A ladder that is 4.6 m long is leaning against a wall at an angle of 66.0º with respect to the ground. If the base of the ladder is moved 0.31 m away from the wall, how far will the top of the ladder go down?

Answers

Answer 1

Answer:

The top of the ladder go down by 0.15 m.

Explanation:

Here we have right angled triangle.

Hypotenuse = 4.6 m

Bottom angle = 66º

Length from ladder bottom to wall bottom = 4.6 cos66 = 1.87 m

Length from ladder top to wall bottom = 4.6 sin66 = 4.20 m

New length from ladder bottom to wall bottom =  1.87 + 0.31 = 2.18 m

By Pythagoras theorem

New length from ladder top to wall bottom is given by

          [tex]\sqrt{4.6^2-2.18^2}=4.05m[/tex]

Distance  the top of the ladder go down = 4.20 - 4.05 = 0.15 m


Related Questions

When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round your answers to two decimal places.) (a) What is the temperature of the drink after 45 minutes? °C (b) When will its temperature be 16°C? min

Answers

(a) The temperature of the drink after 45 minutes is 13.85°C.

(b) The temperature of the drink will be ( 16°C ) after ( 67.75) minutes.

To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature. The formula can be written as:

[tex]\[ \frac{dT}{dt} = -k(T - T_a) \][/tex]

(a) To find the temperature of the drink after 45 minutes, we first need to determine the constant (k). We can do this using the given data points:

1. At ( t = 0 ), ( T = 5°C).

2. At ( t = 25) minutes, ( T = 10°C), and [tex]\( T_a[/tex] = 20°C.

Using these values, we can write the equation as:

[tex]\[ 10 = 20 - (20 - 5)e^{-25k} \][/tex]

Solving for (k):

[tex]\[ e^{-25k} = \frac{20 - 10}{20 - 5} \] \[ e^{-25k} = \frac{10}{15} \] \[ e^{-25k} = \frac{2}{3} \] \[ -25k = \ln\left(\frac{2}{3}\right) \] \[ k = -\frac{\ln\left(\frac{2}{3}\right)}{25} \][/tex]

Now that we have (k), we can find the temperature after 45 minutes:

[tex]\[ T = 20 - (20 - 5)e^{-45k} \] \[ T = 20 - 15e^{45\left(\frac{\ln\left(\frac{2}{3}\right)}{25}\right)} \] \[ T = 20 - 15e^{1.8\ln\left(\frac{2}{3}\right)} \] \[ T = 20 - 15\left(\frac{2}{3}\right)^{1.8} \] \[ T \approx 20 - 15(0.41) \] \[ T \approx 20 - 6.15 \] \[ T \approx 13.85C \][/tex]

So, the temperature of the drink after 45 minutes is approximately ( 13.85°C ).

(b) To find when the temperature of the drink will be (16°C), we use the same formula and solve for ( t ):

[tex]\[ 16 = 20 - (20 - 5)e^{-kt} \] \[ e^{-kt} = \frac{20 - 16}{20 - 5} \] \[ e^{-kt} = \frac{4}{15} \] \[ -kt = \ln\left(\frac{4}{15}\right) \] \[ t = -\frac{\ln\left(\frac{4}{15}\right)}{k} \] \[ t = -\frac{\ln\left(\frac{4}{15}\right)}{-\frac{\ln\left(\frac{2}{3}\right)}{25}} \] \[ t = 25\frac{\ln\left(\frac{4}{15}\right)}{\ln\left(\frac{2}{3}\right)} \][/tex]

Using a calculator, we find:

[tex]\[ t \approx 25\frac{\ln(0.2667)}{\ln(0.6667)} \] \[ t \approx 25 \times 2.71 \] \[ t \approx 67.75 \][/tex]

So, the temperature of the drink will be ( 16°C ) after ( 67.75) minutes.

A cylinder 8 inches in diameter and 3 ft long is concentric with a pipe of 8.25 inches i.d. Between cylinder and pipe there is an oil film. What force is required to move the cylinder along the pipe at a constant velocity of 3 fps? The kinematic viscosity of the oil is 0.006 ft2 /s; the specific gravity is 0.92.

Answers

Answer:

Required force equals 623.498 lb

Explanation:

We shall use newton's law of viscosity to calculate the shear force that acts on the cylinder

By Newton's law of viscosity we have

[tex]\tau =\mu \frac{dv}{dy}\\\\where \tau[/tex] is shear stress that acts on the internal surface

[tex]\mu[/tex] is dynamic viscosity of the fluid

[tex]\frac{dv}{dy}[/tex] is the velocity gradient that exists across the flow

The dynamic viscosity is calculated as follows

[tex]\mu =\rho \nu[/tex]

[tex]\mu =\rho \nu \\\rho[/tex] is density of the fluid

[tex]\nu =[/tex] kinematic viscosity of the fluid

By  no slip boundary condition the fluid in contact with the stationary cylinder shall not have any velocity while as the fluid in contact with the moving cylinder shall have velocity equal to that of the cylinder itself. This implies a velocity gradient shall exist across the gap in between the cylinders.

Applying values of the quantities we can calculate shear stress as follows

The density of fluid is [tex]\rho=G\times \rho_{w}[/tex]

G = specific gravity of fluid

[tex]\rho_{w}[/tex]is density of water

[tex]\tau =\rho \nu \frac{dv}{dy}\\\\\tau=62.42\times 0.92\times 0.006\times \frac{3}{\frac{0.125inches}{12inch}}\\\\\tau=99.23lb/ft^{2}[/tex]

This pressure shall oppose the motion of the internal cylinder hence the force of opposition = [tex]F=\tau\times Area[/tex]

Using the area of internal cylinder we get total force

F=[tex]2\pi rl\times \tau\\\\F=2\pi\times \frac{4}{12}ft\times 3ft\times \tau\\\\ F=623.498lb[/tex]

Final answer:

The force required to move the cylinder along the pipe can be calculated using the drag force formula. F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf

Explanation:

To calculate the force required to move the cylinder along the pipe at a constant velocity, we need to consider the drag force acting on the cylinder.

