If a ball is thrown vertically upward from the roof of a 32 foot tall building with a velocity of 48 ft/sec, its height in feet after T seconds is ______.

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

[tex]\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2[/tex]

In parallel

[tex]\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}[/tex]

[tex]\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0[/tex]

Solving the above quadratic equation

[tex]\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}[/tex]

[tex]\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega[/tex]

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

The two resistors in question, when wired in a series, have a combined resistance of about 36.36 Ohms. When wired in parallel, they share a resistance of roughly 7.5 Ohms. The two resistor values would then most likely be around 28.86 Ohms and 7.5 Ohms.

The nature of your question indicates a focus on the properties and calculations associated with electrical resistors in a circuit. These resistors can either be configured in a series or parallel connection, which will drastically change their behavior and derived readings.

From the context of your question, it is evident that a 12V battery is being used in conjunction with two resistors. In a series connection, the sum of the voltages across each resistor will equal the total voltage, effectively segmenting the 12V battery's output. In a parallel connection, each resistor would experience the full 12V impact, leading to larger current readings. Your values indicate a 0.33A current in a series scenario and a 1.60A current in a parallel situation.

Knowing this, we can apply Ohm's Law, which states Voltage (V) equals Current (I) times Resistance (R). In the series connection, the total resistance can be calculated as 12V divided by 0.33A equals approximately 36.36 Ohms. For the parallel connection, the total resistance would be 12V divided by 1.60A equals approximately 7.5 Ohms. Subtracting these two values, we can find that one of the resistors is around 28.86 Ohms while the other is roughly 7.5 Ohms. This would satisfy both the series and parallel conditions outlined in your question.

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Answer:

Frequency, [tex]f=1.2\times 10^{10}\ Hz[/tex]

Explanation:

It is given that,

Magnetic field, B = 0.43 T

We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

[tex]f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}[/tex]

[tex]f=1.2\times 10^{10}\ Hz[/tex]

Hence, this is the required solution.

Answer:

[tex]A = 0.2875 m^2[/tex]

Explanation:

As we know that

[tex]Q = 4.6 \mu C[/tex]

E = 1.8 kV/mm

now we know that electric field between the plated of capacitor is given as

[tex]E = \frac{\sigma}{\epsilon_0}[/tex]

now we will have

[tex]1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}[/tex]

[tex]\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})[/tex]

[tex]\sigma = 1.6 \times 10^{-5} C/m^2[/tex]

now we have

[tex]\sigma = \frac{Q}{A}[/tex]

now we have area of the plates of capacitor

[tex]A = \frac{Q}{\sigma}[/tex]

[tex]A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}[/tex]

[tex]A = 0.2875 m^2[/tex]

Answer:

The total work done will be zero.

Explanation:

Given that,

Mass = 100 kg

Force = 392 N

Velocity = 20 m/s

Distance s= 10 m

We need to calculate the work done

Using balance equation

The net force will be

[tex]F'=F-\mu mg[/tex]

[tex]F'=392-0.4\times100\times9.8[/tex]

[tex]F'=0[/tex]

The net force is zero.

Hence, The total work done will be zero by all forces on the object.

Answer:

[tex]\dot{W} = 339.84 W[/tex]

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

[tex]\dot{W} =\dot{m}\frac{V^{2}}{2}[/tex]

here [tex]\dot{m}[/tex]is mass flow rate and given as

[tex]\dot{m} = \rho*Q[/tex]

[tex]\dot{W} =\rho*Q\frac{V^{2}}{2}[/tex]

Putting all value to get minimum power

[tex]\dot{W} =1.18*9*\frac{8^{2}}{2}[/tex]

[tex]\dot{W} = 339.84 W[/tex]

The minimum power required to accelerate air to a velocity of 8 m/s at a rate of 9 m^3/s with a density of 1.18 kg/m^3 is roughly 339.84 Watts.

The subject of this question is physics, specifically the topic of power in mechanical systems. To find the minimum power required, we use the formula Power = 0.5 * density * volume flow rate * velocity^2. Plugging in the given values: Power = 0.5 * 1.18 kg/m^3 * 9 m^3/s * (8 m/s)^2, we get approximately 339.84 W. Therefore, the minimum power that must be supplied to the fan is roughly 339.84 Watts.

