A hospital needs 0.100 g of 133 54Xe for a lung-imaging test. If it takes 10 days to receive the shipment, what is the minimal amount mXe of xenon that the hospital should order?

Answer:

a)0.48 m/s

b) 0.583 m/s

Explanation:

As the wagon rolls,

momentum'p'= m x v => 95.8 x 0.530 = 50.774 Kgm/s

(a)Rock is thrown forward,

momentum of rock = 0.325 x 15.1 =  4.9075 Kgm/s

Conservation of momentum says momentum of wagon is given by

50.774 - 4.9075 = 45.8665

Therefore, Speed of wagon = 45.8665 / (95.8-0.325) = 0.48 m/s

(b) Rock is thrown backward,

momentum of wagon = 50.774  + 4.9075 = 55.68  Kgm/s

Therefore, speed of wagon = 55.68 / (95.8-0.325) = 0.583 m/s

Final answer:

The speed of the wagon after the rock is thrown directly forward is 0.529 m/s, and after the rock is thrown directly backward is 1.63 m/s.

Explanation:

To find the speed of the wagon after the rock is thrown, we can use the principle of conservation of momentum. Since there is no external force acting on the system, the total momentum before and after the rock is thrown will be the same.

For part (a) when the rock is thrown directly forward, the momentum of the wagon and rider will remain the same, but the momentum of the rock will change to zero. So, the final momentum of the system will be the same as the initial momentum.

Using the equation: initial momentum = final momentum

(mass of wagon + rider + rock) x initial velocity of wagon = (mass of wagon + rider) x final velocity of wagon

(95.8 kg) x (0.530 m/s) = (95.8 kg + 0.325 kg) x vf

vf = 0.529 m/s

For part (b) when the rock is thrown directly backward, the momentum of the wagon and rider will change to zero, but the momentum of the rock will remain the same. So, the final momentum of the system will be the same as the initial momentum.

Using the equation: initial momentum = final momentum

(mass of wagon + rider + rock) x initial velocity of wagon = (mass of rock) x final velocity of rock

(95.8 kg) x (0.530 m/s) = (0.325 kg) x vf

vf = 1.63 m/s

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x [tex]10^{8}[/tex] m/s)

f= 3 x [tex]10^{8}[/tex] / 2.4

f=1.25 x  [tex]10^{8}[/tex] hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x  [tex]10^{8}[/tex]

T= 8 x [tex]10^{-9}[/tex] s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x [tex]10^{-9}[/tex] => 800 x [tex]10^{-9}[/tex] s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

[tex]x_{min}[/tex]= τc/2 =>  (800 x [tex]10^{-9}[/tex] x 3 x [tex]10^{8}[/tex])/2

[tex]x_{min}[/tex]=120m

the answer would be 2, I just took the test..hope this helps :))

The ray 2 shows the correct direction of the reflected ray. Therefore, the correct option is b.

The ray that depicts the light that is reflected by the surface is the reflected ray that corresponds to a certain incident ray. The angle of reflection is the angle formed between the surface normal and the reflected beam. According to the Law of Reflection, the angle of reflection for a specular surface is always the same as the angle of incidence.

In the given problem, the second ray is the ray that is depicted correctly as a reflected ray as it is bounced back after striking the mirror at a specified angle above the horizontal. Rest all the other rays are not depicted correctly as the ray 1 is reflected below normal, ray 3 is reflected horizontally and the ray 4 is refracted.Therefore, the correct option is b.

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The question is incomplete, but most probably the complete question is,

The diagram below shows a light ray from a pencil hitting a mirror. An upright pencil sits to the left of a mirror, which is convex facing left. An arrow from the tip of the pencil runs to the center of the mirror. Line 1 runs from the point of reflection at an acute angle below the first arrow. Arrow 2 runs at an acute angle above the first arrow. Arrow 3 runs perpendicularly up from the point reflection. Arrow 4 runs through the mirror down at an obtuse angle, pointing toward a distant point on a line running from the base of the pencil. Which ray shows the correct direction of the reflected ray?

a.1

b.2

c.3

d.4

Answer:

a. The energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules

b. The total number of revolutions the particle makes before stopping is 1.43 revolutions

Explanation:

a.

Given

m = mass of particle = 0.350-kg

u = initial speed of 10.00 m/s

v = final speed = 5.50 m/s

r = radius = 0.600 m

We assume that the floor is horizontal;

This means that F = mg.

We also assume the rotational kinetic energy to be negligable.

Having listed the assumptions, we proceed as follows;

Let ∆E represent The energy transformed from mechanical to internal in the particle hoop.