The drag force can be calculated using the formula:

F = 6πηrv

Where F is the force, η is the viscosity of the oil, r is the radius of the cylinder, and v is the velocity of the cylinder.

In this case, the radius of the cylinder is half of the diameter, so the radius is 4 inches or 1/3 ft. The velocity is given as 3 fps.

Plugging in the values, we get:

F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf

Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m.

Answers

Final answer:

To bring a 950-kg car to rest from 90 km/h over 120 m requires an average force of 2473.96 N. If the car hits a concrete abutment and stops within 2 m, the force exerted is much higher, at 148,437.50 N. This illustrates the impact of stopping distance on the force experienced by a vehicle.

Explanation:

Work-Energy Theorem Application

We'll first convert the speed from km/h to m/s by multiplying by 1000/3600. Therefore, 90.0 km/h is 25 m/s. Using the work-energy theorem, we know the work done to stop the car is equal to the change in kinetic energy.

Plugging in the values: W = KE = 1/2mv²

W = 1/2(950 kg)(25 m/s)²

W = 1/2(950 kg)(625 m²/s²)

W = 296,875 J

Since work is also equal to force times distance (W = Fd), the force needed can be found by dividing the work by the distance.

F = W/d = 296,875 J/120 m = 2473.96 N

This is the average force required to stop the car over 120 m. Now let's calculate the force if the car hits a concrete abutment and stops in 2.00 m:

F = 296,875 J/2 m = 148,437.50 N

The force exerted on the car in this case is significantly higher, showing the importance of cushioning distance in reducing impact forces.

Benny wants to estimate the mean lifetime of Energizer batteries, with a confidence level of 97%, and with a margin of error not exceeding ±10 hours. If the standard deviation of the lifetime is known to be 55 hours, how many batteries does Benny need to sample?

Answers

Answer:

143 batteries does Benny need to sample

Explanation:

Given data

confidence level = 97%

error  = ±10 hours

standard deviation SD = 55 hours

to find out

how many batteries does Benny need to sample

solution

confidence level is 97%

so a will be 1 - 0.97 = 0.03

the value of Z will be for a 0.03 is 2.17 from standard table

so now we calculate no of sample i.e

no of sample  = (Z× SD/ error)²

no of sample = (2.16 × 55 / 10)²

no of sample = 142.44

so  143 batteries does Benny need to sample

Three forces act on an object. If the object is in translational equilibrium, which of the following must be true? I. The vector sum of the three forces must equal zero; II. The magnitude of the three forces must be equal; III. The three forces must be parallel.

Answers

Answer:

Option I

Explanation:

When ever the system is in equilibrium, it means the net force on the system is zero.

If the number of forces acting on a system and then net force on the system is zero, it only shows that the vector sum of all the forces is zero.

The half-life of1 is 8.04 days. (a) Calculate the decay constant for this isotope. (b) Find the number of 1311 nuclei necessary to of 0.5 uCi produce a sample with an activity

Answers

Explanation:

Given that,

[tex]T_{\frac{1}{2}}=8.04\ days[/tex]

We need to calculate the decay constant

Using formula of decay constant

[tex]\lambda=\dfrac{0.693}{t_{\frac{1}{2}}}[/tex]

[tex]\lambda=\dfrac{0.693}{8.04\times24\times3600}[/tex]

[tex]\lambda=9.97\times10^{-7}\ sec^{-1}[/tex]

We need to calculate the number of [tex]^{131}I[/tex] nuclei

[tex]N=\dfrac{A\ ci}{\lambda}[/tex]

Where,

A= activity

ci = disintegration

[tex]N=\dfrac{0.5\times10^{-6}\times3.7\times10^{10}}{9.97\times10^{-7}}[/tex]

[tex]N=1.855\times10^{10}[/tex]

Hence, This is the required solution.

Consider a Cassegrain-focus, reflecting telescope. Images recorded at Cassegrain-focus will be:

A. Oriented the same as in the sky

B. Flipped compared to what is in the sky

C. Rotated clockwise compared to what is in the sky

D. Rotated counter-clockwise compared to what is in the sky

Answers

B. Flipped compared to what is in the sky

What is the power of the man who pushes the box 8m with a force of 15N in a 6seconds? A. 20 Watts B. 30 Watts C. 40 Watts D. 40 Watts

Answers

Answer:

  A.  20 Watts

Explanation:

Power is the rate of doing work.

  P = Fd/t = (15 N)(8 m)/(6 s) = 20 N·m/s = 20 W

20 watts. If the power of the man who pushes the box 8m with a force of 15n is 6 seconds, it will force 20 watts in 6 seconds.

First, you do is multiply and then divide to give us the final answer.

[tex]\displaystyle 8m\times15n\div6s[/tex]

[tex]\displaystyle 8\times15=120\div6=20[/tex]

[tex]\Large\boxed{20}[/tex], which is our answer.

At the 18th green of the U.S. Open you need to make a 23.5ft putt to win the tournament. When you hit the ball, giving it an initial speed of 1.45 m/s, it stops 5.10 ft short of the hole. Assuming the deceleration caused by the grass is constant, what should the initial speed have been to just make the putt? 2.What initial speed do you need to make the remaining 5.10 ft putt?

