What factors does the kinetic energy of a body depend
on?
Ans​

Answer:

E(x,t) = Emaxcos(kx - ωt + φ),

B(x,t) = Bmaxcos(kx - ωt + φ).

Explanation:

E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves the electric field E and the magnetic field B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E x B.

Answer:

The answer is d. Twice as great.

Explanation:

The German physicist and mathematician Georg Simon Ohm says in his basic law of electrical circuits or Ohm's law that the potential difference V that is applied at the ends of a conductor is proportional to the intensity I of the current that circulates and that the electrical resistance R is the ratio factor between I and V.

The equation would be I = V / R

If by any chance R is reduced because it is inversely proportional, the current would increase.

So if the current goes from 10 to half 5, its current would double.

Answer:

(1). 9.35 m/s.

(2).V2 = 3.22 m/s to the left.

Explanation:

So, we are given the following data or parameters or infomation in the question above as;

(1). "open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. "

(2). "A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal."

(3). "The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds."

(4). "The package comes to a stop in the cart after 4 seconds."

(a). So, in order to solve this question, we will be making use of the equations below;

(1). Gravitational potential energy = mass × acceleration due to gravity × height.

(2). Kinetic energy = 1/2 × mass × (velocity)^2.

(3). Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).

So, starting from equation (1) above;

Gravitational potential energy = mass × acceleration due to gravity × height.

Gravitational potential energy = 15 × 9.8 × 4 = 588 J.

Kinetic energy = 1/2 × mass × (velocity)^2. = 1/2 × 15 × (3)^2.= 67.5 J.

kinetic energy (final) = 1/2 × 15 × v^2. = 7.5v^2.

Hence, Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).

[potential energy (final) = 0].

67.5 + 588 = 7.5v^2.

v = 9.35 m/s

(b). -3 cos (27°) = -2.67 m/s.

15 × -2.67 + 50 × 5 =( 15 + 50) × v2.

=> 209.5 = 65v2.

=> V2 = 3.22 m/s to the left.

(1) The speed of the package before it lands on the cart is 9.35 m/s.

(2) The final speed of the cart is 3.22 m/s

The gravitational potential energy of a system is given by

PE = mgh

where m is the mass

g is the acceleration due to gravity

and h is the height.

The kinetic energy is given by:

KE = (1/2) mv²

According to the law of conservation of energy:

Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).

The final potential energy will be zero since taking the cart as the reference as ground.

PE(initial) = 15 × 9.8 × 4 = 588 J

KE(initial) = (1/2) mv² = 1/2 × 15 × 3² = 67.5 J

KE(final) = 1/2 × 15 × v² = 7.5v²

So,

67.5 + 588 = 7.5v²

v = 9.35 m/s

(b) The package leave at an angle of 27° down the horizontal with velocity 3 m/s, so the horizontal speed is:

v ' = -3 cos (27°) = -2.67 m/s.

Negative sign, as the package is moving opposite to the cart

After the package lands on the cart, both move together, so from conservation of momentum:

15 × (-2.67) + 50 × 5 =( 15 + 50) × V

where V is the speed of cart and package together

209.5 = 65 V

V = 3.22 m/s to the left.

The question was incomplete. After searching it, it is likely that it was as given below:

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 27∘ from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what is (1) the speed of the package just before it lands in the cart and (2) the final speed of the cart?

Learn more about the conservation of energy:

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Answer:

Do neither of these things ( c )

Explanation:

For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula

L = (l) [tex]\sqrt{1 -\frac{v^{2} }{c^{2} } }[/tex]

where l = Measured distance from object at rest, L =  contracted measured in relation to the observer , v = speed of clock , c = speed of light

you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law,  PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

[tex]V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}[/tex]

Given;

[tex]\frac{dP}{dt}[/tex] (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) =  600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( [tex]\frac{dv}{dt}[/tex]);

(600 x 40) + (150 x [tex]\frac{dv}{dt}[/tex]) = 0

[tex]\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min[/tex]

Therefore, the volume is decreasing at 160 cm³/min

Final answer:

Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C. To find the rate at which the volume is decreasing at a given instant, we differentiate the equation PV = C with respect to time and substitute the given values.