This is given by

∆E = KE1 - KE2

Where KE1 = ½mu²

KE2 = ½mv²

So, ∆E = KE1 - KE2 becomes

∆E = ½mu² - ½mv²

∆E = ½m(u² - v²)

∆E = ½ * 0.350 * (10² - 5.5²)

∆E = 12.20625

∆E = 12.21J (Approximated)

Hence, the energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules

b.

Let n = number of revolutions

The relationship between n and the energy is

1/n = (KE1 - KE2)/KE1

Make n the subject of formula

n = KE1 / (KE1 - KE2)

n = ½mu² / (½mu² - ½mv²) --- Simplify

n = ½mu² / (½m(u² - v²)) ---- Divide through by ½m

n = u² / (u² - v²)

n = 10² / (10² - 5.5²)

n = 1.433691756272401

n = 1.43 rev

Hence, the total number of revolutions the particle makes before stopping is 1.43 revolutions

Final answer:

The energy lost to friction in one revolution is 11.9875 J, and the particle makes a total of 2 revolutions before coming to a stop.

Explanation:

Energy Loss Due to Friction

The energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution (a) is determined by the change in kinetic energy of the particle. The initial kinetic energy is given by ½ mv², where ‘m’ is the mass of the particle and ‘v’ is the initial velocity. The final kinetic energy is likewise given by ½ mv² using the final velocity. The difference in these energies gives the energy lost to friction.

Kinetic Energy Initial = ⅓ (0.350 kg)(10.00 m/s)² = 17.5 J
Kinetic Energy Final = ⅓ (0.350 kg)(5.50 m/s)² = 5.5125 J
Energy Transformed = Kinetic Energy Initial - Kinetic Energy Final = 17.5 J - 5.5125 J = 11.9875 J

Total Number of Revolutions

For part (b), to find the total number of revolutions the particle makes before stopping, assuming a constant frictional force, we can use a deceleration approach or an energy approach. The work done against friction for each subsequent revolution will be the same; thus each revolution results in the same amount of energy loss until the particle comes to a stop. By dividing the total initial kinetic energy by the energy lost per revolution, we find the number of revolutions.

Revolutions = Total Initial Kinetic Energy / Energy Transform per Revolution = 17.5 J / 11.9875 J ≈ 1.46 revolutions
Since it cannot complete a partial revolution after losing all kinetic energy, we round down to get 1 full revolution. Therefore, the particle makes a total of 1 + 1 = 2 revolutions before stopping.

Answer:

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,

C

Explanation:

Resolving the vectors to vertical and horizontal component;

Vertical;

Vector A = 6sin30

Vector B = 4sin60

Resultant vertical = 6sin30 + 4sin60 = 6.464m

Horizontal;

Vector A = 6cos30

Vector B = -4cos60

Resultant horizontal = 6cos30 - 4cos60 = 3.196m

Resultant R = √(6.464^2 + 3.196^2) = 7.2m

Tanθ = 6.464/3.196

θ = taninverse (6.464/3.196) BN

θ = 64° north of East.

Or

26° east of north

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.

Given data:

The magnitude of vector A is, A = 6.0 m.

The direction of vector A is, 30° north of east.

The magnitude of vector B is, B = 4.0 m.

The direction of vector B is, 30° west of north.

The quantity having both the magnitude as well as the magnitude are known as vector quantities.

Resolving the vectors to vertical and horizontal component;

Along the Vertical direction;

Vector A = 6sin30

Vector B = 4sin60

Resultant vertical vector = 6sin30 + 4sin60 = 6.464 m

Along the Horizontal direction;

Vector A = 6cos30

Vector B = -4cos60

Resultant horizontal vector = 6cos30 - 4cos60 = 3.196 m

Now, the resultant vector is calculated as,

[tex]R=\sqrt{6.464^{2}+3.196^{2}}\\\\R = 7.2 \;\rm m[/tex]

And the resultant direction of vectors is,

[tex]tan \theta = 6.464/3.196\\\\\theta = tan^{-1} (6.464/3.196) \\\theta =64 ^{\circ}[/tex]( North of East)

θ = 64° north of East.

Or

26° east of north

Thus, we can conclude that the  resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.

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The average velocity of car A between t₁ and t₂ is greater than the average velocity of car B between t₁ and t₂.

The average velocity is the ratio of the total displacement traveled to the total time taken by the body. Its unit is m/sec.

The given data in the problem is;

t₁ is the time at position 1

xo is the position of car A and car B

v is the speed of cars A and B

t₂ is the time at position 2

3v is the speed of car A

v is the speed of car B

The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.

At time t₁ A moved from Xo to 2Xo, resulting in a displacement of 2Xo.

At time t2, A travels at a speed of 3V, therefore his new position is 3Xo, as opposed to 2Xo, which is at 5Xo. A's displacement from the starting location is 5Xo.

