Because of your knowledge of physics and interest in weapons, you've got a summer job with the FBI, your job is to determine if the weapon that was found at the scene of a crime was precisely the same with which the crime. For this your boss has asked you to determine the speed of exit of the weapon that was in the scene. Design an experiment in detail where you explain the results.

Answer:

2.82 × 10^6 watt

Explanation:

H = 120 m,

Mass per second = 2400 kg/s

Power = work / time

Power = m g H / t

Power = 2400 × 9.8 × 120 / 1

Power = 2.82 × 10^6 Watt

Answer:

[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s

[tex]E = 3.68 * 10^{-21}[/tex] J

Explanation:

given data

uncertainty in proton position [tex]\Delta x[/tex] = 0.015 nm

according to Heisenberg's principle of uncertainty

[tex]\Delta x \Delta P_{x}= \frac{\frac{h}{2\pi }}{2}[/tex]

Where h is plank constant = [tex]6.6260 * 10^{-34}[/tex] j-s

[tex]\Delta P_{x}= \frac{\frac{h}{2\pi }}{2\Delta x}[/tex]

[tex]\Delta P_{x}= \frac{\frac{6.6260 * 10^{-34}}{2\pi }}{2*0.015*10^{-9}}[/tex]

[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s

b) kinetic energy  of proton whose momentum

[tex]P =\Delta p[/tex]

[tex]E =\frac{\Delta p^{2}}{2m}[/tex]

where m is mass is proton

[tex]E =\frac{(3.51*10^{-24})^{2}}{2*1.67*10^{-27}}[/tex]

[tex]E = 3.68 * 10^{-21}[/tex] J

Answer : The total pressure in the container at equilibrium is, 0.4667 atm

Solution :  Given,

Initial pressure of [tex]H_2S[/tex] = 0.259 atm

Equilibrium constant, [tex]K_p[/tex] = 0.841

The given equilibrium reaction is,

                             [tex]H_2S(g)\rightleftharpoons H_2(g)+S(g)[/tex]

Initially                 0.259          0           0

At equilibrium   (0.259 - x)      x           x

Let the partial pressure of [tex]H_2[/tex] and [tex]S[/tex] will be, 'x'

The expression of [tex]K_p[/tex] will be,

[tex]K_p=\frac{(p_{H_2})(p_{S})}{p_{H_2S}}[/tex]

Now put all the values of partial pressure, we get

[tex]0.841=\frac{(x)\times (x)}{(0.259-x)}[/tex]

By solving the term x, we get

[tex]x=0.2077atm[/tex]

The partial pressure of [tex]H_2[/tex] and [tex]S[/tex] = x = 0.2077 atm

Total pressure in the container at equilibrium = [tex]0.259-x+x+x=0.259+x=0.259+0.2077=0.4667atm[/tex]

Therefore, the total pressure in the container at equilibrium is, 0.4667 atm

Answer:

Time required to completely fill this container = 78.43 seconds

Explanation:

Volume of container = 4 m³

Rate of filling of container = 0.051 m³/s

We have the equation

             [tex]\texttt{Time required}=\frac{\texttt{Volume of container}}{\texttt{Rate of filling of container}}[/tex]

Substituting

             [tex]\texttt{Time required}=\frac{4}{0.051}=\frac{4000}{51}=78.43 s[/tex]

Time required to completely fill this container = 78.43 seconds

Answer:

The diameter of the circle at the surface is 1.814 m.

Explanation:

Given that,

Depth = 80 cm

Let the critical angle be c.

From Snell's law

[tex]sin c=\dfrac{1}{n}[/tex]

[tex]c=sin^{-1}\dfrac{1}{n}[/tex]

The value of n for water,

[tex]n = \dfrac{4}{3}[/tex]

Put the value of n in the equation (I)

[tex]c =sin^{-1}\dfrac{3}{4}[/tex]

[tex]c=48.6^{\circ}[/tex]

We know that,

[tex]tan c=\dfrac{r}{h}[/tex]

We calculate the radius of the circle

[tex]r =h tan c[/tex]

[tex]r=80\times10^{-2}\times\tan48.6^{\circ}[/tex]

[tex]r =0.907\ m[/tex]

The diameter of the circle at the surface

[tex]d =2\times r[/tex]

[tex]d =2\times 0.907[/tex]

[tex]d =1.814\ m[/tex]

Hence, The diameter of the circle at the surface is 1.814 m.