Explanation:

Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C, where P is the pressure, V is the volume, and C is a constant. In this scenario, the pressure is increasing at a rate of 40 kPa/min. To find the rate at which the volume is decreasing at this instant, we differentiate the equation PV = C with respect to time. Applying the chain rule, we get:

d(PV)/dt = d(C)/dt

P(dV/dt) + V(dP/dt) = 0

dV/dt = - (V/P)(dP/dt)

We substitute the given values where the volume V = 600 cm3 and pressure P = 150 kPa:

dV/dt = - (600 cm3 / 150 kPa) (40 kPa/min)

dV/dt = -160 cm3/min

Therefore, the volume is decreasing at a rate of 160 cm3/min at this instant.

Answer:

Reactants: Fe, [tex]H_2O[/tex] and [tex]O_2[/tex]

Product: [tex]Fe_2O_3.H_2O[/tex]

Explanation:

Two electrochemical reactions occur during the formation of rust;

The two products from the two electrochemical equations then react together to form iron hydroxide.

[tex]2Fe^{2+} + 4OH^- --> 2Fe(OH)_2[/tex]

The iron hydroxide thus produced slowly reacts with oxygen to form rust which is [tex]Fe_2O_3.H_2O[/tex].

Hence, the reactants are iron (Fe), water ([tex]H_2O[/tex]) and oxygen ([tex]O_2[/tex]) while the product is rust ([tex]Fe_2O_3.H_2O[/tex]).

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

The new angular velocity is computed using the kinematic expression w² = wo² + 2a0, where 'wo' is the original angular velocity 'a' is the angular acceleration and 'w' is the new angular velocity. The calculation is guided by principles of physics primarily involving angular momentum and acceleration.

The calculation for the new angular velocity can be gleaned from the kinematic expression w² = wo² + 2a0. The formula described represents the angular motion of the frame. The angular velocity originally is 20 rad/s and a couple MO of 30 N.m is applied to the frame for 5 seconds. The development of this concept involves principals of angular momentum as well as angular acceleration.

The angular acceleration can be calculated using the relation a = nett

Bearing in mind the initial angular velocity, the applied couple MO, and the duration of its application, the new angular velocity can be computed using the given formula. Note that the laws of physics, specifically the law of conservation of angular momentum, play an essential part in this calculation.

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Answer:

The phase difference is       [tex]\Delta \phi = 1.9995 rad[/tex]  

Explanation:

From the question we are told that

    The distance between the  loudspeakers is [tex]d = 2m[/tex]

     The distance of the listener from the wall  [tex]D = 81.7 \ m[/tex]

     The frequency of the  loudspeakers is  [tex]f = 4450Hz[/tex]

      The velocity of sound is [tex]v_s = 343 m/s[/tex]

The path difference of the sound wave that is getting to the listener is mathematically represented as

        [tex]\Delta z =\sqrt{d^2 + D^2} -D[/tex]

Substituting values

        [tex]\Delta z =\sqrt{2^2 + 81.7^2 } -81.7[/tex]

       [tex]\Delta z =0.0245m[/tex]

The phase difference is mathematically represented as

           [tex]\Delta \phi[/tex] =  [tex]\frac{2 \pi}{\lambda } * \Delta z[/tex]

Where [tex]\lambda[/tex] is the wavelength which is mathematically represented as

          [tex]\lambda = \frac{v_s }{f}[/tex]

substituting value  

          [tex]\lambda = \frac{343 }{4450}[/tex]

        [tex]\lambda = 0.0770 m[/tex]

Substituting value into the  equation for phase difference

      [tex]\Delta \phi[/tex] =  [tex]\frac{2 * 3.142 * 0.0245}{0.0770}[/tex]

      [tex]\Delta \phi = 1.9995 rad[/tex]  

Answer:a

Explanation:

Given

[tex][H^+]=5.3\times 10^{-4}\ M[/tex]

and [tex]pH+pOH=14[/tex]

Also [tex]pH=-\log [H^+][/tex]

therefore [tex]pH=-\log (5.3\times 10^{-3})[/tex]

[tex]pH=-(-4-\log (5.3))[/tex]

[tex]pH=3.275[/tex]

Thus [tex]pOH=14-3.275[/tex]

[tex]pOH=10.275[/tex]

and [tex]pOH=-\log [OH^{-}][/tex]

[tex][OH]^{-1}=10^{-10.25}[/tex]

[tex][OH]^{-1}=1.88\times 10^{-11}\ M[/tex]