Answers

Answer:

1.63 m/s

Explanation:

a = acceleration of the ball

d = stopping distance = 23.5 - 5.10 = 18.4 ft = 5.61 m

v₀ = initial velocity of the car = 1.45 m/s  

v = final velocity of the car = 0 m/s  

Using the equation

v² =  v₀² + 2 a d

0² = 1.45² + 2 a (5.61)

a = - 0.187 m/s²

To win the tournament :

a = acceleration of the ball = - 0.187 m/s²

d = stopping distance = 23.5 ft = 7.1 m

v₀ = initial velocity of the car = ?

v = final velocity of the car = 0 m/s  

Using the equation

v² =  v₀² + 2 a d

0² =  v₀² + 2 (- 0.187) (7.1)

v₀ = 1.63 m/s

Final answer:

The initial speed required can be calculated using the formulas of motion under constant deceleration. Here, we first calculated deceleration from the given initial velocity and distance and then applied that deceleration to find the initial speed required for the remaining distance.

Explanation:

In the given problem, the ball falls short of the target. It indicates that the ball decelerated while it was in motion. The distance covered and the initial speed are given, so the deceleration can be calculated using the formula for motion under constant acceleration (v^2 = u^2 + 2as). In this case, the final velocity (v) is 0, the initial velocity (u) is 1.45m/s, and the distance covered (s) is the total - the short distance, which in meters turns out to be (23.5ft - 5.10ft) * 0.3048 = 5.608m. By substituting these values, we can solve for acceleration (a). For the remaining 5.10ft, which is approximately 1.554m, we can use the found deceleration and the physics equation v = sqrt(u^2 + 2as) to find the initial speed that is required for the remaining distance.

Learn more about the Physics of Motion here:

https://brainly.com/question/13966796

#SPJ3

An aluminum wire has a resistance of 7.00 Ω at 30.0°C. Determine its resistance (in Ω) at 430°C. The temperature coefficient of resistivity for aluminum wire is 3.90 ✕ 10−3 (°C)−1. (Assume that the temperature coefficient of resistivity was measured using the reference temperature 20°C.)

Answers

Answer:

17.92 Ω

Explanation:

R₀ = Initial resistance of the aluminum wire at 30.0°C = 7.00 Ω

R = resistance of the aluminum wire at 430.0°C = ?

α = temperature coefficient of resistivity for aluminum = 3.90 x 10⁻³ °C⁻¹

ΔT = Change in temperature = 430 - 30 = 400 °C

Resistance of the wire is given as

R = R₀ (1 + α ΔT)

R = (7) (1 + (3.90 x 10⁻³) (400))

R = 17.92 Ω

A long straight wire runs up and down vertically and carries current downward. What is the direction of the magnetic field produced by this current at a location nearby and directly east of the wire? A) southward
B) downward
C) upward
D) northward
E) westward

Answers

Answer:

A) southward

Explanation:

Applying the rule of the right hand for the determination of the magnetic field produced by the current flowing through the cable yields the following results: between the wires it is to the north, the west of the wires is to the south and east of the wires , which asks the problem is to the south.

a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient of thermal expansion [in (degrees c)^-1] for this material.

Answers

Answer:

The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹

Explanation:

Li = 0.2m

ΔL = 0.2 mm = 0.0002m

T1 = 21ºC

T2 = 120ºC

ΔT =99ºC

α =ΔL/(Li*ΔT)

α =0.0002m /(0.2m * 99ºC)

α = 1.01 *10⁻⁵   (ºC)⁻¹

The correct value of the linear coefficient of thermal expansion for the material is [tex]1.01 \times 10^{-5} \ (^\circ{C})^{-1}}\)[/tex].

To determine the linear coefficient of thermal expansion [tex](\(\alpha\))[/tex] for the material, we can use the formula:

[tex]\[\alpha = \frac{\Delta L}{L_0 \Delta T}\][/tex]

where:

[tex]\(\Delta L\)[/tex] is the change in length,

[tex]\(L_0\)[/tex] is the original length, and

[tex]\(\Delta T\)[/tex] is the change in temperature.

Given:

The original length [tex]\(L_0\)[/tex] is 0.20 m,

The elongation [tex]\(\Delta L\)[/tex] is 0.20 mm, which we need to convert to meters to match the units of [tex]\(L_0\)[/tex].

Since 1 m = 1000 mm, 0.20 mm = [tex]\(0.20 \times 10^{-3}\)[/tex] m,

The change in temperature [tex]\(\Delta T\)[/tex] is from 21°C to 120°C, so [tex]\(\Delta T = 120^\circ{C} - 21^\circ{C} = 99^\circ{C}\)[/tex].

Now we can plug these values into the formula:

[tex]\[\alpha = \frac{0.20 \times 10^{-3} \ m}{0.20 \ m \times 99^\circ{C}}\][/tex]

[tex]\[\alpha = \frac{0.20}{0.20 \times 99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]

[tex]\[\alpha = \frac{1}{99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]

[tex]\[\alpha \approx 1.0101 \times 10^{-5} \ (^\circ{C})^{-1}\][/tex]

Rounding to two significant figures, we get:

[tex]\[\alpha \approx 1.01 \times 10^{-5} \ (^\circ{C})^{-1}\}[/tex]

On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.87 m/s2. (a) How long does such a car, initially traveling at 23.1 m/s, take to stop? (b) How far does it travel in this time?

Answers

Answer:

(a): the car takes to stop 4.74 seconds.

(b): the car travels 54.78 meters in this time.