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Answer:

2.87 Volt

Explanation:

N = 19

change in magnetic field = 5.6 T

Time, dt = 744 ms = 0.744 s

Induced emf,

e = N × change in flux / time

e = 19 × 3.14 × 0.08 × 0.08 ×5.6/0.744

e = 2.87 Volt

Answer:

energy dissipated = 132 kJ

Explanation:

mass = 61 kg

height drop = 227 m

velocity = 11 m/s

potential energy due to height drop from top to bottom

                         P.E. =  m g h

                         P.E. =  61× 9.8× 227

                         P.E. = 135,700 J

kinetic energy = [tex]\frac{1}{2}mv^2[/tex]

                        = [tex]\frac{1}{2}\times 61 \times 11^2[/tex]

                        = 3690.5 J

energy dissipated = P.E - K.E.

                              = 135,700 J -3690.5 J

                              =132,009.5 J = 132 kJ

Answer:

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

Explanation:

A batter hits a baseball into the air. The formula that models the baseball’s height above the ground, y, in feet, x seconds after it is hit is given by :

[tex]y=-16x^2+64x+5[/tex].........(1)

(a) We need to find the maximum height reached by the baseball. The maximum height reached is calculated by differentiating equation (1) wrt x as :

[tex]\dfrac{dy}{dx}=0[/tex]

[tex]\dfrac{d(-16x^2+64x+5)}{dx}=0[/tex]

[tex]-32x+64=0[/tex]

x = 2 seconds

(b) The height of the ball is given by equation (1) and putting the value of x in it as :

[tex]y=-16(2)^2+64(2)+5[/tex]

y = 69 feets

So, at 2 seconds the baseball reaches its maximum height and its maximum height is 69 feets.

Answer:

wow

Explanation:

Answer:

0.2447 m

Explanation:

Amplitude, A = 0.25 m, T = 5.67 s, t = 29.6 s, y = ?

The general equation of SHM is given by

y = A Sin wt

y = 0.25 Sin (2 x 3.14 t /5.67)

Put t = 29.6 s

y = 0.25 Sin (2 x 3.14 x 29.6 / 5.67)

y = 0.2447 m

The close were the centers of the objects at closest approach 2.7 x 10^-14 m

E =8.8 MeV = 8.8 x 1.6 x 10^-13 J

q = 2 e = 2 x 1.6 x 10^-19 C

Q = 82 e = 82 x 1.6 x 10^-19 C

Let d be the distance of closest approach

E = k Q q / d

Where, K = 9 x 10^9 Nm^2 / C^2

d = k Q q / E

d = (9 x 10^9 x 82 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / (8.8 x 1.6 x 0^-13)

d = 2.7 x 10^-14 m

Answer:

option (a)

Explanation:

Wavelength is defined as the distance traveled by the wave in one complete oscillation.

The number of oscillations completed in one second is called frequency.

The relation for the wave velocity is given by

wave velocity = frequency x wavelength

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor [tex]C= 2.00\mu F[/tex]

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

[tex]I = \dfrac{V}{Z}[/tex]

[tex]Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}[/tex]

Put the value of Z into the formula

[tex]I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}[/tex]

Put the value into the formula

[tex]0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}[/tex]

[tex]L=35.8\ mH[/tex]

Hence, The inductance of the inductor is 35.8 mH

Answer:

Inductance,L:

"The property of the conductor or the solenoid to generate the electromotive force,emf due to the flow of current,I."

Unit:  henry,H as it is equivalent to, kg.m².sec⁻².A⁻².

Explanation:

Data:

Solution:

We need to calculate the inductance, L of the solenoid inside a circuit,

Answer:

0.9

Explanation:

h = 400 mm, h' = 325 mm

Let the coefficient of restitution be e.

h' = e^2 x h

325 = e^2 x 400

e^2 = 0.8125

e = 0.9

The coefficient of restitution between the ball and the surface is 0.9.

The coefficient of restitution (e) between the ball and the surface can be determined using the formula:

[tex]\[ e = \sqrt{\frac{\text{height after collision}}{\text{height before collision}}} \][/tex]

First, we need to convert the heights from millimeters to meters for consistency, since the standard units for height in physics are meters.

The initial height (h_i) before the collision is 400 mm, which is equivalent to 0.4 m (since 1 m = 1000 mm).