As the pH is less than 7 therefore solution is acidic

Answer:4.5 joules

Explanation:

charge=3c

Potential difference=1.5v

Work=potential difference x charge

Work=1.5 x 3

Work=4.5

Work=4.5 joules

Answer:[tex]v=5.529\ m/s[/tex]

Explanation:

Given

number is given as [tex]=\frac{v^2}{gl}[/tex]

and [tex]number=2.6[/tex]

[tex]l=1.2\ m[/tex]

g=acceleration due to gravity[tex](9.8\ m/s^2)[/tex]

calculating for [tex]v[/tex]

[tex]v^2=gl(2.6)[/tex]

[tex]v=\sqrt{2.6\times 9.8\times 1.2}[/tex]

[tex]v=\sqrt{30.576}[/tex]

[tex]v=5.529\ m/s[/tex]

Answer:

potential difference between the defibrillator pads = 2000 V

Explanation:

We are given;

Current;I = 20A

Time;t = 5 m.s = 5 x 10^(-3) s

Electrical energy;W = 200 J

Now let's find the value of the electric charge from the formula;.

q = It

Where q is charge, I is current and t is time.

Thus; q = 20 x 5 x 10^(-3)

q = 0.1 C

Now, the potential difference between the defibrillator pads would be calculated from;

V = W/q

Where V is potential difference and W and q are as stated earlier.

Plugging in the relevant values gotten earlier, we have;

V = 200/0.1

V = 2000 V

Given Information:  

Number of turns = N = 52

Diameter of coil = d = 12 cm = 0.12 m

Time = t = 0.10 seconds

Magnetic field = B = 0.30 T  

Required Information:  

Induced electric field = E = ?  

Answer:

Induced electric field = E = 4.68 V/m

Explanation:

The Maxwell's third equation can be used to find out the induced electric field,

∫E.dl = -dΦ/dt  

Where E is the induced electric field, dl is the circumference of the loop and dΦ/dt  is the rate of change of magnetic flux and is given by

Φ = NABcos(θ)

Where N is the number of turns, A is the area of coil and B is the magnetic field and cos(θ) = 1

Φ = NAB

∫E.dl = -dΦ/dt  

E(2πr) = -d(NAB)/dt

E =1/(2πr)*-d(NAB)/dt

E =NA/(2πr)*-dB/dt

Area is given by

A = πr²

E =Nπr²/(2πr)*-dB/dt

E =Nr/2*-dB/dt

The magnetic field reduce from 0.30 to zero in 0.10 seconds

E =Nr/2*-(0.30 - 0)/(0 - 0.10)

E =Nr/2*-(0.30)/(-0.10)

E = Nr/2*-(-3)

The radius r is given by

r = d/2 = 0.12/2 = 0.06 m

E = (52*0.06)/2*(3)

E = 1.56*3

E = 1.56*3

E = 4.68 V/m

Therefore, the induced electric field in the coil is 4.68 V/m

Answer:

Explanation:

The original frequency of sound  f₀

The apparent frequency of sound fa

For apparent frequency the formula is

fa =  [tex]f_0\times\frac{V+v}{V-v }[/tex]     , v is velocity of jumper which increases as he goes down .

Beat frequency

= fa - f₀

=  [tex]f_0\times(\frac{V+v}{V-v }-1)[/tex]

=  [tex]f_0\times(\frac{2v}{V-v })[/tex]

since v is very small in comparison to V , velocity of sound , in the denominator , v can be neglected.

beat frequency = [tex]f_0\times(\frac{2v}{V })[/tex]

v , the velocity of jumper will go on increasing as long as net force on the jumper is positive or

mg > kx where x is extension in the cord and k is its force constant . Below this point kx or restoring force becomes more than weight of the jumper and then net force on the jumper directs upwards. At this point beat frequency becomes maximum.

Final answer:

The fraction of hydrogen atoms in the 2P states at T=5900 K is calculated using the Boltzmann distribution. The energy of the 2S and 2P states is 10.2 eV, and their degeneracies are factored into the Boltzmann factor to determine the relative populations of these states.

