A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?

Answer:

The rate of current in the solenoid  is 0.398 A/s

Explanation:

Given that,

Electric field [tex]E = 4.0\ \mu V/m[/tex]

Distance = 2.0 cm

Radius = 3.0 cm

Number of turns per unit length = 800

We need to calculate the rate of current

Using formula of electric field for solenoid

[tex]E = \dfrac{x}{2}\mu_{0}n\dfrac{dI}{dt}[/tex]

Where, x = distance

n = number of turns per unit length

E = electric field

r = radius

Put the value into the formula

[tex]4.0\times10^{-6}=\dfrac{2.0\times10^{-2}}{2}\times4\pi\times10^{-7}\times800\times\dfrac{dI}{dt}[/tex]

[tex]\dfrac{dI}{dt}=\dfrac{4.0\times10^{-6}\times2}{2.0\times10^{-2}\times4\pi\times10^{-7}\times800}[/tex]

[tex]\dfrac{dI}{dt}=0.397\ A/s[/tex]

Hence, The rate of current in the solenoid  is 0.398 A/s.

To find the distance from the building where the water will hit the highest possible fire, we analyze the horizontal and vertical components of the water's motion. Using the given information, we calculate the horizontal distance using the horizontal velocity and time, and the vertical distance using the vertical velocity and time. By finding the time at which the water reaches its maximum height, we can determine the horizontal distance.

A fire hose ejects a stream of water at an angle of 35.0° above the horizontal with a speed of 25.0 m/s. To find the distance from the building where the water will hit the highest possible fire, we can analyze the horizontal and vertical components of the water's motion. The horizontal distance can be calculated using the horizontal velocity and time, while the vertical distance can be determined using the vertical velocity and time. By finding the time at which the water reaches its maximum height, we can calculate the horizontal distance.

Using the given information, we can determine that the initial horizontal velocity is 25.0 m/s * cos(35.0°) = 20.4 m/s, and the vertical velocity is 25.0 m/s * sin(35.0°) = 14.3 m/s. The time it takes for the water to reach its maximum height (where the vertical velocity becomes zero) can be found using the equation vy = v0y - gt, where vy is the vertical velocity, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time. Solving for t, we get t = v0y / g. Plugging in the values, we have t = 14.3 m/s / 9.8 m/s^2 = 1.46 seconds. Finally, we can find the horizontal distance by multiplying the initial horizontal velocity by the time: x = v0x * t = 20.4 m/s * 1.46 s = 29.8 m. Therefore, the fire hose should be located approximately 29.8 meters away from the building to hit the highest possible fire.

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The fire hose should be located approximately 59.8 meters from the building from the building to hit a fire at the highest point the water stream can reach.

Projectile Motion in Fire Hose Stream

To determine how far from a building a fire hose should be located to hit the highest possible fire, we need to analyze the projectile motion of the water stream ejected.

The water stream is ejected at an angle of 35.0° above the horizontal with an initial speed of 25.0 m/s. We will assume the water reaches its highest point where its vertical component of velocity becomes zero.

Step-by-Step Calculation

Answer:

(a) 64.58 N

(b) 43125 N/C

Explanation:

(a) q1 = 796 uC = 796 x 10^-6 C, q2 = 481 nC = 481 x 10^-9 C,

r = 23.1 cm = 0.231 m

Force, F = k q1 q2 / r^2

F = (9 x 10^9 x 796 x 10^-6 x 481 x 10^-9) / (0.231)^2

F = 64.58 N

(b) F = 6.9 x 10^-15 N

E = F / q

E = (6.9 x 10^-15) / (1.6 x 10^-19) = 43125 N/C

Answer:

783.63 Watt

Explanation:

T = 111 degree C = 111 + 273 = 384 K

T0 = 45 degree C = 45 + 273 = 318 K

A = 1.6 m^2

e = 0.75

According to the Stefan's law, the energy radiated per second is given by

[tex]E = \sigma eA\left ( T^{4}- T_{0}^{4}\right )[/tex]

[tex]E = 5.67\times 10^{-8}\times 0.75\times 1.6 ( 384^{4}- 318^{4}\right ))[/tex]

E = 783.63 Watt

Answer:

a) 0.41 m/s

b) 0.51 m/s

Explanation:

(a)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (15.4) + (93.1 - 0.292) V'

V' = 0.41 m/s

b)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = - 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (- 15.4) + (93.1 - 0.292) V'

V' = 0.51 m/s

Answer:

Option (B)

Explanation:

In a perfectly inelastic collision, the two bodies collide each other after the collision.