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

Answer:

Charge, [tex]q=3.24\times 10^{-18}\ C[/tex]

Explanation:

A moving particle encounters an external electric field that decreases its kinetic energy from 9650 eV to 8900 eV as the particle moves from position A to position B The electric potential at A is 56.0 V and that at B is 19.0 V. We need to find the charge of the particle.

It can be calculated using conservation of energy as :

[tex]\Delta KE=-q(V_B-V_A)[/tex]

[tex]q=\dfrac{\Delta KE}{(V_B-V_A)}[/tex]

[tex]q=\dfrac{ 9650\ eV-8900\ eV}{(19\ V-56\ V)}[/tex]

q = -20.27 e

[tex]q =-20.27e\times \dfrac{1.6\times 10^{-19}\ C}{e}[/tex]

[tex]q=-3.24\times 10^{-18}\ C[/tex]

Hence, this is the required solution.

Answer:

220.8 m

30.5 m/s

Explanation:

a = acceleration of the car = 2.1 m/s²

t = time of travel = 14.5 s

x = displacement of the car

v₀ = initial velocity of the car = 0 m/s

Displacement of the car is given as

x = v₀ t + (0.5) a t²

Inserting the values

x = (0) (14.5) + (0.5) (2.1) (14.5)²

x = 220.8 m

v = final velocity of the car

Final velocity of the car is given as

v = v₀ + at

v = 0 + (2.1) (14.5)

v = 30.5 m/s

Answer:

19.06 Volt

Explanation:

r =0.25 m, l = 11 m, B = 0.17 T, f = 13 Hz

length for one turn = 2 x 3.14 x r = 2 x 3.14 x 0.25 = 1.57 m

Total number of turns in 11 m

N = 11 / 1.57 = 7

Peak emf, e0 = N x B x A x 2 x 3.14 x f

e0 = 7 x 0.17 x 3.14 x 0.25 x 0.25 x 2 x 3.14 x 13 = 19.06 Volt

Answer:

[tex]i_2 = 7.6 A[/tex]

Explanation:

As we know that the force per unit length of two parallel current carrying wires is given as

[tex]F = \frac{\mu_o i_1 i_2}{2\pi d}[/tex]

here we know that

[tex]F = 3.6 \times 10^{-5} N/m[/tex]

[tex]i_1 = 0.52 A[/tex]

d = 2.2 cm

now from above equation we have

[tex]3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7} (0.52)(i_2)}{2\pi (0.022)}[/tex]

[tex]3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2[/tex]

[tex]i_2 = 7.6 A[/tex]

Answer:

2.63 Hz

Explanation:

m = mass of the object = 8.0 kg

x = stretch in the spring = 3.6 cm = 0.036 m

k = spring constant of the spring

using equilibrium of force

Spring force = weight of object

k x = m g

k (0.036) = (8) (9.8)

k = 2177.78 N/m

frequency of oscillation is given as

[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]

[tex]f = \frac{1}{2\pi }\sqrt{\frac{2177.78}{8}}[/tex]

[tex]f [/tex] =  2.63 Hz

The frequency of the spring is about 2.6 Hz

[tex]\texttt{ }[/tex]

Simple Harmonic Motion is a motion where the magnitude of acceleration is directly proportional to the magnitude of the displacement but in the opposite direction.

[tex]\texttt{ }[/tex]

The pulled and then released spring is one of the examples of Simple Harmonic Motion. We can use the following formula to find the period of this spring.

[tex]T = 2 \pi\sqrt{\frac{m}{k}}[/tex]

T = Periode of Spring ( second )

m = Load Mass ( kg )

k = Spring Constant ( N / m )

[tex]\texttt{ }[/tex]

The pendulum which moves back and forth is also an example of Simple Harmonic Motion. We can use the following formula to find the period of this pendulum.