Explanation:

The fraction of hydrogen atoms in the 2P states at a temperature of 5900 K can be calculated using the Boltzmann distribution. According to quantum mechanics, the energy levels of a hydrogen atom involve principal quantum numbers, with the ground state being 1s and higher energy excited states being designated by higher quantum numbers and corresponding letters (s, p, d, f, etc.) for their angular momentum quantum numbers. Since we are given that the energy of 2S and 2P states are the same at 10.2 eV and the ground state (1S) has an energy we'll call 0, we can use the Boltzmann factor to find the relative populations of these states.

To calculate the fraction of hydrogen atoms in the 2P state, use the following Boltzmann factor equation: fraction = (g2P / gTotal) * exp(-E2/kT), where g2P is the degeneracy of the 2P state, gTotal is the total degeneracy of all states considered, E2 is the energy of the 2P state, k is the Boltzmann constant, and T is the temperature. The degeneracy of the 2P state is 3 (since there are three 2P states) and the only other state considered here is the 2S state, which has degeneracy 1, making gTotal = 4. Plugging in the energy of 10.2 eV for E2 and converting it to joules (multiply by 1.602 x 10-19 J/eV), using k = 1.38 x 10-23 J/K, and T = 5900 K, we can calculate the fraction of hydrogen in the 2P state.

Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

Answer:

The force that is exerted when a shopping cart is pushed:

-Contact

The force that causes a metal ball to move toward a magnet:

-Noncontact

Answer:

Check the explanation

Explanation:

From given data, it can be noted that 95% of given confidently data, means 5% of data is uncertain. According to the question, we have to calculate uncertainty in Cp .

 Kindly check the attached image below for the step by step explanation to the question above.

Complete question;

A plane circular loop of conducting wire of radius r-10 cm which possesses 15 turns is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength is increased at a constant rate from IT to 5T in a time interval of 10 s.

a) What is the emf generated around the loop?

b) If the electrical resistance of the loop is 15Ω, what current flows around the loop as the magnetic field is increased?

Answer:

A) E.M.F generated around loop = 0.163 V

B)Current in loop; I = 0.011 A

Explanation:

A) We are given;

Initial magnetic field strength;B1 = 1T

Final magnetic field strength;B2 = 5T

Number of turns;N = 15 turns

Radius; r = 10cm = 0.1m

Angle;θ = 30°

Time interval; Δt = 10 s

Now, the formula for magnetic flux is: Φ = NABcosθ

Where;

N is number of turns

A is area = πr²

B is magnetic field strength

θ is angle

So, initial magnetic flux is;

Φ1 = NA(B1)cosθ

Plugging in the relevant values to obtain;

Φ1 = 15*(π*0.1²)(1)cos30

Φ1 = 0.4081 Wb

Similarly, final magnetic flux is;

Φ2 = NA(B2)cosθ

Plugging in the relevant values to obtain;

Φ2 = 15*(π*0.1²)(5)cos30

Φ2 = 2.0405 Wb

The time rate of change of the flux is;

dΦ_B/dt = (Φ2 - Φ1)/Δt

So, dΦ_B/dt = (2.0405 - 0.4081)/10

dΦ_B/dt = 0.163 Wb/s

Thus, the emf generated around the loop is; E = dΦ_B/dt = 0.163 V

B) from Ohm's law, the current which flows around the loop in response to the emf is given as;

I = E/R

We are given R =15Ω

Thus; I = 0.1632/15

I = 0.011 A

Answer:

Total resistance=33.3 ohms

Current=0.36 amperes

Explanation:

Voltage=12v

R1=50 ohms

R2=50 ohms

R3=50 ohms

Two connected in series:

R1+R2=50+50=100 ohms

In parallel

1/R=1/50 + 1/100

1/R=(2+1)/100

1/R=3/100

Cross multiplying we get

100=3R

Divide both sides by3

100/3=3R/3

33.3=R

Total resistance=33.3 ohms

Current=voltage ➗ resistance

Current=12 ➗ 33.3

Current=0.36 amperes

Answer:

Explanation:

Coefficient of static friction(μs) = 0.20 Coefficient of kinetic friction (μk) = 0.16

Mass of block (m) = 59kg

acceleration due to gravity (g) = 9.8m/s^2

Normal force = weight of block = mg

Weight of block(W) = 59kg × 9.8m/s^2 = 578.2N

N = 578.2 N

Maximum force of static friction Fs(max) is given by ;