Explanation:

a= 4.87 m/s²

Vi= 23.1 m/s

Vf= Vi - a*t

t= Vi/a

t= 4.74 sec

d= Vi*t - (a*t²)/2

d= 54.78m

The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30% d)-14.02%

Answers

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

[tex]f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]

now it is given that tension is increased by 30%

so here we will have

[tex]T' = T(1 + 0.30)[/tex]

[tex]T' = 1.30T[/tex]

now new value of fundamental frequency is given as

[tex]f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}[/tex]

now we have

[tex]f_o' = \sqrt{1.3}f_o[/tex]

so here percentage change in the fundamental frequency is given as

[tex]change = \frac{f_o' - f_o}{f_o} \times 100[/tex]

% change = 14.02%

The velocity of an object moving along a straight line is given by – v(t) = t 2 − 3t + 2 (a) Find the displacement of the object from t = 0 to t = 3. (b) Find the distance traveled from t = 0 to t = 3.

Answers

Answer:

a) Displacement of the object from t = 0 to t = 3 is 1.5 m

b)  Distance of the object from t = 0 to t = 3 is 1.83 m

Explanation:

Velocity, v(t) = t² - 3t + 2

a) Displacement is given by integral of v(t) from 0 to 3.

   [tex]s=\int_{0}^{3}(t^2-3t+2)dt=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^3=\frac{3^3}{3}-\frac{3^3}{2}+6=1.5m[/tex]

b) t² - 3t + 2 = (t-1)(t-2)

   Between 1 and 2,  t² - 3t + 2 is negative

   So we can write t² - 3t + 2 as -(t² - 3t + 2)

   Distance traveled

             [tex]s=\int_{0}^{1}(t^2-3t+2)dt+\int_{1}^{2}-(t^2-3t+2)dt+\int_{2}^{3}(t^2-3t+2)dt\\\\s=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^1-\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_1^2+\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_2^3\\\\s=\frac{1^3}{3}-\frac{3\times 1^2}{2}+2-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )+\frac{1^3}{3}-\frac{3\times 1^2}{2}+2+\frac{3^3}{3}-\frac{3\times 3^2}{2}+6-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )=1.83m[/tex]

Final answer:

The displacement of the object from t = 0 to t = 3 is -4.5 m, indicating the object moved 4.5 meters in the opposite direction from its initial position. The distance the object traveled during the same period can be found by taking the integral of the absolute value of the velocity function from 0 to 3, and adding the magnitudes for time intervals when the velocity was positive and when it was negative.

Explanation:

The given function indicates the velocity of an object moving along a straight line as a function of time - v(t) = t^2 - 3t + 2. It's a quadratic function so one way to find the displacement from t = 0 to t = 3, is to integrate the velocity function. The integral of v(t) from 0 to 3 gives the total change in position, or displacement, which would be the integral ∫ from 0 to 3 of (t^2 - 3t + 2) dt = [t^3/3 - 1.5t^2 + 2t] from 0 to 3 = 3^3/3 - 1.5 * 3^2 + 2*3 - (0 - 0 + 0) = 3 - 13.5 + 6 = -4.5 m.

On the other hand, distance is a scalar quantity and does not account for the direction, only the magnitude of movement. As such the object's total distance travelled from t = 0 to t = 3 may be calculated by finding the integral of the absolute value of the velocity from 0 to 3. In this case, v(t) is positive from t = 0 to t = 1 and negative from t = 1 to t = 3 (as substantiated by equating the velocity function to 0). Thus, the total distance traveled by the object is the sum of distances in segments (0,1) and (1,3), obtained as the sum of integrals ∫ from 0 to 1 of (t^2 - 3t + 2) dt + ∫ from 1 to 3 of (-t^2 + 3t - 2) dt.

Learn more about the Physics of motion here:

https://brainly.com/question/13966796

#SPJ3

During a tennis serve, a racket is given an angular acceleration of magnitude 155 rad/s2. At the top of the serve, the racket has an angular speed of 20.0 rad/s. If the distance between the top of the racket and the shoulder is 1.40 m, find the magnitude of the total acceleration of the top of the racket.

Answers

Answer:

600.6 m/s^2

Explanation:

α = 155 rad/s^2

ω = 20 rad/s

r = 1.4 m

Tangential acceleration, aT = r x α = 1.4 x 155 = 217 m/s^2

Centripetal acceleration, ac = rω^2 = 1.4 x 20 x 20 = 560 m/s^2

The tangential acceleration and the centripetal acceleration both are perpendicular to each other. Let a be the resultant acceleration.

a^2 = aT^2 + ac^2

a^2 = 217^2 + 560^2

a = 600.6 m/s^2

Final answer:

The total acceleration of the top of the racket during the tennis serve is approximately 580 m/s². This is determined by considering both the centripetal and tangential accelerations as perpendicular components and using the Pythagorean theorem for calculations.

Explanation:

In this physics problem, we're given the angular acceleration, angular speed, and the distance between the top of the racket and shoulder (radius) to determine the total acceleration of the racket top during a tennis serve. To find the total acceleration, we must take into account both the centripetal (or radial) acceleration and the tangential acceleration (due to the change in speed).

First, let's calculate the centripetal acceleration, given by the formula ac=ω²r, where ω is the angular speed and r is the radius of the motion (in this case, the length of the arm). So, ac = (20.0 rad/s)² x 1.4m = 560 rad/s².

The tangential acceleration (at) is simply equal to the angular acceleration, which is 155 rad/s² (as provided in the question).

To find the total acceleration, we consider these two accelerations as perpendicular components and use the Pythagorean theorem: a = sqrt(ac² + at²). Substituting the values, we get a = sqrt((560 m/s²)² + (155 m/s²)²) ≈ 580 m/s².

Therefore, the total acceleration of the top of the racket is approximately 580 m/s².

Learn more about Total Acceleration here:

https://brainly.com/question/21564527

#SPJ3

An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can absorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (m el = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J, ℏ = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s)

Answers

Answer:

The longest wavelength of light is 209 nm.