[tex]T = 2 \pi\sqrt{\frac{L}{g}}[/tex]

T = Periode of Pendulum ( second )

L = Length of Pendulum ( kg )

g = Gravitational Acceleration ( m/s² )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of the object = m = 8.0 kg

extension of the spring = x = 3.6 cm = 3.6 × 10⁻² m

Unknown:

frequency of the spring = f = ?

Solution:

Firstly, we will calculate the spring constant as follows:

[tex]F = kx[/tex]

[tex]mg = kx[/tex]

[tex]k = mg \div x[/tex]

[tex]k = \frac{mg}{x}[/tex]

[tex]\texttt{ }[/tex]

Next, we could calculate the frequency of the spring as follows:

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}[/tex]

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{mg / x}{m}}[/tex]

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{g}{x}}[/tex]

[tex]f = \frac{1}{2 \pi}\sqrt{\frac{9.8}{3.6 \times 10^{-2}}}[/tex]

[tex]f = \frac{35\sqrt{2}}{6\pi} \texttt{ Hz}[/tex]

[tex]f \approx 2.6 \texttt{ Hz}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Simple Harmonic Motion

[tex]\texttt{ }[/tex]

Keywords: Simple , Harmonic , Motion , Pendulum , Spring , Period , Frequency

Final answer:

The magnetic field at a point on the +x-axis near a wire carrying current in the +y-direction will be directed into the page, as determined by the right-hand rule. To cancel the magnetic field at a specific point on the +x-axis, a second wire carrying equal but opposite current must be arranged parallel to the first.

Explanation:

The question concerns the determination of the magnetic field direction around a long, straight wire carrying a current. According to the right-hand rule, if you point your right thumb in the direction of the current, then the direction in which your fingers curl will show the direction of the magnetic field. In this case, the current is running in the +y direction, so if one were to stand at a point on the +x-axis near the wire and apply the right-hand rule, the fingers would curl into the page, indicating that the magnetic field at that point is directed into the page.

An experimenter wanting to make the total magnetic field at the coordinate x = 1.0 m zero must arrange a second wire parallel to the first with current in the opposite direction. This second wire would also need to be carrying the same magnitude of current but in the -y direction to cancel out the magnetic field from the first wire at the point of interest.

Answer:

3.6 x 10⁻⁷ H

Explanation:

C = capacitance of the capacitor = 6.57 x 10⁻¹² F

L = inductance of the inductor = ?

f = frequency of broadcasting system = 103.9 MHz = 103.9 x 10⁶ Hz

frequency is given as

[tex]f = \frac{1}{2\pi \sqrt{LC}}[/tex]

[tex]103.6\times 10^{6} = \frac{1}{2(3.14)\sqrt{(6.57\times 10^{-12})L}}[/tex]

Inserting the values

L = 3.6 x 10⁻⁷ H

Answer:

The initial temperature of copper is 96.3 °C

Explanation:

Specific heat capacity is the energy needed to raise the temperature of one gram of material by one degree celsius. The energy absorbed or released by a material during temperature change can be calculated by the below formula:

[tex]Q=mc(T_{2}-T_{1})[/tex] , where:

Q = energy (J)

m = mass (g)

c = specific heat capacity (J/g·°C)

[tex]T_{1}[/tex] = initial temperature (°C)

[tex]T_{2}[/tex] = final temperature (°C)

In an ideal situation, it can be assumed that all the energy lost by the piece of copper is gained by the water, resulting in its temperature rise and that there is no change of mass/state for either material. Thus, the equation can be written as below:

[tex]+Q_{c}=-Q_{w}[/tex]

[tex]+m_{c}c_{c}(T_{c2}-T_{c1})=-m_{w}c_{w}(T_{w2}-T_{w1})[/tex]

It can also be assumed that the final temperature of both the copper and water are the same. Thus substituting below values in above equation will give:

[tex]m_{w}[/tex] = 400 g

[tex]c_{w}[/tex] = 4.18 J/g·°C

[tex]T_{w1}[/tex] = 20.7 °C

[tex]T_{w2}[/tex] = 22.2 °C

[tex]m_{c}[/tex] = 89.1 g

[tex]c_{c}[/tex] = 0.38 J/g·°C

[tex]T_{c1}[/tex] = ? °C

[tex]T_{c2}[/tex] = 22.2 °C

[tex]+89.1*0.38*(22.2-T_{c1})=-400*4.18(22.2-20.7)[/tex]

Solving for [tex]T_{c1}[/tex] gives:

[tex]T_{c1}[/tex] = 96.3 °C

Here, we are required to determine the initial temperature of the copper.

The initial temperature of the copper is;

T(c1) = 96.27 °C

The energy required to raise the temperature of one gram of a material by one degree celsius is termed the Specific Heat Capacity of that material.

Mathematically, we have;

Q=mc{T(2) -T(1)}

Q=mc{T(2) -T(1)} where:

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

m(c) × c(c) × (T2(c) - T1(c)) = -{m(w) × c(w) × (T2(w) - T1(w))}

However, the final temperature of the copper piece and water can be assumed to be equal

i.e. T2(c) = T1(w)

+89.1*0.38*(22.2-T(c1)) = -400*4.18(22.2-20.7)

(22.2−T(c1) ) = −2508/33.858

(22.2−T(c1) ) = -74.07

T(c1) = 74.07 + 22.2

T(c1) = 96.27 °C.

Therefore, the initial temperature of the copper is; T(c1) = 96.27 °C

Read more:

https://brainly.com/question/19863538&

(a) It takes approximately 0.841 seconds to reach the water from a 3.45 m diving board, with a speed of about 8.24 m/s.

(b) If stepping off a 12.0 m diving board, it takes approximately 1.565 seconds to reach the water, with a speed of about 15.34 m/s.

To solve this problem, we can use the kinematic equations of motion to calculate the time taken to reach the water and the speed upon entering the water.

(a) To find the time taken to reach the water, we can use the kinematic equation:

Height = (1/2) * gravity * time^2

Where:

Height = 3.45 m (height of the diving board)

Gravity = 9.8 m/s² (acceleration due to gravity)

Time is the time taken to reach the water.

Rearranging the equation to solve for time:

time = sqrt((2 * height) / gravity)

time = sqrt((2 * 3.45 m) / 9.8 m/s²)

time ≈ sqrt(0.7061)

time ≈ 0.841 s

(b) To find the speed upon entering the water, we can use the equation:

Speed = gravity * time

Where:

Speed is the speed upon entering the water.

Substituting the known values:

Speed = 9.8 m/s² * 0.841 s

Speed ≈ 8.24 m/s

(c) If the person steps off a 12.0 m diving board, we can repeat the above calculations using height = 12.0 m:

time = sqrt((2 * 12.0 m) / 9.8 m/s²)

time ≈ sqrt(2.449)

time ≈ 1.565 s

Speed = 9.8 m/s² * 1.565 s

Speed ≈ 15.34 m/s

The approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]

To calculate the magnitude of the pressure on the sail due to radiation pressure, we can use the formula for pressure:

[tex]\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \][/tex]

The force due to radiation pressure is given by:

[tex]\[ \text{Force} = 2 \times \text{radiative momentum} \][/tex]

The radiative momentum can be calculated using the energy intensity of sunlight:

[tex]\[ \text{Radiative momentum} = \text{Energy intensity} \times \text{Time} \][/tex]

Given:

- Energy intensity of sunlight [tex](\( I \))[/tex] is approximately [tex]\( 1300 \, \text{W/m}^2 \),[/tex]

- Time [tex](\( t \)) is \( 1 \, \text{s} \)[/tex] (per second),

- Atmospheric pressure is about [tex]\( 105 \, \text{N/m}^2 \).[/tex]

Let's calculate the radiative momentum:

[tex]\[ \text{Radiative momentum} = 1300 \times 1 \, \text{N/s} = 1300 \, \text{N/s} \][/tex]

Now, let's calculate the force:

[tex]\[ \text{Force} = 2 \times 1300 \, \text{N/s} = 2600 \, \text{N/s} \][/tex]

Finally, let's calculate the pressure on the sail:

[tex]\[ \text{Pressure} = \frac{2600 \, \text{N/s}}{1 \, \text{m}^2} = 2600 \, \text{N/m}^2 \][/tex]

So, the approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]

Answer:

The correct option is B. an alpha particle.