A. As the frog and the lily pad stick together so it is a perfectly inelastic collision.

B. The bowling ball and the pin do not stick together so it is not an example of perfectly inelastic collision.

C. A fish swallows the another, so it is a perfectly inelastic collision.

D. Bubble gum sticks to the golf ball so it is a perfectly inelastic collision.

Answer:

144540.16735 hp

Explanation:

Diameter of pipe = 36 in

Specific gravity of oil = 0.89

ρ = Density of oil = 0.89×62.2407

Q = Flow rate of oil = 1×10⁶ bbl/day = 1×10⁶×5.61458/86400 =  64.9836033565 ft³/s

hf = 13 ft/1000ft = 0.013 ft/ft

for 10 miles

hf = 0.013×10×5280 = 686.4 ft

g = Acceleration due to gravity = 32.17405 ft/s²

Power

P = ρQghf

⇒P = 0.89×62.2407×64.9836033565×32.17405×686.4

⇒P = 79497097.342 ft lbf/s

1 ft lbf/s = 0.0018181817 hp

79497097.342 ft lbf/s = 144540.16735 hp

∴ Horsepower which must be delivered to the oil by each pump is  144540.16735 hp

Answer:

Explanation:

The half-life of a radioisotope, in this case carbon-14, is the time that a sample requires to reduce its amount to half, and it is a constant for every radioisotope (it does not change with the amount of sample).

Then, the formula for the remaining amount of a radioisotope is:

Where:

The number of half-lives for carbon-14 elapsed for the dinosaur fossil is:

Then, A / A₀ = (1/2)ⁿ = (1/2)¹¹⁸⁶⁷ ≈ 0.00000 .

The number is too small, and when you round to five decimal places the result is zero. That is why carbon-14 cannot be used to date dinosaur fossils, given that they are too old.

C-14 dating is not applicable to a 68 million years old dinosaur fossil, because the half-life of C-14 (5730 years) only allows for accurate dating up to about 50,000-57,000 years. After such time period, the remaining C-14 would be too small to measure accurately. Therefore, for a 68 million year old fossil, the remaining C-14 would be effectively zero.

The question is asking about the amount of Carbon-14 (C-14) remaining in a dinosaur fossil that is approximately 68 million years old. However, C-14 dating is not applicable to such old samples. This is because C-14 decays back to Nitrogen-14 (N-14) with a half-life of approximately 5,730 years. Because of this half-life, the best-known method for determining the absolute age of fossils with C-14 allows for reliable dating of objects only up to about 50,000 years old. After 10 half-lives, which would be about 57,000 years, any original C-14 existing in a sample would become too small to measure accurately

For very old samples like a 68 million year old dinosaur fossil, isotopes with much longer half-lives are used for dating, such as uranium-235 and potassium-40. Therefore, for a 68 million year old dinosaur fossil, the remaining C-14 would effectively be zero.

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Answer:

64.08 rad

Explanation:

f = 78 rpm = 76 / 60 rps

θ = 0.45 rad, t = 8 s

(θ2 - θ1) / t = ω

θ2 - θ1 = 2 x 3.14 x 76 x 8 / 60 = 63.63

Angle turns from initial position = 0.45 + 63.63 = 64.08 rad

Final answer:

The magnitude of the total acceleration of the top of the racket during the serve, considering both tangential and radial components, is approximately 270.5 m/s^2.

Explanation:

To find the magnitude of the total acceleration of the top of the racket, we need to consider both tangential and radial (centripetal) acceleration components. The tangential acceleration (at) is directly provided by the angular acceleration (α), and is calculated by multiplying the angular acceleration by the radius (r), so at = α × r. Here, α = 150 rad/s2 and r = 1.30 m, giving at = 195 m/s2. The radial acceleration (ar), also known as centripetal acceleration, depends on the angular speed (ω) and the radius (r), and is calculated with the formula ar = ω2 × r. Given ω = 12.0 rad/s and r = 1.30 m, we find ar = 187.2 m/s2. Finally, the total acceleration (a) is the square root of the sum of the squares of at and ar, resulting in a = √(at2 + ar2). This gives us a = √(1952 + 187.22) m/s2, and by calculating, we find the magnitude of the total acceleration to be approximately 270.5 m/s2.