Explanation:

Given that,

Spring constant = 74 N/m

Mass of electron [tex]m= 9.11\times10^{-31}\ kg[/tex]

Speed of light [tex]c= 3\times10^{8}\ m/s[/tex]

We need to calculate the frequency

Using formula of frequency

[tex]f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]

Where, k= spring constant

m = mass of the particle

Put the value into the formula

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}[/tex]

[tex]f=1.434\times10^{15}\ Hz[/tex]

We need to calculate the longest wavelength that the electron  can absorb

[tex]\lambda=\dfrac{c}{f}[/tex]

Where, c = speed of light

f = frequency

Put the value into the formula

[tex]\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}[/tex]

[tex]\lambda=2.092\times10^{-7}\ m[/tex]

[tex]\lambda=209\ nm[/tex]

Hence, The longest wavelength of light is 209 nm.

When the frequency of the driving force equals the natural frequency of the system, the system is said to be in. (a) Resonance (b) Equilibrium (c) Elastic limit (d) None

Answers

Answer:

Option A is the correct answer.

Explanation:

When the frequency of the driving force equals the natural frequency of the system, the system is said to be in resonance. At resonance the system vibrates in maximum amplitude.

Marching soldiers are cautioned to break stride on a bridge is because of resonance, if the frequency of soldiers stride is equal to frequency of bridge, the bridge will vibrate with maximum amplitude. This will in turn collapse bridge.

Option A is the correct answer.

An AC generator consists of eight turns of wire, each of area 0.0775 m2 , and total resistance of 8.53 Ω. The loop rotates in the external magnetic field of strength 0.222 T at a constant frequency of 51 Hz. What is the maximum induced emf? Answer in units of V.

Answers

Answer:

44.08 Volt

Explanation:

N = 8, A = 0.0775 m^2, R = 8.53 ohm, B = 0.222 T, f = 51 Hz

e0 = N B A w

e0 = 8 x 0.222 x 0.0775 x 2 x 3.14 x 51

e0 = 44.08 Volt

Final answer:

The question pertains to the calculation of the maximum induced emf (voltage) in an AC generator. This can be calculated using the generator's specifications and the formula ε_max = NBAω.

Explanation:

The subject of your question pertains to electromagnetic induction in physics. The induction of emf in an AC generator is described by the equation ε = NBAω sin ωt, where ε is the induced emf, N is the number of turns of wire, B is the magnetic field strength, A is the cross-sectional area of the coil, and ω is the angular velocity of the generator. However, considering you are asking specifically for the maximum induced emf, we calculate this using the equation ε_max = NBAω, as sin ωt=1 at its peak. In your case, the generator consists of 8 turns of wire (N=8), an area of 0.0775 m^2 (A=0.0775), the strength of the magnetic field is 0.222 T (B=0.222), and a frequency of 51 Hz (f=51) which converts to angular velocity (ω) using the formula ω = 2πf. Substituting these values into the equation will give you the maximum induced emf.

Learn more about Maximum induced emf here:

https://brainly.com/question/31493122

#SPJ3

A critical part has a manufacturing specification (in cm) of 0.325 ± 0.010. Based on this information, if this measurement is larger than 0.335 or smaller than 0.315, the product fails at a cost of $120. Determine the Taguchi loss function in the given scenario.

Answers

Answer:

[tex]L(y)=12\times 10^{5}(y-0.325)^2[/tex]

Explanation:

We know that Taguchi loss function given as

[tex]L(y)=k(y-m)^2[/tex]

Where L is the loss when quality will deviate from target(m) ,y is the performance characteristics and k is the quality loss coefficient.

Given that 0.325±0.010 ,Here over target is m=0.325 .

When y=0.335 then L=$120,or when y=0.315 then L=$120.

Now to find value of k we will use above condition

[tex]L(y)=k(y-m)^2[/tex]

[tex]120=k(0.335-0.325)^2[/tex]

[tex]k=12\times 10^{5}[/tex]

So Taguchi loss function given as

[tex]L(y)=12\times 10^{5}(y-0.325)^2[/tex]

Answer:

Explanation:

Manufacturing specification

0.325 ± 0.010 I'm

The quality characteristic is 0.325

Functional tolerance is $120

The lost function is given

λ = C (x—t)²

Where, C is a constant

t is quality characteristic

And x is target value

Constant’ is the coefficient of the Taguchi Loss, or the ratio of functional tolerance and customer loss.

Then, C= tolerance / loss²

Measurement loss is

Loss = 0.335-0.315

Loss =0.01cm

Therefore,

C = 120/0.01²

C = 1,200,000

λ = C (x —t)²

λ = 1,200,00 (x—0.325)²

A rectangular coil with sides 0.10 m by 0.25 m has 500 turns of wire. It is rotated about its long axis in a magnetic field of 0.58 T directed perpendicular to the rotation axis. At what frequency must the coil be rotated for it to generate a maximum potential of 110 V?

Answers

Final answer:

The rectangular coil needs to be rotated at a specific frequency in order to generate a maximum potential of 110 V. The angular velocity of the coil can be calculated using the formula ω = V / (nBA), where V is the potential, n is the number of turns, B is the magnetic field strength, and A is the area of the coil.

Explanation:

To generate a maximum potential of 110 V, the rectangular coil needs to be rotated at a specific frequency. The potential generated by a rotating coil is given by the equation

V = nBAω

where V is the potential, n is the number of turns, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity. Rearranging the equation for ω, we have

ω = V / (nBA)

Substituting the given values, we get

ω = 110 V / (500 turns * 0.10 m * 0.25 m * 0.58 T)

Simplifying this expression will give you the required frequency at which the coil should be rotated to generate a maximum potential of 110 V.