The final height (h_f) after the collision is 325 mm, which is equivalent to 0.325 m.

Now, we can plug these values into the formula for the coefficient of restitution:

[tex]\[ e = \sqrt{\frac{0.325 \text{ m}}{0.4 \text{ m}}} \][/tex]

[tex]\[ e = \sqrt{\frac{325}{400}} \][/tex]

[tex]\[ e = \sqrt{0.8125} \][/tex]

[tex]\[ e = 0.9 \][/tex]

Therefore, the coefficient of restitution between the ball and the surface is 0.9. This means that the collision is relatively elastic, with the ball retaining a significant portion of its initial kinetic energy after the rebound.

Answer:

The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]

(c) is correct option.

Explanation:

Given that,

Length of steel wire = 4.7 m

Cross section[tex]A = 3\times10^{-3}\ m^2[/tex]

Length of copper wire = 3.5 m

Cross section[tex]A = 4\times10^{-5}\ m^2[/tex]

We need to calculate the ratio of young's modulus of steel and copper

Using formula of young's modulus for steel wire

[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]

[tex]Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}[/tex]....(I)

The young's modulus for copper wire

[tex]Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}[/tex]....(II)

From equation (I) and (II)

The ratio of the young's modulus of steel and copper

[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}[/tex]

[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}[/tex]

[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}[/tex]

[tex]\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}[/tex]

Hence, The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]

The center of mass of these objects is located at approximately 12.652 meters on the y-axis.

We can find the center of mass (center of gravity) of these objects by calculating the weighted average of their positions along the y-axis. Here's how:

Consider Each Object:

We have four objects with masses m1, m2, m3, and m4 at positions y1, y2, y3, and y4 respectively.

Center of Mass Formula:

The center of mass coordinate (y_cm) is calculated as:

y_cm = (Σ(mi * yi)) / Σ(mi)

Σ (sigma) represents summation over all objects (i = 1 to 4 in this case).

mi is the mass of the i-th object.

yi is the y-axis position of the i-th object.

Apply the formula to our case:

y_cm = [(2.00 kg * 13.00 m) + (3.00 kg * 12.50 m) + (2.50 kg * 0.00 m) + (4.00 kg * 20.500 m)] / (2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg)

Calculate:

y_cm = [26.00 kgm + 37.50 kgm + 0.00 kgm + 82.00 kgm] / 11.50 kg

y_cm ≈ 145.50 kg*m / 11.50 kg ≈ 12.652 m (rounded to four decimal places)

Therefore, the center of mass of these objects is located at approximately 12.652 meters on the y-axis.

Answer:

23.98 rpm

Explanation:

d = diameter of merry-go-round = 2.4 m

r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m

m = mass of merry-go-round = 270 kg

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²

M = mass of john = 34 kg

Moment of inertia of merry-go-round and john together after jump is given as

  I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²

w = final angular speed

w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s

v = speed of john before jump

using conservation of angular momentum

Mvr + I w₀ = I' w

(34) (5) (1.2) + (194.4) (2.093) = (243.36) w

w = 2.51 rad/s

w = 23.98 rpm

The maximum speed at which the truck can briefly move around the curve without making the toolbox slide can be found by setting the sliding acceleration equal to the centripetal acceleration (v²/r) and solving for v. Factors like tire friction and flat terrain are also taken into account.

To determine how fast the truck can go before the toolbox starts to slide, we need to calculate the maximum static friction (the force that prevents sliding).

The formula to identify the maximum speed that didn't result in sliding, we would use centripetal acceleration, which is the product of tangential speed squared divided by the curve's radius.

The acceleration of the crate on the truck bed must equal centripetal acceleration to prevent sliding. Given that the sliding acceleration is 2.06 m/s², setting this value to equal centripetal acceleration (v²/r where r is 22 m) and solving for v (velocity/speed), would provide the maximum speed at which the truck can move without the toolbox sliding.Tire friction also plays a role, as it allows vehicles to move at higher speeds without sliding off the road.

This also assumes that the road is relatively flat and the truck moves uniformly, similar to the situation illustrated in Figure 24.7 with a motorist moving in a straight direction.

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The speed at which a toolbox in a truck will start to slide when the truck is turning around a curve depends on the balance of the centripetal force and static friction force. These depend on the truck's speed, the curve's radius, the toolbox's mass and the friction coefficient. Without specific values, a numerical answer can't be given.