Answer:

When the hunter pulls the polar bear to him, the polar bear will move 3.3 meters.

Explanation:

It is given that, a 640-N hunter gets a rope around a 3200-N polar bear. They are stationary, 20 m apart, on friction less level ice.

The given system is an isolated system which is initially at rest and they both will meet at the center of mass. Then finding the center of mass of the system as :

[tex]X_{com}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}[/tex]

Multiplying "g" to the numerator and denominator both as :

[tex]X_{com}=\dfrac{gm_1x_1+m_2x_2g}{m_1g+m_2g}[/tex]

[tex]X_{com}=\dfrac{640(0)+3200\times 20}{640+3200}=16.67\ m[/tex]

i.e. 640 N hunter moves at a distance of 16.67 meters and when the hunter pulls the polar bear to him, the polar bear will move, 20 m - 16.67 m = 3.3 meters. Hence, this is the required solution.

Answer and Explanation:

1). Electric Potential:

Electric Potential can be defined as the amount of work required to move a unit  charge from infinity or reference point to a particular or specific point inside the electric field (without any accelerated motion of charge particle). The other terms for electric potential are 'electrostatic potential, potential difference(if a charge is moved from one point to another inside the field) or electric field potential.

2). Inductance:

It can be define as that property of electric circuits due to which it opposes any change in the voltage under the influence of changing magnetic field.

We can also say that the property of an electric conductor due to which it opposes any change in electric current passing through it. Any change in electric current results in the production of magnetic field which when varied results in the generation of Electro Motive Force (EMF) which according to Lenz law is produced in a way that it opposes its cause of production.

Answer:

this is my basic understanding o

. Electric Potential:

Electric Potential can be defined as the amount of work required to move a unit  charge from infinity or reference point to a particular or specific point inside the electric field (without any accelerated motion of charge particle). The other terms for electric potential are 'electrostatic potential, potential difference(if a charge is moved from one point to another inside the field) or electric field potential.

2). Inductance:

It can be define as that property of electric circuits due to which it opposes any change in the voltage under the influence of changing magnetic field.

We can also say that the property of an electric conductor due to which it opposes any change in electric current passing through it. Any change in electric current results in the production of magnetic field which when varied results in the generation of Electro Motive Force (EMF) which according to Lenz law is produced in a way that it opposes its cause of production.

Explanation:

The answer will be 137 mL

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum  theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work  and amount of entropy  and magnitude

solution

first we calculate work i.e heat transfer - work =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C    .................1

so first we get some value from steam table with the help of 100 bar @360°C and  1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change =  specific entropy ( 100 bar @360°C)  - specific entropy ( 1 bar @160°C )

put these value we get

entropy change =  7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum  theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum  theoretical work = specific internal energy - 2038.3

maximum  theoretical work = 2728 - 2038.3

maximum  theoretical work is 689.4 kJ/kg

The work and entropy produced in the adiabatic expansion of steam can be calculated using the First Law of Thermodynamics and thermodynamic principles. Maximum theoretical work corresponds to an isentropic process, which is an ideal, reversible adiabatic process that occurs without an increase in entropy.

The question regards the calculation of work and entropy of an adiabatic expansion process of steam in a piston-cylinder assembly. We'd need specific heat capacity, initial and end states temperatures data and the thermodynamic equations for adiabatic processes. The First Law of Thermodynamics equation dEint = CyndT will be used to calculate internal energy changes. Final work, W, can be calculated using the equation W = pΔV. Entropy change can be obtained using the formula ΔS=Q/T, where Q represents heat transfer, which is 0 for adiabatic processes.

As for the maximum theoretical work obtainable, this corresponds to a reversible adiabatic process or an isentropic process - entropy remains constant. This requires a quasi-static or reversible deformation. The work obtained in this case is at its maximum because there are no irreversibilities or entropy generation, which are responsible for decreasing the performance of real processes and engines.