A young man walks daily through a gridded city section to visit his girlfriend, who lives m blocks East and nblocks North of where the young man resides. Because the young man is anxious to see his girlfriend, his route to her never doubles back—he always approaches her location. In terms of m and n, how many different routes are there for the young man to take?

Answers

Answer:

The man ate eggs.

Explanation:

He should brush his teeth before seeing his girlfriend.

A proton moves perpendicular to a uniform magnetic field B S at a speed of 1.00 3 107 m/s and experiences an acceleration of 2.00 3 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.

Answers

Explanation:

It is given that,

Speed of proton, [tex]v=10^7\ m/s[/tex]

Acceleration of the proton, [tex]a=2\times 10^{13}\ \ m/s^2[/tex]

The force acting on the proton is balanced by the magnetic force. So,

[tex]ma=qvB\ sin(90)[/tex]

[tex]B=\dfrac{ma}{qv}[/tex]

m is the mass of proton

[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 2\times 10^{13}\ \ m/s^2}{1.6\times 10^{-19}\times 10^7\ m/s}[/tex]

B = 0.020875

or

B = 0.021 T

So, the magnitude of magnetic field is 0.021 T. As the acceleration in +x direction, velocity in +z direction. So, using right hand rule, the magnitude of  B must be in -y direction.

Two​ vehicles, a car and a​ truck, leave an intersection at the same time. The car heads east at an average speed of 20 miles per​ hour, while the truck heads south at an average speed of 50 miles per hour. Find an expression for their distance apart d​ (in miles) at the end of t hours.

Answers

Answer:

[tex]d = 10\times t\sqrt{29}miles[/tex]

Explanation:

Given:

't' hour be the time taken for travel by  both the vehicles and 'd'  be the distance between then

then

Distance traveled by the car = 20 × t miles

and

Distance traveled by the truck  = 50 × t miles

now, using the Pythagoras theorem

[tex]d = \sqrt{(20t)^2+(50t)^2}[/tex]

or

[tex]d = \sqrt{400t^2+2500t^2}[/tex]

or

[tex]d = \sqrt{2900t^2}[/tex]

or

[tex]d = 10\times t\sqrt{29}[/tex]

thus, the equation relating the distance 'd' with the time 't' comes as

[tex]d = 10\times t\sqrt{29}miles[/tex]

Final answer:

To find the distance between the car and truck after t hours, apply the Pythagorean theorem. Calculate the hypotenuse of the right triangle formed by their paths. The distance apart in miles is represented by the expression 50√1.16t.

Explanation:

The question involves finding an expression for the distance d between two vehicles traveling perpendicularly away from an intersection, with one vehicle going east and another going south at different speeds. To solve this, we can apply the Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. If the car travels east at 20 miles per hour and the truck travels south at 50 miles per hour, after t hours, the car will have traveled 20t miles and the truck 50t miles. These distances represent the two legs of a right triangle, and the distance d between the vehicles is the hypotenuse.

So, the distance d (in miles) at the end of t hours is given by:

d = √{(20t)² + (50t)²}

You can simplify this further to find:

d = t * √{20² + 50²}

d = t *√{400 + 2500}

d = t *√2900

d = t *50√1.16

d = 50√1.16t

Therefore, the distance apart in miles at the end of t hours is given by the expression 50√1.16t.

A runner of mass 56.1 kg starts from rest and accelerates with a constant acceleration of 1.2 m/s^2 until she reaches a velocity of 5.3 m She then continues running at this constant velocity. How long in seconds does the runner take to travel 118 m? A) 19.08 sec B) 24.47 sec C) 53.9 sec D) 15.733333 sec E) 31.152 sec

Answers

Answer:

Option B is the correct answer.

Explanation:

Final velocity = 5.3 m/s

Acceleration till 5.3 m/s = 1.2 m/s²

Time taken for this

           [tex]t_1=\frac{5.3}{1.2}=4.42s[/tex]

Distance traveled in 4.42 s can be calculated

          s = ut + 0.5 at²

          s = 0 x 4.42 + 0.5 x 1.2 x 4.42² = 11.72 m

Remaining distance = 118 - 11.72 = 106.28 m

Uniform velocity = 5.3 m/s

Time taken

       [tex]t_2=\frac{106.28}{5.3}=20.05s[/tex]

Total time, t = t₁ + t₂  = 4.42 + 20.05 = 24.47 s

Option B is the correct answer.

The runner takes approximately 24.47 seconds to travel 118 meters, considering the time spent accelerating and then running at constant velocity. Therefore, the correct answer is option B.

To determine the time it takes for the runner to travel 118 meters, we need to consider two phases of her motion: acceleration and constant velocity.

Phase 1: Acceleration

Initially, the runner starts from rest (initial velocity, u = 0) and accelerates at a constant rate of 1.2 m/s² until she reaches a velocity of 5.3 m/s.

Step 1: Calculate the time (t1) taken to reach the velocity of 5.3 m/s using the formula v = u + at.

v = 5.3 m/s, u = 0, a = 1.2 m/s²

t1 = (v - u) / a = (5.3 - 0) / 1.2 ≈ 4.417 s

Step 2: Calculate the distance (s1) covered during this acceleration phase using the formula s = ut + 0.5at².

s1 = 0 + 0.5 * 1.2 * (4.417)² ≈ 11.7 m

Phase 2: Constant Velocity

After reaching 5.3 m/s, the runner continues at this constant velocity. We need to find the distance she covers in this phase and the total time taken.

Step 3: Calculate the remaining distance (s2) that needs to be covered at constant velocity.

s1 = 11.7 m, Total distance = 118 m

s2 = 118 - 11.7 = 106.3 m

Step 4: Calculate the time (t2) taken to cover the distance s2 at the constant velocity using the formula t = s / v.

t2 = 106.3 m / 5.3 m/s ≈ 20.075 s

Total Time

Step 5: Add the time taken in both phases to find the total time.