Coefficient of static friction (μs) × N

0.20 × 578.2N = 115.64

As long as P is < 115.64, then the block will remain stationary and thus static frictional force = P (Fs = P)

magnitude and direction of the friction force exerted by the surface on the block if;

(A.) P = 0, Fs = 0

B.) P = 198N;

Fk = μk × N = 0.16 × 578.2 = 92.512N

Fs =

C.) P = 288;

Fk = μk × N = 0.16 × 578.2 =

D.) P value required to initiate motion;

0.2 × 59kg × 9.8m/s^2 = 115.64N

The correct answer is option (a) 0 N (indeterminate direction), (b) 198 N (negative direction, down the incline), (c) 92.6064 N (negative direction, down the incline) and (d) The value of [tex]\( P \)[/tex] required to initiate motion up the incline is 115.758 N.

Let's solve the problem step by step for each case:

(a) When [tex]\( P = 0 \)[/tex]:

The magnitude of the static friction force is equal to the applied force required to start motion, which is the maximum static friction force. This force is given by [tex]\( f_{s,\text{max}} = \mu_s N \)[/tex], where [tex]\( N = mg \)[/tex].

Given m = 59 kg and g = 9.81 m/s², we have:

[tex]\( N = 59 \text{ kg} \times 9.81 \text{ m/s}^2 \)[/tex]

[tex]\( N = 578.79 \text{ N} \)[/tex]

The maximum static friction force is:

[tex]\( f_{s,\text{max}} = 0.20 \times 578.79 \text{ N} \)[/tex]

[tex]\( f_{s,\text{max}} = 115.758 \text{ N} \)[/tex]

Since the block is at rest and no force is applied, the frictional force is exactly what is needed to balance any other forces acting on the block, which in this case is zero.

(b) When [tex]\( P = 198 \)[/tex] N:

The frictional force will exactly oppose the applied force [tex]\( P \)[/tex], so, the magnitude of the frictional force is 198 N, and it acts in the direction opposite to [tex]\( P \)[/tex], which is negative (down the incline).

(c) When [tex]\( P = 288 \)[/tex] N:

If the applied force [tex]\( P \)[/tex] exceeds the maximum static friction force, the block will start to move, and the frictional force will be kinetic. The magnitude of the kinetic friction force is given by [tex]\( f_k = \mu_k N \)[/tex].

[tex]\( f_k = 0.16 \times 578.79 \text{ N} \)[/tex]

[tex]\( f_k = 92.6064 \text{ N} \)[/tex]

(d) To initiate motion up the incline:

To move the block up the incline, the applied force [tex]\( P \)[/tex] must be equal to the maximum static friction force plus any additional force to overcome other forces such as gravity along the incline (which is not provided in the problem).

[tex]\( P = f_{s,\text{max}} \)[/tex]

[tex]\( P = 115.758 \text{ N} \)[/tex]

The angular speed of the merry-go-round would increase when the boy jumps off due to the conservation of angular momentum. However, without specific values like the weights and distances involved, the precise speed cannot be calculated.

The speed of the merry-go-round after the boy jumps off cannot be determined from the given information. However, we can use the principle of conservation of angular momentum to analyze the situation; this principle states that the initial angular momentum (while the boy was still on the merry-go-round) should be equal to the final angular momentum (after the boy jumps off). The angular momentum is the product of moment of inertia and angular speed. When the boy was sitting on the edge of the merry-go-round, he had a certain moment of inertia due to his mass and the distance from the center. Once he jumps off, the moment of inertia decreases, hence according to the conservation law, the angular speed must increase to keep the momentum constant.

However, the exact increase in angular speed could only be calculated if we knew the weight of the boy and of the merry-go-round, as well as the radius of the merry-go-round (to calculate moments of inertia), and the initial angular speed. Another simplification made in this situation is that the friction between the merry-go-round and its axle is negligible, which might not be true in real life scenarios.

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When the boy jumps off the merry-go-round, the total angular momentum of the system should conserve. As the boy's leaving reduces the moment of inertia, the angular speed of the merry-go-round must increase to preserve angular momentum. To calculate the exact increase, we would need specific values for the moment of inertia before and after the jump.

The question asked is associated with angular velocity and moment of inertia in the context of a merry-go-round. As the boy jumps off the merry-go-round, the system's total angular momentum must be conserved, because there are no external torques acting on it.