B moved from Xo to 2Xo at time ti, and the displacement is 2Xo.

At time t₂ proceeds with speed V in the opposite direction, returning to his starting location, and therefore his new position is at Xo. A's the displacement from his starting location is zero.

[tex]\rm V_{avg}= \frac{total \ displacement}{total \ time}[/tex]

The displacement of car A between t₁ and t₂ is greater than the displacement of car B between t₁ and t₂.

Hence the average velocity of car A between t₁ and t₂ is greater than the average velocity of car B between t₁ and t₂.

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Final answer:

A leading voltage across a device at 500 Hz compared to a resistor suggests inductive elements within the device. Voltage leads current in inductors, which can be determined by comparing phase differences using a voltmeter parallel to the device.

Explanation:

When measuring the voltage across a device and comparing it to the voltage across a resistor at a specific frequency, such as 500 Hz, the phase difference provides information about the type of elements in the device. If the voltage across the device leads the voltage across the resistor, this indicates the presence of inductive reactance, suggesting that there are inductive elements within the unknown circuitry.

Reactance and impedance play key roles in AC circuits. The voltage drop across a resistor is in phase with the current, meaning it neither leads nor lags the current. However, in an inductor, the voltage leads the current, and in a capacitor, the voltage lags the current. Therefore, measuring a leading voltage implies inductive characteristics. To evaluate this, a voltmeter would be placed in parallel with the device, ensuring minimal disturbance to the circuit as high resistance in the voltmeter minimizes current flow through it.

For components like inductors and capacitors, the relationship between voltage and current is not linear; thus, they are not ohmic devices like resistors. This is evident in an RLC series circuit, where the impedance at a given frequency depends on the resistance (R), inductance (L), and capacitance (C) of the circuit.

The voltage leading the resistor's voltage at 500 Hz suggests the device contains capacitive elements. A phase difference measurement can confirm this. Capacitors cause the voltage to lead the current in AC circuits.

If you observe that the voltage across a device leads the voltage across a resistor at 500 Hz, it is likely that the device includes a capacitive element. In AC circuits, capacitors cause the voltage to lead the current, which means that the voltage across the capacitor will peak before the voltage across a purely resistive element.

To confirm the presence of a capacitive component, you can perform a more detailed analysis:

This phase difference can be analyzed using an oscilloscope to visualize the waveform and confirm the capacitive nature of the device.

Answer:

The amplitude of the motion of the spring is 1 m.

Explanation:

From the conservation of energy;

[tex](PE_{spring})_{max}= PE+KE[/tex]

which is

[tex]\frac{1}{2} kx^2_{max}=\frac{1}{2} kx^2+\frac{1}{2} mv^2[/tex]

Make [tex]x_{max}[/tex] the subject of formula

[tex]x^2_{max}= \frac{kx^2\times mv^2}{k}[/tex]

[tex]x_{max}= \sqrt{\frac{kx^2\times mv^2}{k} }[/tex]

Substitute 320 N/m for k, 0.017 m for x, 3.7 kg for m, and 0.60 m/s for v.

[tex]x_{max}= \sqrt{\frac{(320)(0.017)^2+ (3.7)(0.60)^2}{320} } \\\\=1m[/tex]

Thus, The amplitude of the motion of the spring is 1 m.

Answer: 0.0667m

Explanation:

given

Mass of the object, m = 3.7kg

Force constant of the spring, k = 320 N/m

Speed of the object, v = 0.6 m/s

Distance from equilibrium position, x = 0.017 m

Using the law of conservation of energy. The total energy conserved is

= 1/2kx² + 1/2mv²

= 1/2 * 320 * 0.017² + 1/2 * 3.7 * 0.6²

= 0.046 + 0.666

= 0.712 J

When v = 0, the maximum deflection is Xmas. This Xmas, is also the amplitude. Such that,

Energy = 1/2kx²

0.712 = 1/2 * 320 * x²

1.424 = 320 x²

x² = 1.424 / 320

x² = 0.00445

x = 0.0667

x = 6.67*10^-2 m

Thus, the amplitude is 0.0667 m

Answer: 1.91*10^8 N/m²

Explanation:

Given

Radius of the steel, R = 10 mm = 0.01 m

Length of the steel, L = 80 cm = 0.8 m

Force applied on the steel, F = 60 kN

Stress on the rod, = ?

Area of the rod, A = πr²

A = 3.142 * 0.01²

A = 0.0003142

Stress = Force applied on the steel/Area of the steel

Stress = F/A

Stress = 60*10^3 / 0.0003142

Stress = 1.91*10^8 N/m²

From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²

Answer:

Explanation:

wavelength of sound = velocity / frequency

= 340 / 780

= .4359 m

Let the observer be at equal distance d from in phase source .

let it moves by distance x for destructive interference .

path difference from source

= d + x - (d - x )

=  2x

for destructive interference

path difference = wave length / 2

2x = .4359 / 2 m

x = .4359 / 4

= .108975 m

10.9 cm

so observer must move by distance 10.9 cm towards on of the centers.

