv² + F d

1/2 mv² = 1/2 mv² + mgμ d

1/2 v² = 1/2 v² + gμ d

1/2 (29.9)² = 1/2 v² + (9.8)(0.102)(10)

v = 29.6 m/s

c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.

∑F = ma

-mg sin θ = ma

a = -g sin θ

Therefore, the final height reached is:

v² = v₀² + 2a(x - x₀)

(0)² = (29.6)² + 2(-9.8 sin 30°) (h / sin 30° - 0)

h = 44.7 m

Using energy:

KE = PE

1/2 mv² = mgh

h = v² / (2g)

h = (29.6)² / (2×9.8)

h = 44.7 m

Answer:

[tex]1.69\cdot 10^4 N/C[/tex]

Explanation:

The relationship between electric field strength and potential difference is:

[tex]E=\frac{V}{d}[/tex]

where

E is the electric field strength

V is the potential difference

d is the distance between the plates

Here we have

V = 218 V

d = 1.29 cm = 0.0129 m

So, the electric field is

[tex]E=\frac{218 V}{0.0129 m}=16900 N/C = 1.69\cdot 10^4 N/C[/tex]

Answer:

because the troposphere has more oxygen than the stratosphere so when he felt like he was floating even though he was free falling it was due to the absence of air.

Answer:

Mass of water, [tex]m=1.47\times 10^{12}\ kg[/tex]

Explanation:

Given that,

Surface area of the reservoir, A = 30 km² = 3 × 10⁷ m²

Average depth, d = 49 m

The volume of the reservoir is V such that,

[tex]V=A\times d[/tex]

[tex]V=3\times 10^7\ m^2\times 49\ m[/tex]

[tex]V=1.47\times 10^9\ m^3[/tex]

We have to find the mass of water is held behind the dam. It can be calculated using the expression for density. We know that density of water, d = 1000 kg/m³

Density, [tex]d=\dfrac{m}{V}[/tex]

[tex]m=d\times V[/tex]

[tex]m=1000\ kg/m^3\times 1.47\times 10^9\ m^3[/tex]

[tex]m=1.47\times 10^{12}\ kg[/tex]

Hence, this is the required solution.

Answer:

The emf and internal resistance of the battery are 9 volt and 2.04 Ω

Explanation:

Given that,

First resistor[tex]R = 100\ \Omega[/tex]

Power[tex]P = 0.794\ W[/tex]

Second resistor[tex]R=200\ \Omega[/tex]

Power[tex]P=0.401\ W[/tex]

We need to calculate the emf and internal resistance

Using formula of emf

[tex]P=\dfrac{e^2}{R+r}[/tex]

For first resistor

[tex]0.794=\dfrac{e^2}{100+r}[/tex]....(I)

For second resistor

[tex]0.401=\dfrac{e^2}{200+r}[/tex].....(II)

From equation (I) and (II)

[tex]\dfrac{795}{401}=\dfrac{200+r}{100+r}[/tex]

Using component of dividend rule

[tex]\dfrac{393}{401}=\dfrac{100}{100+r}[/tex]

[tex]r =2.04\ \Omega[/tex]

Now, Put the value of internal resistance in equation (I)

[tex]0.794=\dfrac{e^2}{100+2.04}[/tex]

[tex]e^2=0.794(100+2.04)[/tex]

[tex]e=\sqrt{0.794(100+2.04)}[/tex]

[tex]e=9\ volt[/tex]

Hence, The emf and internal resistance of the battery are 9 volt and 2.04 Ω

The emf and internal resistance of a battery can be calculated using Ohm's Law (V=IR) and the power formula (P = I²R). By evaluating the current and emf for two different resistors, we can arrive at two equations that, when solved simultaneously, give the emf and internal resistance.

The question is asking for the calculation of the EMF (electromotive force) and internal resistance of the battery. First, we need to understand the relationship between power, voltage, current, and resistance, which is described by Ohm's Law and the power formulas. For the 100 Ω resistor, we can use the formula for power P = I²R to find the current, which comes out to be √P/R = √0.794/100 = 0.0281 A. Then, we can use Ohm's Law V = I(R+r) to find the voltage of the battery (emf), plugging in the values we get V = 0.0281(100+r).

When the resistor is changed to 200 Ω, we repeat the process, giving us the equation V = 0.0281(200+r). Solving these two equations will give us the EMF and internal resistance of the battery.

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Answer:

it will move by d = 8.00 m in x direction before it will turn back

Explanation:

Here the initial velocity of particle is along +x direction

it is given as

[tex]v_i = 4.90 m/s[/tex]

now its acceleration is given as

[tex]a_x = -1.50 m/s^2[/tex]

[tex]a_y = 3.00 m/s^2[/tex]

now when it turns back then the velocity in x direction will become zero

so we will say

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 4.90^2 = 2(-1.50) d[/tex]

[tex] d = 8.00 m[/tex]

so it will move by d = 8.00 m in x direction before it will turn back

Final answer:

The particle moves 8.01 m in the x direction before it turns around, determined by using kinematic equations with the given initial velocity and constant acceleration.