Explanation:

When ²³Na is bombarded with protons [tex]_{1}^{1}\textrm{H}[/tex], ²⁰Ne and one alpha particle [tex]_{2}^{4}\textrm{He}[/tex] is released.

The reaction is as follows:

[tex]_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}[/tex]

Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.

Answer:

Linear momentum, p = 50 kg-m/s

Explanation:

It is given that,

Mass of an object, m = 5 kg

Velocity, v = 10 m/s

We need to find the linear momentum of that object. It is given by :

[tex]p=m\times v[/tex]

[tex]p=5\ kg\times 10\ m/s[/tex]

p = 50 kg-m/s

So, the linear momentum of that object is 50 kg-m/s. Hence, this is the required solution.

The linear momentum of an object with a mass of 5 kg and velocity of 10 m/s is calculated using the formula P = mv, which results in 50 kg*m/s.

The student has asked to calculate the linear momentum of an object that has a mass of 5 kg and a velocity of 10 m/s. To find the linear momentum, we use the formula:

P = mv

Where P is the momentum, m is the mass, and v is the velocity. By plugging in the numbers:

P = 5 kg \\* 10 m/s

P = 50 kg\\*m/s

Therefore, the linear momentum of the object is 50 kg\\*m/s, which corresponds to option (C).

Answer:

From lowest to highest acceleration:

3rd train

2nd train

1st train

Explanation:

The acceleration of an object can be found by using Newton's second law:

[tex]a=\frac{F}{m}[/tex]

where

a is the acceleration

F is the net force on the object

m is the mass of the object

We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.

So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.

The acceleration of the three trains can be ranked as follows: the single empty freight car train, the 50 empty freight cars train, and the 50 equally loaded freight cars train.

The accelerations of the three trains can be ranked as follows:

When an equal force is applied to each train, the train with fewer cars will experience a greater acceleration. This is because the force is distributed over fewer cars, resulting in a larger net force acting on each car.

For example, consider two cars:

Therefore, the single empty freight car train will have the highest acceleration, followed by the 50 empty freight cars train, and finally the 50 equally loaded freight cars train.

Explanation:

It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them

[tex]F=k\dfrac{q_1q_2}{1^2}[/tex]...............(1)

We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.

[tex]F'=k\dfrac{q_1q_2}{0.5^2}[/tex]...............(2)

On dividing equation (1) and (2) we get :

[tex]\dfrac{F}{F'}=\dfrac{k\dfrac{q_1q_2}{1^2}}{k\dfrac{q_1q_2}{0.5^2}}[/tex]

[tex]\dfrac{F}{F'}=0.25[/tex]

F' = 4 F

So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.

Final answer:

The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. The force is inversely proportional to the square of the distance, so if the distance is increased, the force decreases. In this case, if the electron moves 0.5 m away from the proton, the new force can be calculated using Coulomb's Law.

Explanation:

The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. Coulomb's Law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

So, if the distance between the charges is increased to 1 m, the force between them will decrease. The force is inversely proportional to the square of the distance, which means that if the distance is doubled, the force will decrease by a factor of four.

In this case, if the electron moves 0.5 m away from the proton, the new distance becomes 1.5 m. Using Coulomb's Law, we can calculate the new force:

F = k * (q1 * q2) / r^2

Where F is the force, k is the Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges (in this case, both 1 C), and r is the distance (1.5 m). Plugging in these values, we get:

F = (9 x 10^9 Nm^2/C^2) * (1 C * 1 C) / (1.5 m)^2

F ≈ 4 x 10^9 N

Answer:

The length of the simple pendulum is 2.4 meters.