Total time = t1 + t2 ≈ 4.417 s + 20.075 s ≈ 24.492 s

Therefore, the runner takes approximately 24.47 seconds to travel 118 meters. The correct answer is option B.

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that the pendulum completes 99.0 full swing cycles in a time of 128 s . What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per second.

Answers

The magnitude of the gravitational acceleration (g) on this planet is equal to 12.52 [tex]m/s^2[/tex].

Given the following data:

Length = 53.0 cm to m = 0.53 m.

Number of cycle, n = 99.0 cycles.

Time = 128 seconds.

How to determine the magnitude of the gravitational acceleration.

First of all, we would calculate the period for a full swing cycle as follows:

[tex]T=\frac{time}{n} \\\\T=\frac{128}{99.0}[/tex]

Period, T = 1.293 seconds.

Mathematically, the time taken (period) by a pendulum is given by this formula:

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

Making g the subject of formula, we have:

[tex]g=\frac{4\pi^2L}{T^2} \\\\g=\frac{4(3.142)^2 \times 0.53}{1.293^2} \\\\g=\frac{20.929}{1.672}[/tex]

g = 12.52 [tex]m/s^2[/tex].

Read more on period here: https://brainly.com/question/20070798

Final answer:

To determine the gravitational acceleration on an unfamiliar planet, the period of a pendulum was used. After calculating the period per swing, the pendulum formula T = 2π√(l/g) was rearranged to solve for g, yielding approximately 9.826 m/s² as the gravitational acceleration.

Explanation:

The student's question involves calculating the magnitude of the gravitational acceleration on an unfamiliar planet using the period of a simple pendulum. The pendulum's length is 53.0 cm and it completes 99.0 swings in 128 seconds. The formula to calculate the period of a pendulum (T) is T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity.

First, calculate the period (T) of one full swing by dividing the total time by the number of cycles: T = 128 s / 99.0 = 1.29292929 s approximately.Next, use the formula T = 2π√(l/g) to solve for g, where l = 0.53 m. Rearrange the formula to solve for g: g = 4π²(l/T²).Plug in the values: g = 4π²(0.53 m / (1.29292929 s)²) = 9.826 m/s² approximately.

Therefore, the gravitational acceleration on the planet is around 9.826 m/s².

What is the frequency of oscillation for a mass on the end of spring with a period of motion of 2.6 seconds? Calculate answer to one decimal place.

Answers

Answer:

Frequency, f = 0.38 Hz

Explanation:

Time period of the spring, T = 2.6 seconds

We need to find the frequency of oscillation for a mass on the end of spring. The relation between the time period and the frequency is given by :

Let f is the frequency of oscillation. So,

[tex]f=\dfrac{1}{T}[/tex]

[tex]f=\dfrac{1}{2.6\ s}[/tex]

f = 0.38 Hz

or

f = 0.4 Hz

So, the frequency of oscillation for a mass on the end of a spring is 0.38 hertz. Hence, this is the required solution.

A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is: A) 3600 C B) 3000C C) 2400C D) 120 C E) 1800 C

Answers

Answer:

Total charge, q = 1800 C

Explanation:

It is given that,

Power of light bulb, P = 60 watts

It carries a current, I = 0.5 A

Time, t = 1 hour = 3600 seconds

We need to find the total charge passing through it in one hour. We know that current through an electrical appliance is defined as the charge flowing per unit time i.e.

[tex]I=\dfrac{q}{t}[/tex]

[tex]q=I\times t[/tex]

[tex]q=0.5\ A\times 3600\ s[/tex]

q = 1800 C

So, the total charge passing through it is 1800 C. Hence, this is the required solution.

Final answer:

The current of 0.5 amperes means that 0.5 coulombs of charge pass every second. Therefore, over an hour (3600 seconds), a total charge of 1800 Coulombs would pass. So, the answer to the question is E) 1800 C.

Explanation:

The question is about understanding the relationship between electric current, time and charge. The electric current is the charge passing per unit time. In this case, we know that our current is 0.5 amperes, which conveys that 0.5 coulombs of electric charge is passing every second.

 Therefore, when you want to find out how much charge passes in an hour, you would multiply by the number of seconds in an hour which is equal to 3600 (60 seconds per minute x 60 minutes).

 Hence, the total charge passing for an hour is 0.5 amperes x 3600 seconds = 1800 Coulombs, so the answer is (E) 1800 C.

Learn more about Electric Charge here:

https://brainly.com/question/9194793

#SPJ3

A copper wire is 1.6 m long and its diameter is 1.1 mm. If the wire hangs vertically, how much weight (in N) must be added to its free end in order to stretch it 2.8 mm?

Answers

Answer:

Weight required = 194.51 N

Explanation:

The elongation is given by

            [tex]\Delta L=\frac{PL}{AE}[/tex]

Length , L= 1.6 m

Diameter, d = 1.1 mm

Area

   [tex]A=\frac{\pi d^2}{4}=\frac{\pi \times (1.1\times 10^{-3})^2}{4}=9.50\times 10^{-7}m^2[/tex]

Change in length, ΔL = 2.8 mm = 0.0028 m

Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa

Substituting,

      [tex]\Delta L=\frac{PL}{AE}\\\\0.0028=\frac{P\times 1.6}{9.50\times 10^{-7}\times 117\times 10^9}\\\\P=194.51N[/tex]

Weight required = 194.51 N

A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg The ball approaches the cushion with a velocity of +2.20 m/s and rebounds with a velocity of -1.70 m/s. The ball remains in contact with the cushion for a time of 3.40 x 10^-3 s. What is the average net force (magnitude and direction) exerted on the ball by the cushion?