Explanation:

To determine how far the particle moves in the x direction before turning around, we need to find the point at which its velocity in the x direction is zero. Since the particle starts with an initial velocity of 4.90 m/s in the x direction and has a constant acceleration of ax = -1.50 m/s², we can use the kinematic equation v = u + at to find the time when the velocity becomes zero:

v = 0 m/s (the velocity at the turnaround point)
u = 4.90 m/s (initial velocity)
a = -1.50 m/s² (constant acceleration)

Solving for the time (t) gives:

0 = 4.90 + (-1.50)t
t = 4.90 / 1.50
t = 3.27 s

Now using the equation s = ut + (1/2)at² to find the displacement in the x direction:

s = 4.90 m/s ⋅ t + (1/2)(-1.50 m/s²) ⋅ t²
s = 4.90 ⋅ 3.27 + (0.5) ⋅ (-1.50) ⋅ (3.27)²
s = 8.01 m

The particle moves 8.01 m in the x direction before turning around.

Answer:

28.23 years

Explanation:

I = 1100 A

L = 230 km = 230, 000 m

diameter = 2 cm

radius, r = 1 cm = 0.01 m

Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2

n = 8.5 x 10^28 per cubic metre

Use the relation

I = n e A vd

vd = I / n e A

vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)

vd = 2.58 x 10^-4 m/s

Let time taken is t.

Distance = velocity x time

t = distance / velocity = L / vd

t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second

t = 28.23 years

Answer:

28.23 years

Explanation:

It takes 28.23 years to take one electron to travel the full length of the cable.

3.156 107 = number of seconds in a year.

230000 / (2.58 x 10^-4) = 8.91 x 10^8 second.

Answer: [tex]V=\sqrt{\frac{2GM}{R}}[/tex]

Explanation:

Taking into account what is stated in this problem and considering there is no friction during the takeoff of the rocket of the planet, the rocket will escape the gravitational attraction of the massive body when its kinetic energy [tex]K[/tex] and its potential energy [tex]P[/tex] are equal in magnitude.

Written mathematically is:

[tex]K=P[/tex]  (1)

Where:

[tex]K=\frac{1}{2}mV^{2}[/tex] (2)

Being [tex]m[/tex] the mass of the rocket

And:

[tex]P=-\frac{GMm}{R}[/tex]   (3)

Being  [tex]M[/tex] the mass of the planet,  [tex]G[/tex] the gravitational constant and  [tex]R[/tex] the radius of the planet.

Substituting (2) and (3) in (1):

[tex]\frac{1}{2}mV^{2}=-\frac{GMm}{R}[/tex] (4)

Finding [tex]V[/tex], which is the escape velocity:

[tex]V=\sqrt{\frac{2GM}{R}}[/tex]  this is the velocity the rocket must have in order to escape from the surface of the planet

Answer:

It's a long explanation

Explanation:

The Planck mass is approximately the mass of the Planck particle, a hypothetical minuscule black hole whose Schwarzschild radius equals the Planck length. In physics, the Planck mass, denoted by mP, is the unit of mass in the system of natural units known as Planck units. It is approximately 0.02 milligrams. Plank length as the minimum distance at which space can be thought. In physics, the Planck length, denoted ℓP, is a unit of length that is the distance light travels in one unit of Planck time. It is equal to 1.616255×10⁻³⁵ m. It is a base unit in the system of Planck units, developed by physicist Max Planck...I miss you, Landon.

Answer:

90 N

Explanation:

The electrostatic force between two charges is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

[tex]F=(9\cdot 10^9 N m^2 C^{-2})\frac{(0.005 C)^2}{(50 m)^2}=90 N[/tex]

Answer:

8.4 ft-lb

Explanation:

Work = change in energy

W = ½ kx²

When x = 3 ft, W = 15 ft-lb:

15 ft-lb = ½ k (3 ft)²

k = 30/9 lb/ft

When x = 27 in = 2.25 ft:

W = ½ kx²

W = ½ (30/9 lb/ft) (2.25 ft)²

W = 8.4375 ft-lb

Rounding to 2 sig-figs, it takes 8.4 ft-lb of work.

Final answer:

The work required to stretch the spring 27 in. beyond its natural length is 12.728 N-m.

Explanation:

To find the work required to stretch the spring 27 in. beyond its natural length, we can use the work done formula:

Work = (1/2)kx²

Where k is the spring constant and x is the displacement of the spring. We know that the work required to stretch the spring 3 ft beyond its natural length is 15 ft-lb. Let's convert the given measurements into consistent units:

3 ft = 36 in.

15 ft-lb = 12.728 N-m

Now, we can solve for the work required to stretch the spring 27 in. beyond its natural length:

(1/2)k(27)² = (1/2)k(729) = 12.728 N-m

We can rearrange the equation to solve for k:

k = (2 * 12.728) / 729 = 0.0349 N/m