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Answer:

Explanation:

Number of oscillations in 100 s is 68.

time taken to complete one oscillation = 100 / 68 s = 1.47 s

Use the formula for the time period of an oscillating body

[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex]

[tex]T^{2} = 4\pi^{2} \frac{L}{g}}[/tex]

In this equation substitute the value of T amd L and then find the value of g.

Answer:

Final speed in reverse direction will be 11 m/s

Explanation:

As per impulse momentum theorem we know that

Change in momentum of the object is product of force and time

so we have

[tex]P_f - P_i = F\Delta t[/tex]

now we have

[tex]P_f = mv_f[/tex]

[tex]P_i = mv_i[/tex]

[tex]\Delta t = 250 ms[/tex]

F = 21 N

now we have

[tex]0.38(v_f) - 0.38(-2.8) = 21(250\times 10^{-3})[/tex]

[tex]v_f = 13.82 - 2.8[/tex]

[tex]v_f = 11 m/s[/tex]

Answer:

the path of the moving charge will be circular path now

Radius = 1312.5 m

Explanation:

Force on a moving charge due to constant magnetic field is given by

[tex]\vec F = q(\vec v \times \vec B)[/tex]

since here force on the moving charge is always perpendicular to the velocity always as it is vector product of velocity and magnetic field so here magnitude of the speed is always constant

Also the force is since perpendicular to the velocity always

so here the path of the moving charge will be circular path now

now to find out the radius of this circular path

[tex]F = \frac{mv^2}{R}[/tex]

[tex]qvBsin90 = \frac{mv^2}{R}[/tex]

[tex]R = \frac{mv}{qB}[/tex]

[tex]R = \frac{9.1\times 10^{-31}(1.50 \times 10^7)}{1.6 \times 10^{-19}(6.50 \times 10^{-8})}[/tex]

[tex]R = 1312.5 m[/tex]

Answer:

a) Maximum height reached = 1878.90 m

b) Time of flight = 39.14 seconds.

Explanation:

Projectile motion has two types of motion Horizontal and Vertical motion.  

Vertical motion:  

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.  

Considering upward vertical motion of projectile.  

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g [tex]m/s^2[/tex] and final velocity = 0 m/s.  

0 = u sin θ - gt  

t = u sin θ/g  

Total time for vertical motion is two times time taken for upward vertical motion of projectile.  

So total travel time of projectile , [tex]t=\frac{2usin\theta }{g}[/tex]

Vertical motion (Maximum height reached, H) :

We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.  

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

[tex]0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}[/tex]

In the give problem we have u = 192 m/s, θ = 90° we need to find H and t.

a) [tex]H=\frac{u^2sin^2\theta}{2g}=\frac{192^2\times sin^290}{2\times 9.81}=1878.90m[/tex]

Maximum height reached = 1878.90 m

b) [tex]t=\frac{2usin\theta }{g}=\frac{2\times 192\times sin90}{9.81}=39.14s[/tex]

Time of flight = 39.14 seconds.

answer

50°C

Explanation:

Mw*©=Mi*©

250*20=100*©

©=250*20/100

©=50°C//

To find the final temperature after mixing ice and water, one must calculate the heat required to melt the ice and compare it to the heat the water loses in the process, using the specific heat of water and the enthalpy of fusion of ice.

To determine the final temperature of water after adding 100g of ice at 0°C to 250g of water at 20°C, we must consider the heat transfer involved in the process of melting ice and the subsequent equilibrium between the melted ice and the initial water. The specific heat of water (4.184 J/(g°C)) and the enthalpy of fusion of ice (6.01 kJ/mol) are crucial to this calculation.

The process involves calculating the heat required to melt the ice (Q = mfusion * ΔHfusion) and equating it to the heat lost by the water as it cools (Q = mcwater*ΔT), and solving for the final temperature. However, in this scenario, it's important to note that without the specific values or further calculations included in the question details, we can't provide a precise numerical answer but can outline the method to find it.

The transmission axis of the second sheet makes a right angle (90 degrees) with the vertical given that no light passes through it. This is based on Malus's Law for polarization.

The phenomenon described in your question is polarization, which is a characteristic of transverse waves such as light. The sheet of polarizing material allows only the light polarized in the same direction as the sheet's axis of transmission to pass through it. Following Malus's Law for polarization, we can determine that if no light passes through the second polarizer, its transmission axis must be at a right angle (90 degrees) to the initial polarization direction, which is vertical in this case.