Answer:

(A) Velocity of car in order just to clear the river and land safely on the opposite side is [tex]29.68 \frac{m}{s}[/tex]

(B) The speed of the car is [tex]35.14 \frac{m}{s}[/tex]

Explanation:

Given:

Car distance from river [tex]y = 20.5[/tex] m

Mere distance from river [tex]y_{o} = 2.4[/tex] m

River length [tex]x= 57[/tex] m

(A)

For finding the velocity of car,

Using kinematics equation we find velocity of car

  [tex]y - y_{o} = v_{o}t + \frac{1}{2} gt^{2}[/tex]

Where [tex]v_{o} = 0[/tex], [tex]g = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex]                                     ( given in question )

[tex]20.5 - 2.4 = \frac{1}{2} \times 9.8 \times t^{2}[/tex]

   [tex]t= 1.92[/tex] sec

The speed of the car before it lands safely on the opposite side of the river is given by,

   [tex]v_{x} = \frac{x-x_{o} }{t}[/tex]

   [tex]v_{x} = \frac{57-0}{1.92}[/tex]                        ( [tex]x_{o} = 0[/tex] )

   [tex]v_{x} = 29.68[/tex] [tex]\frac{m}{s}[/tex]

(B)

For finding the speed,

The horizontal distance travel by car,

  [tex]v_{y} = v_{o} + at[/tex]

Where [tex]a = 9.8[/tex] [tex]\frac{m}{s^{2} }[/tex]

  [tex]v_{y} = 9.8 \times 1.92[/tex]

  [tex]v_{y} = 18.81[/tex] [tex]\frac{m}{s}[/tex]

For finding the speed,

  [tex]v = \sqrt{v_{x}^{2} + v_{y} ^{2} }[/tex]

  [tex]v = \sqrt{(29.68)^{2}+ (18.81)^{2} }[/tex]

  [tex]v = 35.14[/tex] [tex]\frac{m}{s}[/tex]

Therefore, the speed of the car is [tex]35.14 \frac{m}{s}[/tex]

Answer:

11.07Hz

Explanation:

Check the attachment for diagram of the standing wave in question.

Formula for calculating the fundamental frequency Fo in strings  is V/2L where;

V is the velocity of the wave in string

L is the length of the string which is expressed as a function of its wavelength.

The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)

Therefore L = 1.5λ

If L = 3.0m

1.5λ = 3.0m

λ = 3/1.5

λ = 2m

Also;

V = √T/m where;

T is the tension = 0.98N

m is the mass per unit length = 2.0g = 0.002kg

V = √0.98/0.002

V = √490

V = 22.14m/s

Fo = V/2L (for string)

Fo = 22.14/2(3)

Fo = 22.14/6

Fo = 3.69Hz

Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo

Frequency of the wave = 3×3.69

Frequency of the wave = 11.07Hz

Answer:

Explanation:

Given that,

One fragment is 7 times heavier than the other

Let one fragment mass be M

Let this has a velocity v

And the other 7M

And this a velocity V

Initially the fragment is at rest u = 0

Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

Po = Pf

(M+7M) × 0 = 7M •V − Mv

0 = 7M•V - Mv

Divide both sides by M

0 = 7V -v

v = 7V

Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

Work done by M mass

W(m) = −µk•Mg×d'

Since it is a wordone by friction, that is why we have a negative sign.

Using conservation of energy

Work done by 7M mass is equal to work done by M mass

W(7m) = W(m)

−47.6 µk•Mg = −µk•Mg×d

Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

Answer:

Distance the lighter fragment slides before stopping= 333.2 m

Explanation:

Gravitational acceleration = g

Mass of the lighter fragment = m

Mass of the heavier fragment = 7m

Velocity of the lighter fragment after the explosion = V1

Velocity of the heavier fragment after the explosion = V2

The object is at rest before the explosion hence the total momentum of the system is zero.

mV1 + 7mV2 = 0

V1 = -7V2

Coefficient of friction = [tex]\mu[/tex]

Friction force on the lighter fragment = f1 =[tex]\mu mg[/tex]

Friction force on the heavier fragment = f2 =[tex]7\mu mg[/tex]

Deceleration of the lighter fragment due to friction = a1

ma1 = -f1

ma1 = [tex]-\mu mg[/tex]

a1 =[tex]-\mu g[/tex]