The question is asking about the speed at which the toolbox in the truck would start sliding due to the forces acting upon it when the truck turns around a curve. This is related to centripetal force and frictional force.

When the truck turns, the toolbox experiences a centripetal force which pushes it towards the center of the curve. At the same time, friction between the toolbox and the truck bed is resisting this pushing force, keeping the box in place. At some point, if the truck is going too fast, the centripetal force will overcome friction, and the toolbox will start to slide.

We can use the formula for centripetal force, which is F = mv²/r, where m is the mass of the toolbox, v is the velocity of the truck and r is the radius of the curve. And we know that the maximum static friction force (F_max) is f_s * m * g, where f_s is the static friction coefficient and g is the acceleration due to gravity.

The toolbox starts to slide when F = F_max, so we can find the maximum speed (v_max) before sliding happens by making these two equations equal to each other and solving for v. Without the specific values for the mass of the toolbox and the static friction coefficient, we cannot calculate a numerical answer.

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Answer:

Option C is the correct answer.

Explanation:

We equation for elongation

      [tex]\Delta L=\frac{PL}{AE}[/tex]

Here we need to find load required,

We need to double the wire, that is ΔL = 2L - L = L

A = 5 x 10⁻⁵ m²

E = 2 x 10¹¹ N/m²

Substituting

     [tex]L=\frac{PL}{5\times 10^{-5}\times 2\times 10^{11}}\\\\P=10^7N[/tex]

Option C is the correct answer.

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

[tex]m=\dfrac{F}{g}[/tex]

[tex]m=\dfrac{160\ lb}{32\ ft/s^2}[/tex]

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

[tex]m=\dfrac{F}{g}[/tex]

[tex]m=\dfrac{1.9\ lb}{32\ ft/s^2}[/tex]

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

[tex]W=2300\ kg\times 32\ ft/s^2=73600\ lb[/tex]

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

[tex]W=0.022\ kg\times 32\ ft/s^2=0.704\ lb[/tex]

Hence, this is the required solution.

Final answer:

The mass of a 160 lb human being is 72.576 kg, and the mass of a 1.9 lb cockatoo is 0.86184 kg. A 2300 kg rhinoceros weighs 5060 lbs in English units, while a 22 g song sparrow weighs 0.0484 lbs.

Explanation:

To calculate the mass (in SI units) of a 160 lb human being, we use the conversion factor 1 lb = 0.4536 kg. So, the calculation is:160 lb x 0.4536 kg/lb = 72.576 kg

(a) The mass of a 160 lb human being in SI units is 72.576 kg.

For a 1.9 lb cockatoo, the conversion to kilograms is:1.9 lb x 0.4536 kg/lb
 = 0.86184 kg

(b) The mass of a 1.9 lb cockatoo in SI units is 0.86184 kg

To convert the mass of a 2300 kg rhinoceros to weight in English units, knowing that 1 kg = 2.2 lbs (where weight in pounds is considered the gravitational force on the mass), the weight is calculated as:

2300 kg x 2.2 lbs/kg= 5060 lbs

(c) The weight of a 2300 kg rhinoceros in English units is 5060 lbs.

Finally, for converting a 22 g song sparrow to weight in English units:

22 g x (1 kg/1000 g) x (2.2 lbs/kg)= 0.0484 lbs

(d) The weight of a 22 g song sparrow in English units is 0.0484 lbs.

Answer:

72 m/s

Explanation:

D1 = 3 cm, v1 = 2 m/s

D2 = 0.5 cm,

Let the velocity at narrow end be v2.

By use of equation of continuity

A1 v1 = A2 v2

3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2

v2 = 72 m/s

Answer:

speed, distance  and direction of motion of the object can be determined by analyzing the radio wave.

Explanation:

We know that radar operates by transmitting radio waves to a destination and these waves are comes back to the receiver station. By Considering this transmission and receiver process, we can measure the distance, velocity and path of an object's movement.

Distance can be assessed by taking following consideration,  the velocity of the waves is V. It can help to assess the time made for the waves to be emitted by the radar and felt by the receiver, let the time be t.

