If a mass on a spring has a frequency of 11 Hz, what is its period?

By using similar triangles concept and rates of change in Mathematics, we can find the man's shadow length is decreasing at a rate of 0.5 m/s when he is 4 meters away from the building.

The problem you're describing is typically solved using similar triangles concept in Mathematics and rates of change. As the man walks towards the building, his shadow becomes shorter. We can set up a ratio between the man's height and his distance from the building, and the length of the shadow and the wall. When the man is 4 meters away from the building, a right-triangle is formed with the man's height (2m), his distance from the building (4m), and his shadow's length.

Let s gives the length of the shadow, then the similar triangles imply 12/s = 2/4 which gives us s = 6 meters.

To find the rate at which the man's shadow is decreasing, we can differentiate the similar triangles ratio with respect to time to get -12/s^2 ds/dt =-1/2, sub in s = 6 into the equation to find ds/dt = -0.5 m/s.

So the length of the man's shadow on the building is decreasing at a rate of 0.5 m/s when he is 4 m from the building.

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Final answer:

To find how fast the man's shadow is decreasing, we use similar triangles and related rates in calculus. When the man is 4 m from the building, his 2 m tall shadow is shortening at 3.8 m/s, twice his walking speed of 1.9 m/s.

Explanation:

The question involves determining how fast the length of a man's shadow on a building is decreasing when he walks toward the building and away from the spotlight. This requires the application of similar triangles and the concept of related rates in calculus to solve. If we let the distance between the man and the building be x, and the length of the shadow be y, as the man walks towards the building, both x and y are changing with respect to time.

Given that the man is 2 m tall and the spotlight is 12 m away from the building, we can set up a proportion between the heights of the man and his shadow, and their respective distances from the spotlight. When he is 4 m from the building, the similar triangles can be represented as 2/y = (12-x)/(12). Differentiating both sides with respect to time t yields the related rate of change for the length of the shadow y when x is 4 m and the speed of the man is known, which is 1.9 m/s.

By solving this proportion and applying the derivative, we can find that the rate at which the length of his shadow decreases is exactly double the speed at which the man is walking, because in this scenario, the man and the tip of his shadow form a straight line with the light source. Thus, the length of the shadow decreases at a rate of 3.8 m/s when the man is 4 m from the building.

Answer:

555

Explanation:

The scenario is shown in the image attached below. A Right Angled Triangle is formed. We have an angle of 6 degrees, the side adjacent to angle which measures 1 mile and we need to find the side opposite to the angle.

Since 1 mile = 5280 feet, the side adjacent to the angle has a measure of 5280 feet.

Tangent ratio relates the opposite and adjacent side by following formula:

[tex]tan(\theta)=\frac{Opposite}{Adjacent}[/tex]

Using the given values, we get:

[tex]tan(6)=\frac{x}{5280}\\\\ x=tan(6) \times 5280\\\\ x = 555[/tex]

Thus, rounded to nearest foot, the height of the building is 555 feet

To find the height of the building, we can use trigonometry and the given angle of elevation. The height of the building is approximately 555 feet.

To find the height of the building, we can use trigonometry and the given angle of elevation. Let x be the height of the building. From the given information, we have the opposite side (x) and the adjacent side (1 mile = 5280 feet). The tangent function is used to relate these sides: tan(6°) = x/5280 feet. Solving for x, we get x = 5280 feet × tan(6°).

Using a calculator, we find that tan(6°) is approximately 0.1051. Multiplying this by 5280 feet gives us the height of the building in feet: 0.1051 × 5280 = 554.928 feet.

Rounding to the nearest foot, the height of the building is approximately 555 feet.

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Answer:

(a) 83.73 rad/s

(b) 251.2 m/s

(c) 3505.4 rad/s^2

(d) 5024 m

Explanation:

R = 50 cm = 0.5 m, f = 800 rpm = 800 / 60 rps

(a) Angular velocity, w = 2 x 3.14 x f = 2 x 3.14 x 800 / 60 = 83.73 rad / s

(b) The relation between linear speed and the angular speed is

V = r w

Here, r = 30 cm = 0.3 m

V = 0.3 x 83.73 = 25.12 m/s

(c) Radial acceleration = R w^2 = 0.5 x 83.73 x 83.73 = 3505.4 rad/s^2

(d) Time period T = 2 x 3.14 / w

T = 2 x 3.14 / 83.73 = 0.075 sec

In 0.075 second, angle turn = 360 degree

In 120 second, the angle turn = 360 x 120 / 0.075 = 576000 degree

In 360 degree, the distance traveled = 2 x pi x R

In 576000 degree, the distance traveled = 2 x 3.14 x 0.5 x 576000 / 360

                                                                     = 5024 m

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, [tex]L=50\ mH=0.05\ H[/tex]