Explanation:

Time period of simple pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]

L is the length of pendulum

The time period of the rope is given by :

[tex]T=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]

L' is the length of the rod, L' = 3.6 m

It is given that, the rod have the same period as a simple pendulum and we need to find the length of simple pendulum i.e.

[tex]2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]

On solving the above equation as :

[tex]\dfrac{L}{g}=\dfrac{2L'}{3g}[/tex]

L = 2.4 m

So, the length of the thin rod that is hung vertically from one end and set into small amplitude oscillation 2.4 meters. Hence, this is the required solution.

Answer:

1.87 s

Explanation:

d = distance traveled by the water wave = 64 m

t = time taken to travel the distance = 14 s

[tex]v[/tex] = speed of water wave

Speed of water wave is given as

[tex]v=\frac{d}{t}[/tex]

[tex]v=\frac{64}{14}[/tex]

[tex]v[/tex] = 4.6 m/s

[tex]\lambda[/tex] = wavelength of the wave = 859 cm = 8.59 m

T = period of the wave

period of the wave is given as

[tex]T = \frac{\lambda }{v}[/tex]

[tex]T = \frac{8.59 }{4.6}[/tex]

T = 1.87 s

The power drawn by the air conditioner to cool the house from 35°C to 20°C in 38 minutes, given a COP of 2.8, is calculated to be 1.35 kW.

To calculate the power drawn by the air conditioner, we first determine the energy needed to cool the house from 35°C to 20°C. The energy required (Q) can be calculated using the formula Q = m⋅cv⋅ΔT, where m is the mass of the air, cv is the specific heat capacity at constant volume, and ΔT is the change in temperature.

Plugging the given values into the formula: Q = 800 kg ⋅ 0.72 kJ/kg°C ⋅ (35°C - 20°C) = 8,640 kJ.

Next, we use the Coefficient of Performance (COP) of the air conditioner to find the work input (W) required for this cooling process. The formula relating these quantities is COP = Q/W, rearranging for W gives W = Q/COP.

Therefore, W = 8,640 kJ / 2.8 = 3,085.71 kJ. To find the power drawn, we need the work in kilowatts (kW), knowing that 1 kW = 1 kJ/s, and that the total duration is 38 min or 2,280 seconds. Thus, the power drawn is Power = W / time = 3,085.71 kJ / 2,280 s = 1.35 kW.

Power drawn by the air conditioner is approximately 6.76 kW, calculated from COP, heat transfer, and time taken for cooling.

To solve this problem, we can use the energy balance equation for the air within the house:

[tex]\[ Q = mc\Delta T \][/tex]

where:

- [tex]\( Q \)[/tex] is the heat transfer,

- [tex]\( m \)[/tex] is the mass of air,

- [tex]\( c \)[/tex] is the specific heat capacity of air,

- [tex]\( \Delta T \)[/tex] is the change in temperature.

We also know that the Coefficient of Performance (COP) is defined as:

COP = Q_cooling/W_input

where:

- [tex]\( Q_{\text{cooling}} \)[/tex] is the heat removed from the house (in this case),

- W_input is the work input (power consumed) by the air conditioner.

We are given that the COP [tex](\( \text{COP} = 2.8 \))[/tex] and the initial and final temperatures [tex](\( T_{\text{initial}} = 35^\circ C \)[/tex], [tex]\( T_{\text{final}} = 20^\circ C \))[/tex].

First, let's calculate the heat transfer using the energy balance equation:

[tex]\[ Q = mc\Delta T \][/tex]

[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(35 - 20) \][/tex]

[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(15) \][/tex]

Q = 12000kJ

Now, we can find the work input using the COP formula:

COP = Q_cooling/{W_input

W_input = [tex]\frac{Q_{\text{cooling}}}{\text{COP}}[/tex]

W_input = [tex]\frac{12000 \, \text{kJ}}{2.8}[/tex]

W_input ≈ 4285.71 kJ

To find the power drawn by the air conditioner, we need to convert the energy to power. Since the time taken for cooling is 38 minutes, or [tex]\( \frac{38}{60} \)[/tex] hours:

Power = W_input /Time

[tex]\[ \text{Power} = \frac{4285.71 \, \text{kJ}}{\frac{38}{60} \, \text{hours}} \][/tex]

Power ≈ 6762.45 W

So, the power drawn by the air conditioner is approximately [tex]\( 6762.45 \, \text{W} \)[/tex] or [tex]\( 6.76 \, \text{kW} \)[/tex].