Therefore  distance can be determine as D= v*t/2,

here 2 signifies that the distance travelled by the wave in either direction ( from transmitter to receiver and vice verse)

Using the source wave frequency, speed can be computed. In a specific frequency, the radar starts sending out the frequencies and the reflected wave will have a distinct frequency. The velocity can be determine by

[tex]v= (\Delta f/f)(c/2),[/tex]

where[tex]\Delta f[/tex]is the change in frequency and

c is the speed of light (the wave).

Direction can be determine by applying above principle. change in frequency is used to determine the direction in the following way:

When the frequency transition is very low, the object moves away from the radar and vice verse.

Answer:

A)

0.8 c

B)

0.87 c

Explanation:

A)

L₀ = Original length of the meter stick = 1 m

L = Length observed = 0.6 m

[tex]v[/tex] = speed of the meter stick

Using the equation

[tex]L = L_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.6 = 1 \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.36 = 1 - \left ( \frac{v}{c} \right )^{2}[/tex]

[tex]v[/tex] = 0.8 c

B)

T₀ = Time of the clock at rest = t

T = Time of the clock at motion = (0.5) t

[tex]v[/tex] = speed of the clock

Using the equation

[tex]T = T_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.5 t = t \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

[tex]0.5  =  \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]

v = 0.87 c

Answer:

Not Changed

Explanation:

To know what happened with the volume you need to know the Ideal gas Law

[tex]\frac{P_{1}V_{1}}{T1} =\frac{P_{2}V_{2} }{T2}[/tex]

This law is a combination of the other four laws: Boyles's, Charles's, Avogadro's, and Guy-Lussac's.

The initial state is represented by P1, V1, T1 and the final by P2, V2, T2.

In this case:

[tex]T_{2} =3T_{1} \\P_{2} =3P_{1}[/tex]

Replacing on the equation

[tex]\frac{P_{1}V_{1}}{T1} =\frac{3P_{1}V_{2}}{3T_{1} }[/tex]

If we clear from the equation V2

[tex]\frac{P_{1}V_{1}3T_{1}}{T_{1} 3P_{1}} ={V_{2}}[/tex]

Then cancel both P1 and T1

[tex]\frac{3V_{1}}{3} =V_{2}[/tex]

You will found that

[tex]V_{1} =V_{2}[/tex]

Answer:

99 dB

Explanation:

We have given that sound intensity with a damaged muffler =8× [tex]10^{-3}[/tex]

we have to find the intensity level of the this sound in decibels

for calculating in decibels we have to use the formula

β=[tex]10log\frac{I}{I_0}[/tex]

  =[tex]10log\frac{.008}{10^{-12}}[/tex]

   =10 log8+10 log[tex]10^{9}[/tex]

   =10 log8+90 log10

   =10×0.9030+90

   =99 dB

Final answer:

To calculate the intensity level of a sound in decibels, use the formula β = 10 log10(I/I0). Given the intensity of [tex]8.0 \times 10^{-3[/tex] threshold of hearing of [tex]10^{-12[/tex] the sound intensity level is approximately 99 decibels.

Explanation:

The student asks about calculating the intensity level of sound in decibels, given the sound intensity at the ear of a passenger in a car with a damaged muffler (8.0 × 10-3 W/m2). To find the intensity level in decibels (dB), we use the formula:

β = 10 log10(I/I0)

where:

Plugging in the values:

β = 10 log10(8.0 × 10-3 / 10-12)

β = 10 log10(8.0 × 109)

β = 10 (log10(8) + log10(109))

β = 10 (0.903 + 9)

β = 10 × 9.903

β = 99.03 dB

The intensity level of the sound in the car with the damaged muffler is approximately 99 decibels.

Answer:

111 N

Explanation:

weight (downwards) = 600 N

reaction force (upwards) = 900 N

t = 0.37 s

Net force

F = reaction force - weight = 900 - 600 = 300 N

Impulse = force x time = 300 x 0.37 = 111 N

Answer:

The wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]

Explanation:

The kinetic energy of the electron is, [tex]E=0.5\ MeV=0.5\times 10^6\ eV[/tex]

We need to find the wavelength of this electron. It can be calculated using the concept of DE-broglie wavelength as :

[tex]\lambda=\dfrac{h}{\sqrt{2mE} }[/tex]

h is Plank's constant

m is the mass of electron

[tex]\lambda=\dfrac{6.67\times 10^{-34}\ J-s}{\sqrt{2\times 9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ eV} }[/tex]        

[tex]\lambda=6.99\times 10^{-22}\ m[/tex]

So, the wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]. Hence, this is the required solution.