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

[tex]X_L=2\pi fL[/tex]

[tex]X_L=2\pi \times 400\ Hz\times 0.05\ H[/tex]

[tex]X_L=125.66\ \Omega[/tex]

[tex]X_L=125.7\ \Omega[/tex]

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

To calculate the inductive reactance for a 50 mH inductor with a 400 Hz source, apply the formula XL = 2πfL, resulting in an inductive reactance of 125.6 ohms.

To determine the inductive reactance of a 50 mH inductor across a 15 volt, 400 Hz source, we use the formula for inductive reactance, which is:

L = 2πfL

where:

XL is the inductive reactance

f is the frequency of the AC source (400 Hz in this case)

L is the inductance of the coil (50 mH or 0.050 H)

Substituting the given values:

XL = 2π x 400 Hz x 0.050 H

XL = 125.6 Ω

Therefore, the inductive reactance is 125.6 ohms (Ω).

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

[tex]V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}[/tex]

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

[tex]V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}[/tex]

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

Answer:

There  will be no voltage across  secondary coil because DC voltage source is used in primary coil so there is no electromotive force induced in secondary coil.

Explanation:

In this question we have given

AA battery is used to supplies a constant voltage of 1.5 volts to primary coil of  transformer. In this case, voltage source is DC source which is providing constant voltage and for a transformer to work it is necessary to use an AC source.

Therefore, no EMF will induce in the secondary coil

and we know that in a transformer,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1} }=\frac{N_{2} }{N_{1} }[/tex]...........(1)

hence from above equation it is clear that,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1}}[/tex]............(2)

Here ,

[tex]E_{2}=0[/tex]

Put [tex]E_{2}=0[/tex] in equation (2)

We got

[tex]V_{2}=0[/tex]

There  will be no voltage across  secondary coil because there is no electromotive force induced in secondary coil.

Answer:

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

Explanation:

We start from:

[tex]\frac{dT}{dt}=\frac{-1}{50}(T-21)[/tex]

Separating variables:

[tex]-50dT=(T-21)dt[/tex]

[tex]-50\frac{dT}{T-21}=dt[/tex]

Integrating with initial conditions:

[tex]-50\int\limits^{T}_{91} {\frac{1}{T-21} } \, dT= \int\limits^t_{0} {} \, dt[/tex]

[tex]-50ln(\frac{T-21}{91-21})=t[/tex]

[tex]ln(\frac{T-21}{71})=\frac{-t}{50}[/tex]

Isolating T:

[tex]\frac{T-21}{70} =e^\frac{-t}{50} }[/tex]

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

You may note that when t is zero the temperature is 91 ºC, as is specified by the problem. As well, when t is bigger (close to infinite), the temperature tends to be the room temperature (21 ºC)

The given differential equation can be solved to yield the function y(t) = 70 e^(-t/50) + 21, which describes the temperature of the coffee as a function of time.

In your given differential equation, dy/dt = (− 1/50)(y − 21), y represents the temperature of the coffee at time t, and the equation describes how the temperature changes over time. This is a type of first-order linear differential equation, which can be solved using an integrating factor. The general solution of such equation is given by y(t) = [integral(t, e^(-t/50)*(-1/50)dt] + C e^(t/50), where C is a constant. To solve for C, we use the initial condition: at t = 0, y = 91°C, which yields C = (91 - 21), or C = 70. Substituting back into the original equation provides the final formula for the temperature of the coffee at given time t: y(t) = 70 e−t/50 + 21. It states that, the temperature of coffee decreases over time from its initial temperature, and it will eventually cool to the same temperature as the room (21°C).

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Answer:

Option B is the correct answer.

Explanation:

Period of simple pendulum is given by the expression [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex], where l is the length of pendulum and g is the acceleration due to gravity value.

So if we have a simple pendulum with length l we can find its period. Using the above formula we can calculate  acceleration due to gravity value of that place.

So using simple pendulum we can measure acceleration due to gravity value.

Option B is the correct answer.

Answer:

24.2 degree

Explanation:

n = 2.419

Let the minimum angle of incidence is i.