When the boy was still on the merry-go-round, he was part of the system, contributing to the total angular momentum. But when he departs, he no longer contributes to the system's moment of inertia, effectively reducing it. As a result, the angular speed of the merry-go-round must increase to conserve angular momentum (since angular momentum = moment of inertia x angular speed).

Unfortunately, without specific values for the moment of inertia both before and after the boy jumps off, we can't calculate the exact increase in angular speed. However, the key concept is the conservation of angular momentum: when the moment of inertia decreases, the angular speed must increase to maintain constant total angular momentum.

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Answer:u=-22 m/s

Explanation:

Given

mass of puck [tex]m=0.06\ kg[/tex]

Average force [tex]f_{avg}=1.5\times 10^3\ N[/tex]

time of contact [tex]t=1.2ms=1.2\times 10^{-3}\ s[/tex]

puck leaves with a velocity of [tex]v=8\ m/s[/tex]

We know impulse is [tex]F_{avg}\Delta t[/tex][tex]=\text{change in momentum}[/tex]

therefore

[tex]1.5\times 10^3\times (1.2\times 10^{-3})=P_f-P_i[/tex]

[tex]P_i=0.06\times 8-1.8[/tex]

[tex]P_i=0.48-1.8=-1.32\ kg-m/s[/tex]

Final momentum [tex]P_f=m\times v_f[/tex]

[tex]P_f=0.06\times 8[/tex]

[tex]P_f=0.48\ kg-m/s[/tex]

Impulse on the ball [tex]=F_{avg}\Delta t[/tex]

Impulse[tex]=1.5\times 10^3\times 1.2\times 10^{-3}=1.8\ N-s[/tex]

Initial velocity is given by

[tex]u=\frac{P_i}{m}=\frac{-1.32}{0.06}[/tex]

[tex]u=-22\ m/s[/tex]

i.e. initially ball is moving towards -x-axis

Answer:

mass m= 100 g or .1 kg

volume V= 50 cm3 or 50×[tex]10^{-6[/tex] m3

density ρ= [tex]\frac{m}{V}[/tex]

So,

ρ=[tex]\frac{.1}{50*10^{-6} }[/tex]= 2000 kgm-3 or 2 gcm-3

Explanation:

The density is calculated as 2 g / cm³.

To calculate the density, mass = 100 g

volume = 50 cm ³

      The principle of density was discovered by the Greek scientist Archimedes.

The mass of a unit volume of a material substance.

   Then, formula for density

                              ρ = M/V,

where d is density, M is mass, and V is volume.

Unit of density is grams per cubic centimeter (g / cm³) .

Substituting all the values,

                            ρ = 100 / 50

                               = 2

Hence, the density of an object is ρ = 2 g / cm³.

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Answer:

a) 20.81 J

b) 8.29 J

Explanation:

V = iR + L di/dt

where

i = a(1-e^-kt)

for large t

i = V/R

i = 24 / 9.4

i = 2.55 A

so

i = 2.55(1-e^-kt)

di/dt = 2.55 k e^-kt

24 = 24-24e^-kt + 6.4(2.55)k e^-kt

24 = 6.4(2.55) k

k = 24 / (6.4 * 2.55)

k = 24 / 16.32

k = 1.47 = R/L

so

i = 2.55(1-e^-(Rt/L))

current is maximum at great t

i max = 2.55 - 0

energy = (1/2) L i^2

E = (1/2)(6.4)2.55^2

E = 20.81 Joules

one time constant T = L/R and e^-(Rt/L) = 1/e = .368

i = 2.55 (1 - 0.368)

i = 2.55 * 0.632

i = 1.61 amps

energy = (1/2)(6.4)1.61^2

E = 8.29 Joules

The energy stored in the inductor when the current reaches its maximum value is 20.80 J. After one time constant, the energy stored is approximately 8.25 J.

we need to determine the energy stored in the inductor at two different points in time.

a) The energy stored in an inductor is given by the formula:

U = (1/2) × L × I²

Maximum Current

In an RL circuit, the maximum current is achieved when the circuit reaches a steady state. The maximum current is given by:

I_max = V/R, where V is the voltage and R is the resistance.