Deceleration of the heavier fragment due to friction = a2

7ma2 = -f2

7ma2 = -7[tex]\mu mg[/tex]

a2 = [tex]-\mu g[/tex]

Final velocity of the lighter fragment = V3 = 0 m/s

Final velocity of the heavier fragment = V4 = 0 m/s

Distance traveled by the lighter fragment before coming to rest = d1

Distance traveled by the heavier fragment before coming to rest= d2 =6.8 m

V4² = V2² + 2a₂d₂

[tex],[/tex]

[tex](0)^2 = V_2^2 + (2)(-\mu g)d_2[/tex]

[tex]d_2 = \frac{V_2^2}{2 \mu g}[/tex]

V₃² = V₁² + 2a₁d₁

[tex](0) = (-7V_2)^2 + 2(-\mu g)d_1[/tex]

[tex]d_1 = \frac{49V_2^2}{2 \mu g}[/tex]

d₁ = 49d₂

d₁ = (49)(6.8)

d₁ = 333.2 m

Distance the lighter fragment slides before stopping = 333.2 m

Answer:

-2.2394181 J amount of kinetic energy is lost.

Explanation:

Let Vf be the final speed

(48.7) (9.8) = (48.7 + 1100) Vf

477.26 = 1148.7 Vf

=>   Vf = 0.4155 m/s

KE initiali = 1/2 (0.0487) (9.8)²

KE initial = 1/2 (0.0487) (96.04)

KE initial = 1/2 (4.677148)

KE initial = 2.338574 J

KE final = 1/2 (1.1 + 0.0487) (0.4155)²

KE final = 1/2 (1.1487‬) (0.17264025)

KE final = 1/2 (0.198311855175)

KE final = 0.0991559 J

KE (lost) = KEf - KEi

             = 0.0991559 J - 2.338574 J

             = -2.2394181 J

Answer:force equals to rate of change of momentum

Explanation:

F=force

t=time

m=mass

v=final velocity

u=initial velocity

(mv-mu)/t=rate of change of momentum

Force=rate of change of momentum

F=(mv-mu)/t

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the netforce, and inversely proportional to the mass of the object.

The charge on the capacitor after 2 s is 0.43 C.

Explanation:

The formula for finding the voltage while charging a capacitor is

[tex]V = V_{0}(1-e^{\frac{-t}{RC} })[/tex]

Here , V₀ is the initial potential before charging and t is the time at which we have to determine the voltage, R is the resistance and C is the capacitance for the given circuit.

The given problem have given us the values for V₀ = 50 V and maximum current I is given as 500 mA.

Then, resistance can be determined using Ohm's law: [tex]R = \frac{V}{I} =\frac{50}{500*10^{-3} } = 100 Ohm[/tex]

The capacitance is defined as the ratio of charge to the unit voltage.

[tex]C = \frac{Q}{V} = \frac{IT}{IR} = \frac{T}{R}[/tex]

Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be [tex]\frac{1}{100} = 10 mF[/tex]

So, the values for parameters like V₀ = 50 V, R = 100 Ω, C = 10 mF and t = 2 s.

Then, [tex]V =50*(1-e^{\frac{-2}{100*10*10^{-3} } }) =50*(1-e^{-2}) = 50*(1-0.1353)[/tex]

V= 43 V.

Then, [tex]Q = CV = 10*10^{-3} * 43 V = 0.43 C[/tex]

Thus, the charge on the capacitor after 2 s is 0.43 C.

The unit of charge is Coulomb.

The charge on the capacitor after 2 second is 0.43 Coulomb.

To finding the voltage while charging a capacitor is given as,

      [tex]V=V_{0}(1-e^{-\frac{t}{RC} } )[/tex]

Here , V₀ is the initial potential, R is the resistance and C is the capacitance

It is given that V₀ = 50 V and maximum current I is given as 500 mA.

So,  [tex]R=\frac{V}{I}=\frac{50}{500*10^{-3} } =100ohm[/tex]

Time constant = RC = 1

 So,     [tex]C=\frac{1}{R} =\frac{1}{100}F[/tex]

Now, we have  V₀ = 50 V, R = 100 Ω, C = 1/100 F and t = 2 s.

 [tex]V=50(1-e^{-\frac{2}{100*\frac{1}{100} } } )=50(1-e^{-2} )=50(1-0.1353)=43V[/tex]

We know that,  [tex]Q=CV=\frac{1}{100}*43=0.43C[/tex]

Thus, the charge on the capacitor after 2 s is 0.43 C.

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Answer:

Light travels fast in air but i dont know about water

Answer:

light slower in water than in Air

Explanation:

Air's refractive index is about 1.0003, while water's is about 1.3. This means that light is “slower” in water than air. This is because it's more likely to hit a molecule and then get re-emitted, lengthening the amount of time the light takes to get through a certain distance of the medium.