The Malus's Law is an equation that describe the intensity of the light after passing through a polarizer as: I = I0 cos^2 θ where I0 is the initial intensity of the light, θ is the angle between the light polarization direction and the transmission axis of the polarizer. If I = 0, then cos^2 θ = 0, which indicates that θ is 90 degrees. This concept comes under the area of wave optics in Physics.

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The transmission axis of the second sheet of polarizing material makes a 90-degree angle with the vertical because it must be perpendicular to the first sheet's axis for no light to pass through. Hence, with the first polarizer vertical, the second must be horizontal.

The question asks about the angle that the transmission axis of the second sheet of polarizing material makes with the vertical, given that no light passes through. According to the principles of light polarization, if no light passes through a second sheet of polarizing material, it means the transmission axis of the second sheet is perpendicular to the axis of the first one.

Since the first sheet polarizes light vertically, this implies the transmission axis of the second sheet must be horizontal. Therefore, as per common geometric considerations where vertical and horizontal lines are perpendicular, the angle between the vertical (first filter axis) and horizontal (second filter axis) is 90 degrees.

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Answer:

Average speed, v = 25 m/s

Explanation:

It is given that,

Distance travelled by the object, d = 50 meters

Time taken, t = 2 seconds

Average speed is defined as the total distance divided by total time taken i.e.

[tex]v=\dfrac{d}{t}[/tex]

[tex]v=\dfrac{50\ m}{2\ s}[/tex]

v = 25 m/s

So, the sped of the object is 25 m/s. Hence, this is the required solution.

The average speed of an object that travels 50 meters in 2 seconds is calculated by dividing the distance by time, resulting in 25 m/s.

The average speed of an object is calculated by dividing the total distance travelled by the total time taken to travel that distance. In this case, the object travels 50 meters in 2 seconds, so the average speed is calculated as follows:

Average speed = Total distance \/ Total time = 50 meters \2 seconds = 25 meters per second (m/s).

Therefore, the correct answer to the question is b. 25 m/s.

Answer:

7.4 x 10^-14 N

Explanation:

m = 0.012 kg, M = 0.3 kg, r = 1.8 m

Let the gravitational force be F.

F = G M m / r^2

F = (6.67 x 10^-11 x 0.3 x 0.012) / (1.8)^2

F = 7.4 x 10^-14 N

Answer:

Float is travelling at speed of 3.86 m/s

Explanation:

As per Doppler's effect we know that when source and observer moves relative to each other then the frequency is different from actual frequency

so here when source moves closer to the observer

[tex]f_1 = f_o(\frac{v}{v - v_s})[/tex]

now we have

[tex]356 = f_o(\frac{v}{v - v_s})[/tex]

again when source moves away from the observer then we have

[tex]f_2 = f_o(\frac{v}{v + v_s})[/tex]

[tex]348 = f_o(\frac{v}{v + v_s})[/tex]

now divide above two equations

[tex]\frac{356}{348} = \frac{v + v_s}{v - v_s}[/tex]

[tex]1.023 = \frac{v + v_s}{v - v_s}[/tex]

[tex]1.023( v - v_s) = (v + v_s)[/tex]

here we know that the speed of sound in air is 340 m/s

so we have

[tex]1.023(340 - v_s) = (340 + v_s)[/tex]

[tex]7.816 = 2.023 v_s[/tex]

now we have

[tex]v_s = 3.86 m/s[/tex]

The speed of the float in the parade is determined using the observed frequencies of the flute when the band is approaching and moving away informed by the Doppler Effect physics theory. You generate equations for both scenarios and solve for the speed of the source, using the average speed of sound in the air.

This problem relates to the Doppler Effect, which explains why the frequency of a sound changes when the source of the sound is moving in relation to the observer. The formula to calculate a Doppler Shift when the source is moving away (as in the float in the parade) is given by: f' = f [(v + v0) / v] and when the source is approaching is given by: f' = f [(v - v0) / v]. Where, f' is the observed frequency, f is the source frequency, v is the sound speed (340 m/s), and v0 is the speed of the source.