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]s=0+\dfrac{1}{2}at^2[/tex]

[tex]a=\dfrac{2s}{t^2}[/tex]

[tex]a=\dfrac{2\times 12\ m}{(1.2\ s)^2}[/tex]

a = 16.67 m/s²

Now put the value of a in equation (1) as :

[tex]q=\dfrac{ma}{E}[/tex]

[tex]q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}[/tex]

q = 0.0000249 C

or

[tex]q=2.49\times 10^{-5}\ C[/tex]

Hence, this is the required solution.

Answer:

10.99 m

Explanation:

m = mass of the block = 0.245 kg

k = spring constant of the vertical spring = 4975 N/m

x = compression of the spring = 0.103 m

h = height to which the block rise

Using conservation of energy

Potential energy gained by the block = Spring potential energy

mgh = (0.5) k x²

(0.245) (9.8) h = (0.5) (4975) (0.103)²

h = 10.99 m

Answer:

The S wave arrives 6 sec after the P wave.

Explanation:

Given that,

Distance of P = 45 km

Speed of p = 5000 m/s

Speed of S = 3000 m/s

We need to calculate the time by the P wave

Using formula of time

[tex]t = \dfrac{D}{v}[/tex]

Where, D = distance

v = speed

t = time

Put the value in to the formula

[tex]t_{p} =\dfrac{45\times1000}{5000}[/tex]

[tex]t_{p} = 9\ sec[/tex]

Now, time for s wave

[tex]t_{s}=\dfrac{45000}{3000}[/tex]

[tex]t =15\ sec[/tex]

The required time is

[tex]\Delta t=t_{s}-t_{p}[/tex]

[tex]\Delta t=15-9[/tex]

[tex]\Delta t =6\ sec[/tex]

Hence, The S wave arrives 6 sec after the P wave.

Answer : The height of the unit cell is, [tex]4.947\AA[/tex]

Explanation : Given,

Density of zinc = [tex]7.14g/cm^3[/tex]

Atomic radius = [tex]1.332\AA[/tex]

Atomic weight of zinc = 65.37 g/mole

As we know that, zinc has haxagonal close packed crystal structure. The number of atoms in unit cell of HCP is, 6.

Formula used for density :

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]      .............(1)

where,

[tex]\rho[/tex] = density  of zinc

Z = number of atom in unit cell  = 6 atoms/unit cell (for HCP)

M = atomic mass

[tex](N_{A})[/tex] = Avogadro's number  = [tex]6.022\times 10^{23}atoms/mole[/tex]

a = edge length of unit cell

[tex]a^3[/tex] = volume of unit cell

Now put all the values in above formula (1), we get

[tex]7.14g/cm^3=\frac{(6\text{ atoms per unit cell})\times (65.37g/mole)}{(6.022\times 10^{23}atoms/mole)\times a^{3}}[/tex]

[tex]V=a^{3}=9.122\times 10^{-23}cm^3[/tex]

[tex]V=9.122\times 10^{-29}m^3[/tex]

Now we have to calculate the height of the unit cell.

Formula used :

[tex]V=6\sqrt {3}r^2h[/tex]

where,

V = volume of unit cell

r = atomic radius = [tex]1.332\AA=1.332\times 10^{-10}m[/tex]

conversion used : [tex](1\AA=10^{-10}m)[/tex]

h = height of the unit cell

Now put all the given values in this formula, we get:

[tex]9.122\times 10^{-29}m^3=6\sqrt {3}\times (1.332\times 10^{-10}m)^2h[/tex]

[tex]h=4.947\times 10^{-10}m=4.947\AA[/tex]

Therefore, the height of the unit cell is, [tex]4.947\AA[/tex]

To find the height of the zinc unit cell, we apply the geometric relationship of the hexagonal close-packed structure to the given atomic radius. After performing the calculation, we get the height of the unit cell. Density provides context for the metal's packing efficiency but is not directly used in the calculation.