The value of minimum angle of incidence so that all the light reflects back into the diamond is the critical angle for air diamond interface.

The relation for the critical angle and the refractive index of diamond is given by

Sin ic = 1 / n

where, ic is the critical angle for air diamond interface and n be the refractive index for diamond.

Sin ic = 1 / 2.419 = 0.4133

ic = 24.4 degree

Thus, the value of minimum value of angle of incidence for which all the light reflects back into diamond is 24.4 degree.

Explanation:

A light ray is traveling in a diamond (n = 2.419). If the ray approaches the diamond-air interface, what is the minimum angle of incidence that will result ...

Answer:

Length of wire=2.16 m

Diameter of wire=0.232 mm

Explanation:

m= mass of copper wire= 0.82 g

R= Resistance of copper wire= 0.87 ohms

D= Density of copper= 8.96 g/cm^3

ρ= Resistivity= 1.7×10^-8 Ωm

[tex]Density=\frac{mass}{volume}\\\Rightarrow volume=\frac{mass}{density}\\\Rightarrow volume=\frac {0.82}{8.96}\\\Rightarrow volume=0.091\ cm^3\\ volume = \pi r^2 l\\\Rightarrow \pi r^2=\frac{volume}{l}\\ \Rightarrow \pi r^2=\frac {0.091}{l}\\[/tex]

[tex]\rho=R\frac{A}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{\pi r^2}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{0.091\times 10^{-6}}{l^2}\\\Rightarrow l^2=\frac {0.87\times 0.091\times 10^{-6}}{1.7\times 10^{-8}}\\\Rightarrow l^2=0.046\times 10^2\\\Rightarrow length=2.16\ m[/tex]

[tex]\pi r^2=\frac {0.091}{l}\\\Rightarrow r^2=\frac {0.091\times 10^{-6}}{2.16 \pi}\\\Rightarrow r^2=1.34\times 10^{-8}\\\Rightarrow r=0.00011\ m\\\Rightarrow d=0.000232\ m\\\therefore diameter=0.232\ mm[/tex]

Answer:

The change in momentum of an object is 1.6 kg-m/s.  

Explanation:

It is given that,

Mass of the object, m = 0.3 kg

Force acting on it, F = 8 N

Time taken, t = 0.2 s

We need to find the change in momentum of an object. The change in momentum is equal to the impulse imparted on an object i.e.

[tex]\Delta p=J=F.\Delta t[/tex]

[tex]\Delta p=8\ N\times 0.2\ s[/tex]

[tex]\Delta p=1.6\ kg-m/s[/tex]

So, the change in momentum of an object is 1.6 kg-m/s. Hence, this is the required solution.

The change in the momentum of the object is 1.6 kg-m/s.

Change in momentum is the product of force acting on a body and time

To calculate the change in momentum of the object, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The change in the momentum of the object is 1.6 kg-m/s.

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Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

         v² = u² + 2as

         v² = 44.50² + 2 x (-9.81) x 71.32

         v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

        v = u + at

         24.10 = 44.50 - 9.81 t , t = 2.07 s

         -24.10 = 44.50 - 9.81 t , t = 6.99 s      

Since the second ball throws after 2.3 seconds  we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

           s = ut + 0.5 at²

           71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

           u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.  

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.  

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

Answer:

The ball remain in air for 3.06 seconds.

Maximum height reached = 22.94 m

Explanation:

We have equation of motion v = u + at

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0 = 15 -9.81 t

        t = 1.53 s

Time of flight = 2 x 1.53 = 3.06 s

The ball remain in air for 3.06 seconds.

We also have equation of motion v² = u² + 2as

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0² = 15² - 2 x 9.81 x s

        s = 22.94 m

Maximum height reached = 22.94 m

Answer:

122241.02 J

Explanation:

Work Done = 230000 J

[tex]T_H=\text {High Temperature Reservoir}=830\ K\\T_L=\text {Low Temperature Reservoir}=288\ K\\\text{In case of Carnot Cycle}\\\text{Efficiency}=\eta\\\eta=1-\frac{T_L}{T_H}\\\Rightarrow \eta =1-\frac{288}{830}\\\Rightarrow \eta=0.65\\[/tex]

[tex]\eta=\frac{\text{Work}}{\text{Heat}}\\\Rightarrow \eta=\frac{W}{Q_H}\\\\\Rightarrow 0.65=\frac{230000}{Q_H}\\\Rightarrow Q_H=\frac{230000}{0.65}\\\Rightarrow Q_H=352214.02[/tex]

[tex]Q_L=Q_H-W\\\Rightarrow Q_L=352214.02-230000\\\Rightarrow Q_L=122241.02\ J[/tex]

∴Heat exhausted into the air to produce 230,000 J of work is 122241.02 Joule

Correct Answer:

122000 J

The period of the vertical block-spring system on the moon is approximately 2.45 seconds.

