What is the best coefficient of performance for a refrigerator that cools an environment at -26.0°C and has heat transfer to another environment at 50.0°C?

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.

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

Answer:

55 ft/s

Explanation:

A₁ = Area of rectangular cross-section at input side = 1.5 x 11 = 16.5 ft²

A₂ = Area of rectangular cross-section at far end = 1.5 x 6 = 9 ft²

v₁ = speed of water at the input side of channel = 30 ft/s

v₂ = speed of water at the input side of channel = ?

Using equation of continuity

A₁ v₁ = A₂ v₂

(16.5) (30) = (9) v₂

v₂ = 55 ft/s

Answer:

L / 16

Explanation:

Mass = m, Radius = R, angular momentum = L

Now, new radius, R' = R/4, mass = m, angular momentum, L' = ?

By the law of conservation of angular momentum

If there is no external torque is applied, the angular momentum of the system remains conserved.

L = I x w

Moment of inertia I depends on the mass and the square of radius of the star.

If the radius is one fourth, the angular momentum becomes one sixteenth.

So, L' = L / 16

The weight of an 8 kg substance can be calculated in various units using the weight equation w = mg and the appropriate conversion factors. The weight is 78.4 N, 0.0784 kN, 78.4 kg·m/s², 8 kgf, 17.64 lbf, and 10.83 lbm·ft/s².

To calculate the weight of an object in different units, we need to use the equation for weight: w = mg, where m is the mass of the object and g is the acceleration due to gravity. In this case, the mass (m) of the substance is given as 8 kg, and the value of g on Earth is approximately 9.80 m/s².

Therefore, the weight of the substance in various units is:

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Answer:

The man ate eggs.

Explanation:

He should brush his teeth before seeing his girlfriend.

Answer:

Explanation:

Answer:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ      .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ  = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)²     .....(2)

T Cosθ = mg    .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ =  0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

d = 2 x 0.5 x 0.632 = 0.632 m

Answer:

[tex]F_{net} = 4.22 N[/tex]

Explanation:

Since charge 1 and charge 2 are positive in nature so here we will have repulsion type of force between them

It is given as

[tex]F_{12} = \frac{kq_1q_2}{r^2}[/tex]

[tex]F_{12} = \frac{(9\times 10^9)(5 \mu C)(24 \mu C)}{0.23^2 + 0.69^2}\frac{-0.23\hat i + 0.69 \hat j}{\sqrt{0.23^2 + 0.69^2}}[/tex]

[tex]F_{12} = 2.81(-0.23\hat i + 0.69\hat j)[/tex]

Since charge three is a negative charge so the force between charge 1 and charge 3 is attraction type of force

[tex]F_{13} = \frac{(9\times 10^9)(5 \mu C)(5 \mu C)}{0.27^2 + 0^2} (-\hat i)[/tex]

[tex]F_{13} = 3.1(- \hat i)[/tex]

Now we will have net force on charge 1 as

[tex]F_{net} = F_{12} + F_{13}[/tex]

[tex]F_{net} = (-0.65 \hat i + 1.94 \hat j) + (-3.1 \hat i)[/tex]

[tex]F_{net} = (-3.75 \hat i + 1.94 \hat j)[/tex]

now magnitude of total force on the charge is given as

[tex]F_{net} = 4.22 N[/tex]

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.

(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.

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, [tex]p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s[/tex]

After the collision, final momentum [tex]p_f=0.21\ kg\times 42\ m/s+0.046v[/tex]

Using the conservation of momentum as :

[tex]p_i=p_f[/tex]

[tex]11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v[/tex]

v = 63.91 m/s

So, the speed of the  golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

Answer:

The velocity of the block is 22 m/s.

Explanation:

Given that,

Mass = 2.50 kg

Velocity = 8 .00 m/s

Force = 7.00 N

Time t = 5.00

We need to calculate the change in velocity it means acceleration

Using newton's law

[tex]F = ma[/tex]

Where,

m = mass

a = acceleration

Put the value into the formula

[tex]a=\dfrac{F}{m}[/tex]

[tex]a = \dfrac{7.00}{2.50}[/tex]

[tex]a= 2.8m/s^2[/tex]

We need to calculate the velocity of the block

Using equation of motion

[tex]v = u+at[/tex]

Where,

v = final velocity

u = initial velocity

a = acceleration

t =time

Put the value in the equation

[tex]v= 8.00+2.8\times5.00[/tex]

[tex]v=22\ m/s[/tex]

Hence, The velocity of the block is 22 m/s.

The final velocity of the block after applying a force of 7.00 N for 5.00 s is approximately -6.00 m/s.

To calculate the velocity of the block after a force of 7.00 N directed to the left has been applied for 5.00 s, we can use Newton's second law of motion.

Newton's second law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration.

In this case, the mass of the block is given as 2.50 kg and the force applied is 7.00 N. We can calculate the acceleration using the formula:

acceleration = force/mass

Substituting the given values, we get:

acceleration = 7.00 N / 2.50 kg

Calculating, the acceleration is approximately 2.80 m/s² to the left. Since the block initially had a velocity of 8.00 m/s to the right, we subtract the acceleration from the initial velocity to get the final velocity:

final velocity = initial velocity - acceleration * time

Substituting the given values:

final velocity = 8.00 m/s - 2.80 m/s² * 5.00 s

Calculating, the final velocity is approximately 8.00 m/s - 14.00 m/s = -6.00 m/s.

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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.

Answer:

The speed is [tex]7.07\times10^{4}\ m/s[/tex]

Explanation:

Given that,

Speed of proton [tex]v= 10^{5}\ m/s[/tex]

Final potential = 10 v

Initial potential = 5 V

We need to calculate the speed

Using formula of energy

[tex]\dfrac{1}{2}mv^2=eV[/tex]

[tex]v^2=\dfrac{2eV}{m}[/tex]

The speed of the particle is directly proportional to the potential.