In the problem, it's given that the observed frequencies when the band is approaching and moving away are 356 Hz and 348 Hz, respectively. By setting up two equations using the Doppler Shift formula for both observed frequencies, we can solve for v0. In this case, we use the average speed of sound in the air, which is about 340 m/s.

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Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

[tex]F = ma[/tex]...(I)

Using formula of friction force

[tex]F= \mu m g[/tex]....(II)

Put the value of F in the equation (II) from equation (I)

[tex]ma=\mu mg[/tex]....(III)

[tex]a = \mu g[/tex]

Put the value in the equation (III)

[tex]a=0.24\times9.8[/tex]

[tex]a=2.352\ m/s^2[/tex]

We need to calculate the distance,

Using equation of motion

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

[tex]s=0+\dfrac{1}{2}2.352\times(3.0)^2[/tex]

[tex]s=10.584\ m\ approx\ 11\ m[/tex]

Hence, The distance is 11 m.

The truck can travel a maximum distance of 10.584 meters in 3.0 seconds without the box sliding.

To determine the maximum distance the truck can travel in 3.0 s without the box sliding, we start by calculating the maximum acceleration the box can experience before it starts to slide.

This is given by the static friction force, which is the product of the coefficient of static friction  and the normal force (N). Assuming the box’s weight is W = mg, the normal force is N = mg.

The maximum static friction force (f) is:

f = μ N = μ mg

Given μs = 0.24, we need to ensure:

f > ma

Therefore, 0.24mg = ma, solving for a gives us:

a = 0.24g = 0.24 × 9.8 m/s² ≈ 2.352 m/s²

Next, we need to calculate the maximum distance using the kinematic equation for constant acceleration:

d = ut + (1/2)at²

Where:

Substituting these values, we get:

d = 0 + (1/2) × 2.352 m/s² × (3.0 s)² = 0.5 × 2.352 × 9 = 10.584 m

Thus, the maximum distance the truck can travel without the box sliding is approximately 10.584 meters.

Answer:

[tex]\vec \tau = 31.9 N[/tex]

Explanation:

As we know that torque due to a force is given by the formula

[tex]\vec \tau = \vec r \times \vec F[/tex]

here we know that force is exerted due to weight of the mass hold in his hand

so we have

F = mg = 59 N

Now Lever arm is the perpendicular distance on the line of action of force from the axis about which the system is rotated

so here we can say

[tex]r = Lcos\theta [/tex]

[tex]r = 0.56 cos15 = 0.54 m[/tex]

now we have

[tex]\vec \tau = (0.54)(59)[/tex]

[tex]\vec \tau = 31.9 Nm[/tex]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

[tex]mu=(m+M)v[/tex]

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

[tex]v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s[/tex]

Answer:

[tex]\Delta v = 4.41 \times 10^{-37} cm/s[/tex]

Explanation:

As per Heisenberg's uncertainty principle we know that

[tex]\Delta P \times \Delta x = \frac{h}{4\pi}[/tex]

so here we have

[tex]\Delta P = m\Delta v[/tex]

[tex]\Delta x = 8.54 m[/tex]

now from above equation we have

[tex]m\Delta v \times (8.54) = \frac{h}{4\pi}[/tex]

[tex]1400(\Delta v) \times (8.54) = \frac{6.626 \times 10^{-34}}{4\pi}[/tex]

[tex]\Delta v = 4.41 \times 10^{-39} m/s[/tex]

[tex]\Delta v = 4.41 \times 10^{-37} cm/s[/tex]

0.005366 seconds is the time will the current flowing in this circuit be 55.0% of the peak current.

In a simple AC circuit with a resistor connected to an AC voltage source, the current flowing through the circuit can be calculated using Ohm's law:

I(t)=ΔV(t)/R

Where: I(t) is the current at time

ΔV(t) is the instantaneous voltage at time t

R is the resistance of the resistor.

Given that the source voltage is [tex]\bigtriangleup V=\bigtriangleup V_{max}sin(2 \pi ft)[/tex], and you want to find the time  t at which the current is 55% of the peak current, you need to find the time when [tex]I(t)=0.55\times I_{peak}.[/tex]

Find the peak current [tex]I_{peak}[/tex].

The peak current corresponds to the current when the source voltage is at its peak value  [tex]\bigtriangleup V_{max}[/tex] ​ .