To calculate the height of the unit cell of zinc, we must first determine the type of crystal structure zinc has. Zinc crystallizes in a hexagonal close-packed (hcp) structure. Given that its atomic radius is 1.332 Å (or 0.1332 nm), we can use this information along with the geometry of the hcp structure to calculate the height (c) of the unit cell.

In an hcp structure, the height (c) can be found using the relationship between the atomic radius (r) and the height, which is c = 2\sqrt{\frac{2}{3}}\cdot r. Substituting the given value for the atomic radius:

c = 2\sqrt{\frac{2}{3}}\cdot (0.1332 \text{ nm})

Calculating this gives us the height of the unit cell for zinc.

While this question does not require use of the density directly, it's worth noting that the density of zinc is 7.14  g/cm³, which indicates a relatively high atomic packing efficiency typical for metals. Using the atomic weight (65.37 grams/mole) and the Avogadro's number, we could further explore the relationship between the density and the unit cell parameters if needed.

Answer:

Magnetic field at given point is along + x direction

Explanation:

As we know that position of the wire is along Y axis

current is flowing along +y direction

so here we will have

[tex]\vec B = \frac{\mu_0 i (\vec {dl} \times \hat r)}{4\pi r^2}[/tex]

now the direction of length vector is along +Y direction

also the position vector is given as

[tex]\hat r = \hat k[/tex]

now for the direction of magnetic field we can say

[tex]\vec B = (\hat j \times \hat k)[/tex]

[tex]\vec B = \hat i[/tex]

so magnetic field is along +X direction

Using the right-hand rule for a wire carrying current in the +y direction along the y-axis, the magnetic field at the point (0,0,1.045 m) on the x-axis would be directed in the +x-direction.

When a current runs through a wire in the +y direction along the y-axis, we can use the right-hand rule to determine the direction of the magnetic field at a point near the wire. For a point on the x-axis such as (0,0,1.045 m), you would point your thumb in the direction of the current (upward along the y-axis) and curl your fingers.

Your fingers will curl in the direction of the magnetic field lines, which at the given point on the x-axis would circle around the wire. This means the magnetic field at that point would be directed in the +x-direction.

Answer:

The bullet will have been dropped vertically h= 0.54 meters by the time it hits the target.

Explanation:

d= 100m

V= 300 m/s

g= 9.8 m/s²

d= V*t

t= d/V

t= 0.33 s

h= g*t²/2

h=0.54 m

Answer:

a) [tex]k_{avg}=6.22\times 10^{-21}[/tex]

b) [tex]k_{avg}=8.61\times 10^{-21}[/tex]

c)  [tex]k_{mol}=3.74\times 10^{3}J/mol[/tex]

d)   [tex]k_{mol}=5.1\times 10^{3}J/mol[/tex]

Explanation:

Average translation kinetic energy ([tex]k_{avg} [/tex]) is given as

[tex]k_{avg}=\frac{3}{2}\times kT[/tex]    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

[tex]k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8[/tex]  

[tex]k_{avg}=6.22\times 10^{-21}J[/tex]

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

[tex]k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416[/tex]  

[tex]k_{avg}=8.61\times 10^{-21}J[/tex]

c ) The translational kinetic energy per mole of an ideal gas is given as:

       [tex]k_{mol}=A_{v}\times k_{avg}[/tex]

here   [tex]A_{v}[/tex] = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        [tex]k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}[/tex]

          [tex]k_{mol}=3.74\times 10^{3}J/mol[/tex]

d) now at T = 143° C

        [tex]k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}[/tex]

          [tex]k_{mol}=5.1\times 10^{3}J/mol[/tex]