Answer:
The number of turns is 64.
Explanation:
Given that,
Magnetic field = 0.050 T
Area of coil = 100 cm²
Frequency = 60 Hz
Output voltage emf= 12 V
We need to calculate the number of turns
Using formula of induced emf
[tex]emf =NAB\omega[/tex]
[tex]N=\dfrac{emf}{A\times B\times2\pi f}[/tex]
[tex]N=\dfrac{12}{0.01\times0.050\times2\times3.14\times60}[/tex]
[tex]N =63.6 = 64\ turns[/tex]
Hence, The number of turns is 64.
Answer:
You need 63.66 turns.
Explanation:
The number of turns of a magnetic field is given by the following formula:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
In which N is the number of turns, V is the maximum output voltage, S is the area of the rotating coil, in square meters and T is the measure of the magnetic field and f is the frequency.
In this problem, we have that:
Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. This means that [tex]V = 12[/tex] and [tex]f = 60[/tex].
A uniform magnetic field of 0.050 T is available. This means that [tex]T = 0.050[/tex].
If the area of the rotating coil is 100 cm2, how many turns do you need?
This means that [tex]S = 0.01[/tex]m². So:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
[tex]N = \frac{12}{0.01*0.05*120\pi}[/tex]
[tex]N = 63.66[/tex]
You need 63.66 turns.
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.
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.
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.
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.
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.
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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?
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.
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
(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 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
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.
The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?
(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]
What will be the acceleration and time of the car?(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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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.
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.
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)
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.
Two small plastic spheres are given positive electrical charges. When they are 30.0 cm apart, the repulsive force between them has magnitude 0.130 N. If one sphere has four times the charge of the other, what is the charge of the least charged sphere? Give the answer in nanocoulomb (nC).
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.
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?
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.
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.
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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.
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.
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.
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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?
Answer:
The man ate eggs.
Explanation:
He should brush his teeth before seeing his girlfriend.
High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 42 m/s. Find the speed of the golf ball just after impact. m/s
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.
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.
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 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.
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
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?
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
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.)
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 new particle, the joelon, has just been discovered! Careful measurements show that the joelon has an average lifetime (at rest) of 37 ns. How fast must an average joelon be moving to travel 24 m (as viewed from the lab frame) before it decays?
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
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
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.
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.
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
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.
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.
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.
An infinite plane of charge has surface charge density 7.2 μC/m^2. How far apart are the equipotential surfaces whose potentials differ by 100 V?
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]
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.
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.
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.
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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?
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%
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%
A block of ice with a mass of 2.50 kg is moving on a frictionless, horizontal surface. At time t = 0, the block is moving to the right with a velocity of magnitude 8.00 m/s. Calculate the velocity of the block after a force of 7.00 N directed to the left has been applied for 5.00 s.
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.
Explanation: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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A star of uniform mass with a radius R rotates about its diameter with angular momentum L. Under the action of internal forces the star collapses until its radius is one fourth of its initial size. The magnitude of its new angular momentum is:
L
16L
L/16
4L
L/4
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
Charge 1 of +5 micro-coulombs is placed at the origin, charge 2 of +24 micro-coulombs is placed at x = +0.23 m, y = -0.69 m, charge 3 of -5 micro-coulombs is placed at x = -0.27 m, y = 0 m. What is the magnitude of the total electric force on charge 1 in Newtons?
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]
An irrigation channel has a rectangular cross section of 1.5 ft deep x 11 ft wide on the input side. On the far end of the channel, the channel expands to 6 ft wide while maintaining the same depth. If the water flowing into the channel has a speed of 30 ft/sec, calculate the velocity of the water flow on the far end of the channel.
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