[tex]v^2\propto V[/tex]

Put the value into the formula

[tex](10^{5})\propto 10[/tex]....(I)

For 5 V,

[tex]v^2\propto 5[/tex].....(II)

From equation (I) and (II)

[tex]\dfrac{(10^{5})^2}{v^2}=\dfrac{10}{5}[/tex]

[tex]v=70710.67\ m/s[/tex]

[tex]v=7.07\times10^{4}\ m/s[/tex]

Hence, The speed is [tex]7.07\times10^{4}\ m/s[/tex]

The speed of the proton in the second place is 74.3 m/s.

To calculate the speed of the proton in the second place, first, we need to find the mass of the proton.

Using,

Where:

Make m the subject of the equation

Given:

Substitute these values into equation 2

Finally, to calculate the speed in the second place, we make v the subject of equation 1

Given:

Substitute these values into equation 3

Hence, The speed of the proton in second place is 74.3 m/s.

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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

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.

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.

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.

Answer:

so the distance between two points are

[tex]d = 0.246 \times 10^{-3} m[/tex]

Explanation:

Surface charge density of the charged plane is given as

[tex]\sigma = 7.2 \mu C/m^2[/tex]

now we have electric field due to charged planed is given as

[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]

now we have

[tex]E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}[/tex]

[tex]E = 4.07 \times 10^5 N/C[/tex]

now for the potential difference of 100 Volts we can have the relation as

[tex]E.d = \Delta V[/tex]

[tex]4.07 \times 10^5 (d) = 100[/tex]

[tex]d = \frac{100}{4.07 \times 10^5}[/tex]

[tex]d = 0.246 \times 10^{-3} m[/tex]

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

(a) [tex]-3.5 m/s^2[/tex]

The car's acceleration is given by

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

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

[tex]a=\frac{0-14}{4}=-3.5 m/s^2[/tex]

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

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

where in this case we have

[tex]a=-3.5 m/s^2[/tex] is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

[tex]t=\frac{v-u}{a}=\frac{0-20}{-3.5}=5.7 s[/tex]

(c) [tex]2.9 s[/tex]

As before, we can use the equation

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

Here we have

[tex]a=-3.5 m/s^2[/tex] is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find

[tex]t=\frac{v-u}{a}=\frac{10-20}{-3.5}=2.9 s[/tex]

(1) The acceleration of the car will be [tex]a=-3.5\frac{m}{s^2}[/tex]

(2) The time taken [tex]t=5.7s[/tex]

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s [tex]t=2.9s[/tex]

(1) The acceleration of the car will be calculated as

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

Here

u= 14 [tex]\frac{m}{s}[/tex]

[tex]a=\dfrac{0-14}{4} =-3.5\dfrac{m}{s^2}[/tex]

(2) The time is taken for the same acceleration to 20[tex]\frac{m}{s}[/tex]

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

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

u=20[tex]\frac{m}{s}[/tex]

[tex]t=\dfrac{0-20}{-3.5} =5.7s[/tex]

(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration

From same formula

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

v=10[tex]\frac{m}{s}[/tex]

u=20[tex]\frac{m}{s}[/tex]

[tex]t=\dfrac{10-20}{-3.5} =2.9s[/tex]

Thus

(1) The acceleration of the car will be [tex]a=-3.5\frac{m}{s^2}[/tex]

(2) The time taken [tex]t=5.7s[/tex]

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s [tex]t=2.9s[/tex]

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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

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.

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.

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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

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.

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.

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Well [tex]s=\dfrac{d}{t}[/tex] where s is speed, d is distance and t is time.

We have distance and time so we can calculate speed.

[tex]s=\dfrac{24}{37\cdot10^{-9}}\approx6.5\cdot10^8\frac{\mathbf{m}}{\mathbf{s}}\approx\boxed{6.5\cdot10^2\frac{\mathbf{Mm}}{\mathbf{s}}}[/tex]

Hope this helps.

r3t40

The acceleration of the object at t = 2.50 s in simple harmonic motion can be found using the equation a = -ω²x, where ω is the angular frequency and x is the displacement from the equilibrium position.

The acceleration of the object at t = 2.50 s can be found using the equation for simple harmonic motion:
a = -ω²x

where ω is the angular frequency and x is the displacement from the equilibrium position.

The period of the oscillation is related to the angular frequency by the equation:
T = 2π/ω

Substituting the given period (T = 4.60 s) into the equation and solving for ω, we get:
ω = 2π/T = 2π/4.60 s

Now, substituting the values we have, ω = 2π/4.60 s and x = 8.30 cm, into the acceleration equation:

a = -ω²x = -(2π/4.60 s)² * 8.30 cm

Calculate the value of a to find the acceleration of the object at t = 2.50 s using the given equation for acceleration.

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.

Answer:

Charge on least sphere, q = 570 nC

Explanation:

It is given that,

Two small plastic spheres are given positive electrical charges. The distance between the spheres, r = 30 cm = 0.3 m

The repulsive force acting on the spheres, F = 0.13 N

If one sphere has four times the charge of the other.

Let charge on other sphere is, q₁ = q. So, the charge on first sphere is, q₂ = 4 q. The electrostatic force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]0.13=9\times 10^9\times \dfrac{q\times 4q}{(0.3\ m)^2}[/tex]

[tex]q^2=\dfrac{0.13\times (0.3)^2}{9\times 10^9\times 4}[/tex]

[tex]q=5.7\times 10^{-7}\ C[/tex]

q = 570 nC

So, the charge on the least sphere is 570 nC. Hence, this is the required solution.

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

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.

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.

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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.