The period of a vertical block-spring system is determined by the square root of the ratio of the mass of the block to the spring constant. Since the mass and spring constant remain the same, the period will only be affected by the acceleration due to gravity.

On the moon, where the acceleration due to gravity is about 1/6 that of Earth, the period of the system will be sqrt(6), or approximately 2.45 seconds.

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Answer:

Distance ran by basketball player = 4.5 m

Explanation:

We have equation of motion, v = u+at

Substituting

           6 = 0 + a x 1.5

           a = 4m/s²

Now we need to find distance traveled, we have the equation of motion

           v² = u² + 2as

Substituting

           6² = 0² + 2 x 4 x s

           8 s = 36

              s = 4.5 m

Distance ran by basketball player = 4.5 m

Answer:57.7

Explanation:

Force on a moving charge in  a magnetic field is given by

[tex]F=q\times V\times Bsin\theta [/tex]

Where F=Force experienced by charge

q=charge of particle

V=velocity of particle

B=magnetic field

F=qVBsin(25)-------1

[tex]2F=qVBsin(\theta )------2[/tex]

Divide (1)&(2)

[tex]\frac{1}{2}[/tex]=[tex]\frac{sin(25)}{sin(\theta)}[/tex]

[tex]Sin(\theta )=2\times sin(25)[/tex]

[tex]\theta =57.7 ^{\circ}[/tex]

Given the constraints of the question and the formula for magnetic force on a moving charge, the situation described isn't possible.

The magnetic force experienced by a moving charged particle is given by the formula F = qvBsin(θ), where q is the charge, v is the speed, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the magnetic field. When the force is F at 25°, to experience a force of 2F, the sin of the new angle must be double that of sin(25°) because all other factors (charge, speed, magnetic field strength) are constant.

In the case given, sin(θ) would need to be 2sin(25°), but since the maximimum value sin(θ) can have is 1 (at 90°), this situation isn't possible within the constraints of the question (angle less than 90°). Hence, the student might have misunderstood the problem or there might be an error the question itself. However, if it is assumed that the question intends to find when the component of the force perpendicular to the field is double, the answer would be when θ is 90 degrees because the sin(90°) = 1. This is because when a charged particle moves perpendicular to a magnetic field, the sin(θ) factor in the force equation is maximized.

Final answer:

The required force is approximately 21.63 newtons.

Explanation:

The question involves determining the average force that a shooter must exert to keep a gun from moving while it fires bullets. To find this force, we can use the concept of impulse, which relates force, time, and change in momentum.

The gun fires 115 bullets per minute, which is 115/60 bullets per second. Multiplying the number of bullets per second by the mass and speed of each bullet gives the total momentum per second, which is the same as the force when mass is in kg and speed is in m/s.

First, we find the momentum of each bullet:

Bullet speed (v) = 342 m/s

F ≈ 21.63 N

The shooter must exert an average force of approximately 21.63 newtons to keep the gun from moving.

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

The resultant force acting on the object is (6.00 i + 15.00 j) N, and the magnitude of the resultant force is 16.16 N.

To find the resultant force acting on a 3.00-kg object with an acceleration given by a = (2.00 i + 5.00 j) m/s2, we use Newton's second law of motion (Force = mass  imes acceleration).

For the magnitude of the resultant force, we calculate it using the Pythagorean theorem:

Final answer:

The magnitude of the current that flows through the wire is approximately 16A.

Explanation:

To calculate the magnitude of the current that flows through the wire, we can use the formula for the magnetic field produced by a current-carrying circular loop: B = μ0 * I / (2 * R), where B is the magnetic field strength, μ0 is the permeability of free space, I is the current, and R is the radius of the loop.

Given that the radius of the loop is 3.0 mm and the magnetic field strength at the center of the loop is 1.1 mT, we can rearrange the formula to solve for I:

I = (2 * B * R) / μ0

Substituting the given values, we get:

I = (2 * 1.1 x 10^-3 T * 3.0 x 10^-3 m) / (4π x 10^-7 T.m/A)

Simplifying the expression, we find that the magnitude of the current is approximately 16 A. Therefore, the correct answer is B) 16A.

