A long wire carries a current toward the south in a magnetic field that is directed vertically upward. What is the direction of the magnetic force on the wire?

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:

[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:

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:

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.

(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.

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

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:

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:

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

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.

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.

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].

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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.

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.

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

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:

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%

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

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:

The man ate eggs.

Explanation:

He should brush his teeth before seeing his girlfriend.

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

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.

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.

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]

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.

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.

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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:

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]

(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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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:

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:

  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.

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

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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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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