Answer:

Explanation:

The guy wire is making a right angled triangle  with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.

base / hypotenuse = cos51

base = hypotenuse x cos51

= 8 x cos51

= 5.03 ft .

The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .

Answer:

Explanation:

Given;

peak electric field, E₀ = 96.9 V/m

time of flow, t = 14.9s

area through which the energy flows, A = 0.0227 m²

The intensity of this wave is calculated using the following formula;

[tex]I = \frac{E_{rms}^2}{c \mu_o}[/tex]

where;

root-mean-square electric field, [tex]E_{rms} = \frac{E_o}{\sqrt{2}} = \frac{96.9}{\sqrt{2} } = 68.5187 \ V/m[/tex]

c is speed of light, c = 3 x 10⁸ m/s

μ₀ is permeability of free space (constant), μ₀ = 1.26 x 10⁻⁶

Substitute these values and calculate the intensity of the wave;

[tex]I = \frac{E_{rms}^2}{c \mu_o} = \frac{(68.5187)^2}{(3*10^8)(1.26*10^{-6})} = 12.42 \ W/m^2[/tex]

Thus, intensity of this wave is 12.42 W/m²

The energy of the wave is calculated as follows;

U = IAt

U = 12.42 x 0.0227 x 14.9

U = 4.2 J

Thus, the energy of this wave is 4.2 J

Answer:

360 Nm

Explanation:

Torque: This is the force that tend to cause a body to rotate or twist. The S.I unit of torque is Newton- meter (Nm).

From the question,

The expression of torque is given as

τ = F×d......................... Equation 1

Where, τ = Torque, F = force, d = distance of the bar perpendicular to the force.

Given: F = 40 N, d = 9 m

Substitute into equation 1

τ  = 40(9)

τ  = 360 Nm.

Answer:

360Nm

Explanation:

Torque is defined as the rotational effect of a force. The magnitude of a torque τ, is given by;

τ = r F sin θ

Where;

r = distance from the pivot point to the point where the force is applied

F = magnitude of the force applied

θ = the angle between the force and the vector directed from the point of application to the pivot point.

From the question;

r = 9m

F = 40N

θ = 90° (since the force is applied perpendicular to the end of the bar)

Substitute these values into equation (i) as follows;

τ = 9 x 40 sin 90°

τ = 360Nm

Therefore the torque is 360Nm

The question is incomplete. The complete question is as follows.

The chart shows the voltage of four electric currents. A two-column table with 2 row titled Voltage of Currents. The first column labeled Current has entries W, X, Y, Z. The second column labeled Volts (volts) has entries 9.0, 1.5, 3.0, 4.5.

Which is best supported by the data in the chart?

A. Current W flows at a higher rate than Current Z.

B. Current Y flows at a lower rate than Current X.

C. Current X has a lower potential difference than Current Y.

D. Current Z has a greater potential difference than Current W.

Answer: C. Current X has a lower potential difference than Current Y.

Explanation: The table described is show below:

Current                        Volts (volts)

   W                                      9.0

   X                                       1.5

   Y                                       3.0

   Z                                       4.5

Current is defined by the flow of charge that passes through a material, called conductor in an amount of time:

I = [tex]\frac{Q}{t}[/tex]

where:

Q is charge in Coulomb

t is time in seconds

I is the current in Ampere: A = C/s

According to the question, the currents are not defined by charge or numbers, so it's difficult to determine its value. So, alternatives A and B aren't correct.

Voltage is the difference in electrical potential between two points. It is because of the voltage that the charges flow.

From the table, Current W has the biggest voltage of all the others. So, alternative D is incorrect.

Analysing the voltages of Current X and Y, we observe that they are 1.5 and 3.0, respectively. So, volts or the potential diference of current X is lower than Y, making alternative C the correct one.

Answer:

Current X has a lower potential difference than Current Y.

Explanation:

Answer:

74.10 cubic meters

Explanation:

The weught of load, balloon and heilium equals the weight of air displaced.

We know that density is mass per unit volume hence mass is product of density and volume, m=dv where d is density and v is volume

Weight is product of mass and acceleration due to gravity, W=mg=mgv where g is acceleration due to gravity and m is mass. We take g to be 9.81 m/s2

Weight of load=790 N

Weight of ballon= mg= 1.8*9.81

Weight of helium=0.179*9.81*v

Weight of air displaced =1.29*9.81*v

790+(1.8*9.81)+(9.81*0.179*v)=1.29*9.81*v

(1.29-0.179)v*9.81=807.658

10.89891v=807.658

V=74.104474667650

Rounding off, the volume is 74.10 cubic meters

Answer:

Part(a): The angular acceleration is [tex]\bf{1410.78~rad/s^{2}}[/tex].