Answer:

A

Explanation:

Iron and gadlinium are both very easily made into magnetic substances.  Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

Answer:

v = 2.43 m/s

Explanation:

As we know that block is placed at rest at distance

d = 2 m

so here the centripetal force on the block to move in the circle is due to static friction force

now when block is just going to slide over the disc

then the friction force is maximum static friction which is given as

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

now we have

[tex]v = \sqrt{\mu_s Rg}[/tex]

now plug in the values in the above equation

[tex]v = \sqrt{0.3(2)(9.81)}[/tex]

[tex]v = 2.43 m/s[/tex]

To determine the maximum speed at which the block can attain before it begins to slip, calculate the force of static friction using the given coefficient and weight of the block. Then, use the equation fs = m(a+g) to determine the maximum speed.

To determine the maximum speed at which the block can attain before it begins to slip, we need to calculate the force of static friction. The formula for static friction is fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is the weight of the block, which is mg. The maximum force of static friction is therefore μs(mg). Since the block is on a horizontal surface, the normal force is equal to the weight of the block, and the maximum force of static friction is μs(mg). We can calculate μs using the given coefficient of static friction and determine the maximum speed using the equation fs = m(a+g), where fs is the force of static friction, m is the mass of the block, a is the acceleration, and g is the acceleration due to gravity.

Plug in the given values: ms = 0.3, m = (mass of the block), a = 2 m/s², and g = 9.8 m/s².

Then, solve for the maximum speed by rearranging the equation to isolate v: v = sqrt((fs - m*g)/m).

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Answer:

it will go up along the inclined plane by d = 9.62 m

Explanation:

As we know that puck is moving upward along the slide

then the net force opposite to its speed is given as

[tex]F_{net} = - mgsin\theta - \mu mgcos\theta[/tex]

so here deceleration is given as

[tex]a = -g(sin\theta + \mu cos\theta)[/tex]

now plug in all values in it

[tex]a = -9.81(sin20 + 0.2cos20)[/tex]

[tex]a = -5.2 m/s^2[/tex]

now the distance covered by the puck along the plane is given as

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 10^2 = 2(-5.2)d[/tex]

[tex]d = 9.62 m[/tex]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

[tex]V_{ab}[/tex] = i R

[tex]V_{ab}[/tex] = (0.1832) (65)

[tex]V_{ab}[/tex] = 11.91 Volts

(c)

Power dissipated in the resister R is given as

[tex]P_{R}[/tex] = i²R

[tex]P_{R}[/tex] = (0.1832)²(65)

[tex]P_{R}[/tex] = 2.18 Watt

Power dissipated in the internal resistance is given as

[tex]P_{r}[/tex] = i²r

[tex]P_{r}[/tex] = (0.1832)²(0.5)

[tex]P_{r}[/tex] = 0.0168 Watt

To compute the values in the question, one can utilize Ohm’s law and the power dissipation formula. The current in the circuit is 0.183 A, the terminal voltage of the battery is 11.908 V, the power dissipated in the 65.0 Ω resistor is 2.18 W, and the energy wasted in the battery's internal resistance is 0.0167 W.

To calculate the current in the circuit, we can use Ohm's law which states that the current is equal to the voltage divided by the resistance. This gives us: I = emf/(R+r) where I is the current, emf is the electromotive force of the battery, R is the load resistance and r is the internal resistance of the battery.

Substituting the values given in the question, we get: I = 12.0 V/(65.0 Ω + 0.5 Ω) = 0.183 A. Therefore, the current flowing through the circuit is 0.183 Amperes.

Now that we have the current, we can calculate the terminal voltage of the battery. The terminal voltage (Vab) is given by the equation: Vab = emf - Ir. Substituting the values, we get: Vab = 12.0V - (0.183A * 0.5Ω) = 11.908 V. Therefore, the terminal voltage of the battery is 11.908 Volts.

Lastly, we can calculate the power dissipated in the resistor R and in the battery’s internal resistance r using the formula: Power = I²R. For the resistor, power dissipation = (0.183A)² * 65.0Ω = 2.18 Watts, and for the battery's internal resistance, power dissipation = (0.183A)² * 0.5Ω = 0.0167 Watts.