The wavelength of an electron with a kinetic energy of 0.50 MeV (relativistic) is approximately 7.28 x 10^-12 m.

The wavelength of an electron with a kinetic energy of 0.50 MeV can be calculated using the relativistic de Broglie equation:

λ = h/(m*c)

Where λ is the wavelength, h is Planck's constant (6.63 x 10^-34 Js), m is the mass of the electron (9.11 x 10^-31 kg), and c is the speed of light (3.00 x 10^8 m/s).

Substituting the values:

λ = (6.63 x 10^-34 Js)/((9.11 x 10^-31 kg)*(3.00 x 10^8 m/s))

λ ≈ 7.28 x 10^-12 m

Therefore, the wavelength of the electron is approximately 7.28 x 10^-12 m.

Answer:

(a) 10.52 A

(b) 11 ohm

(c) 1.77 x 10^7 J

Explanation:

P = 1210 W, V = 115 V,

Let i be the current.

(a) Use the formula of power

P = V i

1210 = 115 x i

i = 10.52 A

(b) Let the resistance of heating coil is R.

Use the formula given below

V = i x R

R = V / i = 115 / 10.52 = 11 ohm

(c) Use the formula for energy

E = V x i x t  

E = 115 x 10.52 x 4.07 x 60 x 60 = 1.77 x 10^7 J

Answer:

volumme =0.36 ml

Explanation:

total heat required can be obtained by using following formula

[tex]q= mC \Delta T[/tex].......(1)

where,

m - mass of water,

C - specific heat capacity of water and = 4.184 j g^{-1} °C

[tex]\Delta T[/tex] - total change in temperature.  = 10°C

The density of water is 1 g/cc. hence, 200 mL of water is equal to 200 g

putting all value in the above equation (1)

q = 200*4.184* 10 ° = 8368 J.

Therefore total number of moles of ethanol required to supply 8368 J of heat is

[tex]\frac {8368}{1368000} = 0.006117 moles.[/tex]

The molar mass of ethanol is 46 g/mol.

The mass of ethanol required is 46* 0.006117 = 0.28138 g

The density of ethanol is 0.78 g/ml.

The volume of ethanol required is

[tex]\frac {0.28138}{0.78} = 0.36 ml[/tex]

To heat 200 mL of water by 10 °C, you would need approximately 0.00785 mL of ethanol. This is calculated by converting the volume of water to its mass, calculating the heat energy using the specific heat capacity of water, and then converting that energy to the amount of ethanol required using the heat of combustion of ethanol.

To calculate the amount of ethanol required to heat 200 mL of water by 10 °C, we need to use the formula: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, we need to convert the volume of water to its mass. Since the density of water is 1 g/mL, the mass of 200 mL of water is 200 g.

Next, we can calculate the heat energy using the formula: Q = mcΔT. The specific heat capacity of water is 4.184 J/g °C, the mass of water is 200 g, and the change in temperature is 10 °C. Plugging in these values, we get: Q = (200 g)(4.184 J/g °C)(10 °C) = 8376 J.

Finally, we can convert this heat energy to the amount of ethanol required using the heat of combustion of ethanol. The heat of combustion of ethanol is -1368 kJ/mol. To convert from J to kJ, we divide the heat energy by 1000 to get: 8376 J / 1000 = 8.376 kJ.

Now we can calculate the volume of ethanol. Since the density of ethanol is 0.78 g/mL, we can use the formula: V = m/ρ, where V is the volume, m is the mass, and ρ is the density. The mass of ethanol can be calculated by rearranging the formula: m = Q/ΔH, where Q is the heat energy and ΔH is the heat of combustion. Plugging in the values, we get: m = 8.376 kJ / -1368 kJ/mol = -0.00611 mol.

Now we can calculate the volume of ethanol using the formula: V = m/ρ. The mass of ethanol is -0.00611 mol and the density of ethanol is 0.78 g/mL. Plugging in these values, we get: V = -0.00611 mol / (0.78 g/mL) = -0.00785 mL.

Since volume cannot be negative, the volume of ethanol required to heat 200 mL of water by 10 °C is 0.00785 mL.

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