Part(b): The linear acceleration is [tex]\bf{28.21~m/s^{2}}[/tex].

Explanation:

Given:

The angular velocity, [tex]\omega = 235.6~rad/s[/tex]

Time taken, [tex]t = 0.167~s[/tex]

Diameter of the disk, [tex]d = 0.12~m[/tex]

Radius of the concerned point, [tex]r_{p} = \dfrac{1}{3}(d/2)[/tex].

Part(a):

The angular acceleration ([tex]\alpha[/tex]) is given by

[tex]\alpha &=& \dfrac{\omega}{t}\\&=& \dfrac{235.6~rad/s}{0.167~s}\\&=& 1410.78~rad/s^{2}[/tex]

Part(b):

The radius ([tex]r_{p}[/tex]) of the concerned point is given by

[tex]r_{p} &=& \dfrac{1}{3}\dfrac{0.12~m}{2}\\&=& 0.02~m[/tex]

Linear acceleration ([tex]a[/tex]) is give by

[tex]a &=& \alpha \times r_{p}\\&=& (1410.78~rad/sec)(0.02~m)\\&=& 28.21~m/s^{2}[/tex]

This question involves the concepts of equations of motion for angular motion and tangential acceleration.

a) The magnitude of the average angular acceleration is "1410.78 rad/s²".

b) The magnitude of the tangential acceleration of a point 1/3 of the way out from the center of the disk is "28.22 m/s²".

a)

We will use the first equation of motion for the angular motion to find out the angular acceleration:

[tex]\alpha = \frac{\omega_f-\omega_i}{t}[/tex]

where,

[tex]\alpha[/tex] = angular acceleration = ?

[tex]\omega _f[/tex] = final angular speed = 0 rad/s

[tex]\omega_i[/tex] = initial angular acceleration = 235.6 rad/s

t = time = 0.167 s

Therefore,

[tex]\alpha=\frac{0\ rad/s-235.6\ rad/s}{0.167\ s}\\\\\alpha = 1410.78\ rad/s^2[/tex]

b)

The tangential acceleration can be found using the following formula:

[tex]a_T=(r)(\alpha)[/tex]

where,

[tex]a_T[/tex] = tangential acceleration = ?

r = distance of point from center = [tex]\frac{1}{3}(\frac{d}{2}) = \frac{1}{3}(\frac{0.12\ m}{2}) = 0.02\ m[/tex]

Therefore,

[tex]a_T=(0.02\ m)(1410.78\ rad/s^2)\\a_T=28.22\ m/s^2[/tex]

Learn more about the angular motion here:

brainly.com/question/14979994?referrer=searchResults

The attached picture shows the angular equations of motion.

Answer:

F=30N

a= 3m/s^2

m=?

F=ma

30=m(3)

30/3=m

m=10kg

The mass of the ball is 10kg

Final answer:

The mass of an object thrown with a force of 30 Newtons resulting in an acceleration of 3 m/s² is calculated using the formula m = F/a, resulting in a mass of 10 kg.

Explanation:

The question is regarding the calculation of the mass of an object when a known force is applied to it and an acceleration is observed as a result. According to Newton's Second Law of Motion, which is typically expressed as F = ma, where F is the force applied to an object in newtons (N), m is the mass of the object in kilograms (kg), and a is the acceleration of the object in meters per second squared (m/s²). To find the mass of the object, we rearrange this formula to get m = F / a. Given the applied force of 30 N and the acceleration of 3 m/s² due to that force, we can calculate the mass of the object.

Mass (m) = F / a = 30 N / 3 m/s² = 10 kg

Therefore, the mass of the object that was thrown with a force of 30 Newtons, resulting in an acceleration of 3 m/s², is 10 kg.

Answer:

b)  the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.

Explanation:

check attachment for solution to A

The torque required to rotate the inner cylinder of a rotating viscometer with a fluid of viscosity μ can be calculated using fluid mechanics. However, this solution is approximated and based on a linear velocity profile in the gap, which can change with a significant increase in the gap.

In a rotating viscometer with a fluid of viscosity μ between two concentric cylinders, the torque required to rotate the inner cylinder at a constant speed can be computed using fluid mechanics principles. The shear stress on the fluid due to the rotating cylinder is the product of the viscosity and the velocity gradient across the gap, given by (ωi*ri)/ (ro - ri). This shear stress relates to the force acting on a unit area of the fluid as F/A, and the torque T is the product of the force and the radius of the inner cylinder. Following the equations, we'll obtain T = 2π * L * μ * ωi * (((ro^2 - ri^2) / ln (ro/ri)).