Given V = 24.0 V and R = 9.40 Ω, we have:

I_max = 24.0 V / 9.40 Ω = 2.55 A

Now, the energy stored in the inductor at the maximum current is:

U_max = (1/2) × L × I_max²

Given L = 6.40 H, we get:

U_max = (1/2) × 6.40 H × (2.55 A)² = 20.80 J

b)   Energy After One Time Constant

The time constant for an RL circuit is defined as:

τ = L/R

Given L = 6.40 H and R = 9.40 Ω, we calculate:

τ = 6.40 H / 9.40 Ω ≈ 0.68 s

At one time constant, the current has reached approximately 63.2% of its maximum value:

I_τ ≈ 0.632 × I_max = 0.632 × 2.55 A ≈ 1.61 A

The energy stored in the inductor at this time is:

U_τ = (1/2) × L × I_τ² = (1/2) × 6.40 H × (1.61 A)² ≈ 8.25 J

complete question:

A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively.

(a) Find the energy stored in the inductor when the current reaches its maximum value. J

(b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch  is closed. j

Answer:

The answer is oatmeal.

Explanation:

Since the oatmeal is the hottest substance, it will spread heat to the other substances.

Answer:

C, oatmeal

Explanation:

Answer:

force is  -y direction

Explanation:

The magnetic force is given by the equation

        F = q v x B

where bold letters indicate vectors.

The direction of this force can be found by the right-hand rule, where the thumb points in the direction of speed, the other fingers extend in the direction of the magnetic field and the palm is in the direction of force for a positive charge

let's apply this to our case

the speed is in the direction + x the thumb

the magnetic field in the + z direction fingers extended

the palm is in the direction - y

force is  -y direction

Final answer:

The force due to the magnetic field on a positively charged particle moving in the +x direction and perpendicular to the magnetic field will be in the +-y direction.

Explanation:

At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If q is positive, the direction of the force on the particle due to the magnetic field will be in the +-y direction.

Answer:

Explanation:

Mirror formation in mirror A

focal length f = - 10 cm ( concave mirror )

Let u be object distance , v is image distance .

In case image is virtual

v / u = 2

v = 2u

Mirror formula

[tex]\frac{1}{v} +\frac{1}{u} =\frac{1}{f}[/tex]

Putting the values

[tex]-\frac{1}{2u} +\frac{1}{u} =-\frac{1}{10}[/tex]

[tex]\frac{1}{2u}[/tex] = [tex]-\frac{1}{10}[/tex]

u = -5 cm

In case image is real

[tex]\frac{1}{v} +\frac{1}{u} =\frac{1}{f}[/tex]

[tex]\frac{1}{2u} +\frac{1}{u} =-\frac{1}{10}[/tex]

[tex]\frac{3}{2u} =-\frac{1}{10}[/tex]

u = [tex]-\frac{30}{2}[/tex]

u = - 15 cm

Answer:

L1L1L_1 = ( LLL × WWW)/ www.

Explanation:

From the question we are given that the children's weight are not equal, thus, there will be a need to balance the seesaw "so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground''.

Also, From the question, we are gven that the weight of the heavier child = WWW, the heavier child's is sitting a distance = LLL, the weight of the second child= www.

Therefore, in order to balance the seesaw, she will need to place her second child of weight www on the right side of the pivot to balance the seesaw at a distance of;

L1L1L_1 = ( LLL × WWW)/ www.

Answer:

1.74x10⁻⁵ V

Explanation:

n = 85.7 turns/cm => 8570 turns/metre

The field inside the long solenoid is given by B = μ₀ni

B =  4πx10⁻⁷ x 8570 x 0.175t² = 1.884x10⁻³ t²

dB/dt = 3.78x10⁻³ t

Cross-sectional Area'A'= 2.16 cm²=> 2.16 x [tex]10^{-4}[/tex] m²

Now, rate of change of flux linkage  '|Emf|' is given by:

|Emf| = d(NAB)/dt = NA dB/dt

|Emf|  = 5 x 2.16 x [tex]10^{-4}[/tex] x 3.78x10⁻³ t

|Emf| = 4.0824x10⁻⁶ t

Considering time 't' at which the current = 3.2A , we have

3.2 = 0.175T²

T² = 3.2/0.175

T = 4.28 s

|emf| = 4.0824x10⁻⁶ t  =>  4.0824x10⁻⁶ x4.28

|emf|= 1.74x10⁻⁵ V

Therefore,the magnitude of the emf induced in the secondary winding is 1.74x10⁻⁵ V