Using Ohm's law:

[tex]I_{peak}=\frac{\bigtriangleup V_{max}}{R}[/tex]

The time t when the current  I(t) is 55% of the peak current:

[tex]I(t)=0.55 \times I_{peak}[/tex]

Substitute the expressions for I(t) and  [tex]I_{peak}[/tex].

[tex]\frac{\bigtriangleup V(t)}{R}=0.55 \times \frac{\bigtriangleup V_{max}}{R}[/tex]

2πft=arcsin(0.55)

t=arcsin(0.55)/2πf

t= 0.5716/2×3.14×16.9

t=0.5716/106.132

t=0.0053s

Hence, at approximately 0.005366 seconds after the start of the AC cycle (or 5.366 ms 5.366ms), the current flowing in the circuit will be 55.0% of the peak current.

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Answer:

Heat flow through walls in 10 hours is

[tex]Q = 6.1 \times 10^6 J[/tex]

Explanation:

Thermal conductivity of all the materials is as follows

1). Concrete = 0.8

2). Brick = 0.6

3). Fibreglass = 0.04

4). Corkboard = 0.04

Area of the wall is given as

[tex]A = (8 \times 4) = 32 m^2[/tex]

now the thermal resistance due to each wall is given as

[tex]R_1 = \frac{0.04}{0.6(32)} = 2.08 \times 10^{-3}[/tex]

[tex]R_2 = \frac{0.08}{0.8(32)} = 3.125 \times 10^{-3}[/tex]

[tex]R_3 = \frac{0.06}{0.04(32)} = 46.87 \times 10^{-3}[/tex]

[tex]R_2 = \frac{0.10}{0.04(32)} = 78.13 \times 10^{-3}[/tex]

Now total thermal resistance of all walls

[tex]R = R_1 + R_2 + R_3 + R_4[/tex]

[tex]R = (2.08 + 3.125 + 46.87 + 78.13) \times 10^{-3}[/tex]

[tex]R = 0.130[/tex]

now rate of heat transfer per second is given as

[tex]\frac{dQ}{dt} = \frac{T_1 - T_2}{R}[/tex]

[tex]Q = \frac{40 - 18}{0.130}(10 \times 3600)[/tex]

[tex]Q = 6.1 \times 10^6 J[/tex]

Answer:

[tex]I_1 = 32.5 kg m^2[/tex]

Explanation:

Here as we know that total angular momentum is always conserved for child + round system

so here by angular momentum conservation we have

[tex]I_1 \omega_1 = (I_1 + I_2)\omega_2[/tex]

here we have

[tex]I_2 = mR^2[/tex] (inertia of boy)

[tex]I_2 = (34.5)(1.45^2)[/tex]

[tex]I_2 = 72.5 kg m^2[/tex]

now we have

[tex]I_1(2\pi 55) = (I_1 + 72.5)(2 \pi 17)[/tex]

[tex]I_1 = (I_1 + 72.5)(0.31)[/tex]

[tex]I_1(1 - 0.31) = 22.4[/tex]

[tex]I_1 = 32.5 kg m^2[/tex]

The total moment of inertia of the merry-go-round with the child can be calculated by considering the child as a point mass at a given distance from the axis of rotation. The decrease in the angular velocity of the merry-go-round when the child jumps on is due to the increase in the moment of inertia. The moment of inertia of the merry-go-round itself is close to the total moment of inertia because it has far more mass distributed away from the axis.

The subject is based on the physics concept of moment of inertia, which depends not only on the object's mass but also on its distribution relative to the axis of rotation. The moment of inertia initially decreases when the child, initially at rest, grabs onto the already spinning merry-go-round. This abrupt decrease in angular velocity is because of the increase in moment of inertia caused by the child.

To find the total moment of inertia (I), we first calculate the child’s moment of inertia (Ic) by considering the child as a point mass at a distance of 1.45 m from the axis of rotation. The formula for this is Ic = mR², where m is the child's mass (34.5 kg) and R is the distance from the center (1.45 m).

Then, we calculate for the angular velocity change from 55 rpm to 17 rpm. The goal is to find the moment of inertia of the merry-go-round itself, which should be close to the total moment of inertia because it has much more mass distributed away from the axis than the child does.

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