Answers

Answer:

Force is 432.94 N along the rebound direction of ball.

Explanation:

Force is rate of change of momentum.

[tex]\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}[/tex]

Final momentum = 0.38 x -1.70 = -0.646 kgm/s

Initial momentum = 0.38 x 2.20 = 0.836 kgm/s

Change in momentum = -0.646 - 0.836 = -1.472 kgm/s

Time = 3.40 x 10⁻³ s

[tex]\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}=\frac{-1.472}{3.40\times 10^{-3}}\\\\\texttt{Force}=-432.94N[/tex]

Force is 432.94 N along the rebound direction of ball.

Final answer:

Using Newton's second law and the change in the ball's momentum, the average net force exerted on the billiard ball by the cushion is 436.47 N, directed away from the cushion.

Explanation:

The question relates to the concept of Newton's second law of motion, which states that the force on an object is equal to the mass of the object multiplied by the acceleration (F = ma). In this situation, a billiard ball strikes the cushion perpendicularly, changing its velocity and hence experiencing acceleration. To calculate the average net force exerted on the ball by the cushion, we can use the change in velocity and the time of impact in the following steps:

Calculate the change in momentum of the ball (Δp = m−v_f - m−v_i), where m is the mass, v_f is the final velocity, and v_i is the initial velocity.

Divide the change in momentum by the time of impact (Δt) to get the average force (F_avg). Use the formula F_avg = Δp / Δt.

Now let's apply these steps to the given values:

The change in momentum is Δp = 0.38 kg * (-1.70 m/s) - 0.38 kg * (+2.20 m/s) = -0.38 kg * (-1.70 - 2.20) m/s = -0.38 kg * (-3.90) m/s = 1.482 kg−m/s.

The average force is F_avg = 1.482 kg−m/s / 3.40 x 10^-3 s = 436.47 N.

The average net force exerted on the ball by the cushion is 436.47 N, directed away from the cushion since the ball rebounded after the collision.

Other Questions
9. Matter is anything thatMsO A. has mass and takes up space.O B. is important to human society.O C. has only one physical stateO D. contains two elements combined chemicallyMark for review (Will be highlighted on the review page)de Previous QuestionNeytas (a) Find the frequency of revolution of an electron with an energy of 114 eV in a uniform magnetic field of magnitude 46.7 T. (b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field. Electric Field of a Point Charge Suppose the electric field 0.500 meters from a positive point charge is 1.25 x 105 Newtons per Coulomb. What is the charge? 0 A. 1.67 ?C (-) B. 3.48 ?C C. 5.14 ?C ? D. 7.29 Even though most corporate bonds in the United States make coupon payments semiannually, bonds issued elsewhere often have annual coupon payments. Suppose a German company issues a bond with a par value of 1,000, 20 years to maturity, and a coupon rate of 7.8 percent paid annually. what is the current price of the bond? Is the answer 4/5 ? Can someone help please ? Dopamine is available as 400 mg in 250 mL of D5W. A 2 year old weighing 12 kg is receiving 10 mcg/kg/min. How many hours will the infusion last? Which chemical reaction model is flawed and why?A) 4C2+S8 -> 4CS2B) 2Fe+3Cl2 -> 2FeCl3C) 2Na+Cl2 -> 2NaClD) P4O10+6H2O -> 4H3PO4 I need help with this. Craigmont uses the allowance method to account for uncollectible accounts. Its year-end unadjusted trial balance shows Accounts Receivable of $146,500, allowance for doubtful accounts of $1,085 (credit) and sales of $1,135,000. If uncollectible accounts are estimated to be 5% of accounts receivable, what is the amount of the bad debts expense adjusting entry? Which term best describes Russia's transition to a market economy Solve the following system:y = x + 34x + y = 18A.(6,3)B.(3,6)C.(-3,6)D.(3,-6) A bag of marbles contains 3 yellow marbles, 4 blue marbles, and 8 red marbles. Which of the following is true of selecting a marble that is a primary color (red, yellow, or blue)? Select all that apply.The probability is This event is certain to happen.This event will not happen.The probability is 1.The probability is 0. During the Pleistocene glaciation, there have been numerous glacial and interglacial stages. About how long on average do these glacial stages reach their peak?Select one:a. 10,000 yearsb. 1,000,000 yearsc. 100,000 yearsd. 1000 yearsQuestion 68Which of the following glacial periods is believed to have occurred just prior to the explosion of life on the planet we call Earth?Select one:a. Karoob. Huronianc. Sturtiand. MarinoanQuestion 69Which one of the following statements concerning glacial deposits is not true?Select one:a. Tills are poorly sorted and the fragments are mostly angularb. Outwash is mainly stratified sand and gravel.c. Glacial erratics are blocks of rock that are too large for the glacier to move.d. Till is deposited directly from the ice; outwash is deposited by melt water streams.Question 70About how long ago did Antarctica begin to freeze over?Select one:a. About 1 billion years agob. Its always been frozenc. About 35 million years agod. 2 Million Years agoQuestion 71All of the following are thought to possibly contribute to the formation of glaciers except for ____________________.Select one:a. precessionb. eccentricityc. obliquityd. icy comets Which president did not live in the white house What does a notational analyst do?A. Helps people design exercise programs.B. Draws and analyzes blood.C. Analyzes sports games in order to help coaches plan out strategy.D. Helps people plan healthy diets. When a mass on a spring is at maximum displacement, what quantity is at a minimum? Which of the following pairs is not correct?a) Skeletal muscle : striatedb) Cardiac muscle : voluntaryc) Cardiac muscle : striatedd) Smooth muscle : involuntary Why was Thomas Aquilas important to the church need help with 5-8 , please help!!!!! Which best explains the inability for life in earths early atmosphere? Check all that apply