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Final answer:

To calculate the power converted to thermal energy due to frictional effects, subtract the ideal mechanical power required to lift water to the higher reservoir (13.2435 kW) from the actual pump power (20 kW), yielding 6.7565 kW.

Explanation:

The student is asking to determine the mechanical power that is converted to thermal energy due to frictional effects when water is pumped from a lower reservoir to a higher reservoir. Given the shaft power of the pump is 20 kW, height difference is 45 m, and flow rate is 0.03 m³/s, we can first calculate the ideal mechanical power required to lift the water to that height.

The gravitational potential energy given to the water per second (which is the power) by pumping it to the height is calculated using the formula P = ρghQ, where ρ (rho) is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²), h is the height difference (45 m), and Q is the flow rate (0.03 m³/s). This results in P = 1000 kg/m³ * 9.81 m/s² * 45 m * 0.03 m³/s = 13243.5 W or 13.2435 kW.

The difference between the actual power supplied by the pump (20 kW) and the ideal power required (13.2435 kW) is the power lost to thermal energy due to friction. Therefore, the power converted to thermal energy is 20 kW - 13.2435 kW = 6.7565 kW.

Answer:

[tex]I(t)=4[1-e^{-2t}][/tex]

Explanation:

For a LR circuit as shown the current at any time t in the circuit is given by

[tex]I(t)=\frac{V}{R}[1-e^{\frac{-Rt}{L}}][/tex]

where

'V' is the voltage

'R' is resistance in the circuit

'L' is the inductance of the circuit

't' is time after circuit is turned on

Applying the given values we get

[tex]I(t)=\frac{32}{8}[1-e^{\frac{-8t}{4}}]\\\\I(t)=4[1-e^{-2t}][/tex]

Answer:

105.8 degree from + X axis

Explanation:

V1 = 46 N at 30 degree

V2 = 63 N at 151 degree

Write the vector form

V1 = 46 (Cos 30 i + Sin 30 j) = 39.84 i + 23 j

V2 = 63 (Cos 151 i + Sin 151 j) = - 55 i + 30.54 j

The resultant of V1 and V2 is given by

V = V1 + V2 = 39.84 i + 23 j - 55 i + 30.54 j = - 15.16 i + 53.54 j

The angle made by resultant of V1 and V2 with X axis is

tan∅ = 53.54 / (- 15.16) = - 3.53

∅ = 105.8 degree from + X axis.

Answer:

acceleration is 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

Explanation:

given data

uniform electric field E  = 54.0 ĵ V/m

uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T

velocity v  = 170 î m/s.

to find out

acceleration

solution

we know magnetic force for proton is

i.e = e (velocity × uniform magnetic field)

magnetic force = e (170 î × (0.200 î + 0.300 ĵ + 0.400 k) )

magnetic force = e (- 68 ĵ + 51 k  )   ..................1

and now for electric force for proton i.e

= uniform electric field × e ĵ

electric force = e (54 ĵ )   ............2

so net force will be  add magnetic force + electric force

from equation 1 and 2

e (- 68 ĵ + 51 k  )  + e (54 ĵ )

e (- 14 ĵ + 51 k  )

so the acceleration (a)  for proton will be

net force = mass × acceleration

a = e (- 14 ĵ + 51 k  )  / 1.6 [tex]10^{-19}[/tex] / 1.67 × [tex]10^{-27}[/tex]

acceleration = 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

The acceleration of the proton can be determined using the equation F = qvBsinθ. Plug in the given values to calculate the acceleration.

The acceleration of the proton can be determined using the equation F = qvBsinθ, where F is the force, q is the charge of the proton, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since the question does not provide the angle, we assume that θ = 0. Therefore, the acceleration of the proton is given by:

a = (q * v * B) / m

where m is the mass of the proton. Plugging in the given values, we can calculate the acceleration.

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Answer:

313.6 m downward

Explanation:

The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.

In fact, we have:

[tex]y(t) = h +u_y t + \frac{1}{2}at^2[/tex]

where

y(t) is the vertical position of the projectile at time t

h is the initial height of the projectile

[tex]u_y = 0[/tex] is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally

t is the time

a = g = -9.8 m/s^2 is the acceleration due to gravity

We can rewrite the equation as

[tex]y(t)-h = \frac{1}{2}gt^2[/tex]

where the term on the left, [tex]y(t)-h[/tex], represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we  find

[tex]y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m[/tex]

So the bullet has travelled 313.6 m downward.