This solution is an approximation as it's based on the assumption that the velocity profile in the gap between the cylinders is linear, which holds true when the gap is thin. If the gap increases disproportionately (i.e., the outer radius R0 were to increase), the linear approximation of the velocity profile may no longer hold. In this scenario, edge effects and non-linear velocity profiles have to be considered, potentially modifying the expected torque value.

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Answer:

0.44 V

Explanation:

Parameters given:

Radius of loop, r = 10.7 cm = 0.107 m

Magnetic field strength, B = 0.803 T

Rate of shrinkage of Radius, dr/dt = 81 cm/s = 0.81m/s

EMF induced is given in terms of magnetic Flux, Φ, as:

EMF = dΦ/dt

Magnetic Flux, Φ, is given as:

Φ = B * A (where A is area of loop)

Therefore, EMF is:

EMF = d(B*A)/dt

The area of the loop is given as A = πr²

EMF = d(Bπr²) / dt = Bπ*d(r²)/dt

=> EMF = Bπ*2r*(dr/dt)

EMF = 0.803 * π * 2 * 0.107 * 0.81

EMF = 0.44 V

Answer:

60833 loops

Explanation:

Given parameters:

Input voltage  = 120V

Output voltage  = 100kV

Primary loops  = 73 loops

Current = 14.1A

Unknown:

Number of loops in the secondary coil  = ?

Solution:

A step up transformer of this nature increases the output voltage. Here, the voltage is raised on the output side.

One thing about a transformer is that the output and input power are always the same.

To find the number of loops in the secondary coil;

             [tex]\frac{N_{p} }{N_{s} } = \frac{V_{p} }{V_{s} }[/tex]

Where N is the number of loops and V is the voltage

            s and p are the secondary and primary parts

Note: 100kV  = 100000V

 Now input the parameters;

          [tex]\frac{73}{N_{s} } = \frac{120}{100000}[/tex]

     120N[tex]_{s}[/tex]   = 73 x 100000

            N[tex]_{s}[/tex]  = 60833 loops

Answer:

0.624 kg

Explanation:

We are given that

Mass of one block,[tex]m_1=1.3 kg[/tex]

[tex]v_1=1.3 m/s[/tex]

[tex]v_2=5 m/s[/tex]

[tex]V=2.5 m/s[/tex]

We have to find the mass of second block.

Mass of second block,[tex]m_2=\frac{m_1V-m_1v_1}{v_2-V}[/tex]

Substitute the values

[tex]m_2=\frac{1.3\times 2.5-1.3\times 1.3}{5-2.5}[/tex]

[tex]m_2=0.624 kg[/tex]

Hence, the mass of second block=0.624Kg

Answer:

The mass of the second block is 0.624 kg.              

Explanation:

Given that,

Mass of the block 1, m₁ = 1.3 kg

Speed of block 1, u₁ = 1.3 m/s

Speed of block 2, u₂ = 5 m/s

The final velocity of the combined blocks is 2.5 m/s, V = 2.5 m/s

It is a case of inelastic collision. Using the conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\1.3\times 1.3+5m_2=(1.3+m_2)2.5\\\\1.69+5m_2=3.25+2.5m_2\\\\m_2=0.624\ kg[/tex]

So, the mass of the second block is 0.624 kg.              

Answer:

Explanation:

An equilibrium is a state in which opposing forces or influences are banned.

An example of equilibrium is in economics when supply and demand are equal. An example of equilibrium is when you are calm and steady. An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.

Answer:

a state in which opposing forces or influences are balanced.

Explanation:

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.

Answer:

The value of path integral is 2.2 [tex]\mu T.m[/tex]

Explanation:

Given:

Current carrying by long wire [tex]I = 2.5[/tex] A

Area of rectangle [tex]A = 18.75[/tex] [tex]m^{2}[/tex]

Angle with surface normal [tex]\theta =[/tex] 45°

According to the ampere's circuital law,

     [tex]\int\limits {B} \, ds = \mu _{o} I_{net}[/tex]

Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex] [tex]ds=[/tex] area element

Here, [tex]I _{net} = I \cos 45[/tex]

 [tex]I_{net} = 2.5 \times \frac{1}{\sqrt{2} }[/tex]

Put value of current in above equation,

  [tex]\int\limits {B} \, ds = 4\pi \times 10^{-7} \times 2.5 \times \frac{1}{\sqrt{2} }[/tex]

  [tex]\int\limits {B} \, ds = 22.2 \times 10^{-7}[/tex]

  [tex]\int\limits {B} \, ds = 2.2 \times 10^{-6}[/tex]

  [tex]\int\limits {B} \, ds = 2.2[/tex] [tex]\mu T.m[/tex]

Therefore, the value of path integral is 2.2 [tex]\mu T.m[/tex]