A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cross section 4 x 10-5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (a) 3.83 x 103 (b) 1.46 x 10-2 (d) 5.85 x 10-3 (c) 1.79 x 10-2 2.

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]

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:

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:

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

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].

Answer:

443.3 N

567.4 N

Explanation:

Consider the triangle ABC

AC = hypotenuse = magnitude of force vector = F = 720 N

AB = adjacent = Component of force along east-west line = [tex]F_{x}[/tex]

BC = Opposite = Component of force along north-south line = [tex]F_{y}[/tex]

θ = Angle = 38 deg

In triangle ABC

[tex]Sin38 = \frac{BC}{AC}[/tex]

[tex]Sin38 = \frac{F_{y}}{F}[/tex]

[tex]0.616 = \frac{F_{y}}{720}[/tex]

[tex]F_{y}[/tex] = 443.3 N

Also, In triangle ABC

[tex]Cos38 = \frac{AB}{AC}[/tex]

[tex]Cos38 = \frac{F_{x}}{F}[/tex]

[tex]0.788 = \frac{F_{x}}{720}[/tex]

[tex]F_{x}[/tex] = 567.4 N

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 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:

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

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 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:

[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:

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:

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

Explanation:

Joule -Thompson effect

 Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process  

Let

 [tex]P_1,T_1 [/tex] Pressure and temperature at inlet and

 [tex]P_2,T_2 [/tex] Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

[tex]\mu _j=\left(\frac{\partial T}{\partial p}\right)_h[/tex]

Now from T-ds equation

dh=Tds=vdp

So

[tex]Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp[/tex]

⇒[tex]dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

So Joule -Thompson coefficient

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

[tex]\dfrac{\partial v}{\partial T}=\dfrac{v}{T}[/tex]

So by putting the values in

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

[tex]\mu _j=0[/tex] For ideal gas.

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:

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]

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:

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

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.

We can determine the angular velocity using the formula ω = v/r, where v is the tangential speed (2.2 m/s) and r is the radius of the circle (510 mm). After converting the radius to meters, we can then plug these values into the formula and solve for ω, yielding an angular velocity of  3.9636 rad/s.

To determine the resulting angular velocity ωOA of rod OA when point B has a displacement s = 220 mm, we can use the following formula:

ωOA = v / r

Where:

ωOA = Angular velocity of rod OA

v = Linear velocity of point B

r = Radius or distance from point O to point B

Given:

v = 2.2 m/s

s = 220 mm = 0.22 m

d = 510 mm = 0.51 m

First, we calculate the radius (r) using the Pythagorean theorem because the motion is along a right-angled triangle:

r² = d² + s²

r² = (0.51 m)² + (0.22 m)²

r² = 0.2601 m² + 0.0484 m²

r² = 0.3085 m²

Now, we calculate the square root of r² to get r:

r = √0.3085 m

r ≈ 0.555 m

Now, we can calculate the angular velocity ωOA:

ωOA = v / r

ωOA = 2.2 m/s / 0.555 m

ωOA ≈ 3.9636 rad/s

The angular velocity ωOA is approximately 3.9636 rad/s. Since the motion is counterclockwise (as indicated by the positive linear velocity of point B), the angular velocity is also positive.

Learn more about Angular Velocity here:

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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:

156.3 J

Explanation:

E = Amplitude of the electric field = 25000 V/m

t = time interval = 1 minute = 60 sec

d = diameter of the spot = 2 mm = 2 x 10⁻³ m

Area of the spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Energy delivered to the spot is given as

U = (0.5)ε₀ E² c A t

Inserting the values

U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)

U = 156.3 J

To find the energy delivered by the laser beam in 1 minute, we need to calculate the power of the laser beam and then multiply it by the duration.

E = Amplitude of the electric field = 25000 V/m

t = time interval = 1 minute = 60 sec

d = diameter of the spot = 2 mm = 2 x 10⁻³ m

Area of the spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Energy delivered to the spot is given as

U = (0.5)ε₀ E² c A t

Inserting the values

U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)

U = 156.3 J

Answer:

A. Both spheres land at the same time.

Explanation:

The horizontal motion doesn't affect the vertical motion.  Since the two spheres have the same initial vertical velocity and same initial height, they land at the same time.

Answer:

A. Both spheres land at the same time.

Explanation:

If air resistance and friction are ignored, then both spheres land at the same time. Falling objects fall toward the center of the Earth with the same constant acceleration, independent of their mass, when they are defined to be in free-fall. For an object to be in free-fall, it has to be in a situation in which both air resistance and friction are considered negligible. This is true regardless of the direction of the fall.

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:

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

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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.

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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 )

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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 !

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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]

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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]

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Grade: High School

Subject: Physics

Chapter: Simple Harmonic Motion

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

To answer this question, let us break down the initial velocity into its horizontal and vertical components.

v_{ix} = vcos(θ)

v_{iy} = vsin(θ)

v_{x} is the horizontal component of the initial velocity, and this value will stay constant over time because we assume only gravity acts on the stone, therefore only the vertical component of the stone's velocity will change over time

v_{iy} is the vertical component of the initial velocity, and this will change over time due to gravity.

v is the magnitude of the initial velocity.

θ is the angle the velocity vector is oriented at with respect to the horizontal.

Given values:

v = 19.6m/s

θ = 40.8°

Plug in these values and solve for v_{ix} and v_{iy}:

v_{ix} = 19.6cos(40.8°) = 14.8m/s

v_{iy} = 19.6sin(40.8°) = 12.8m/s

To find both the horizontal and vertical displacement at any time, we will use this kinematics equation:

D = X + Vt + 0.5At²

When solving for the horizontal displacement, the following values are:

t = elapsed time

D = horizontal displacement

X = initial horizontal displacement

V = initial horizontal velocity

A = horizontal acceleration

There is no initial horizontal displacement, so X = 0m

The initial horizontal velocity V = v_{ix} = 14.8m/s

Assuming we don't care about air resistance, no force has a component acting horizontally on the stone, so A = 0m/s²

Therefore the equation for the stone's horizontal displacement is given by:

D = 14.8t

When solving for the vertical displacement, the following values are:

t = elapsed time

D = vertical displacement

X = initial vertical displacement

V = initial vertical velocity

A = vertical acceleration

There is no initial vertical displacement, so X = 0m

The initial vertical velocity V = v_{iy} = 12.8m/s

Gravity acts downward on the stone, therefore A = -9.81m/s²

Therefore the equation for the stone's vertical displacement is given by:

D = 12.8t - 4.905t²

Now we just plug in various values of t...

a) At t = 1.03s, the horizontal displacement is D = 14.8(1.03) = 15.2m

b) At t = 1.03s, the vertical displacement is D = 12.8(1.03)-4.905(1.03)² = 7.98m

c) At t = 1.73s, the horizontal displacement is D = 14.8(1.73) = 25.6m

d) At t = 1.73s, the vertical displacement is D = 12.8(1.73)-4.905(1.73)² = 7.46m

Before you write down the following results, read the following explanation.

e) At t = 5.05s, the horizontal displacement is D = 14.8(5.05) = 74.7m

f) At t = 5.05s, the vertical displacement is D = 12.8(5.05)-4.905(5.05)² = -60.4m

Notice that we have a negative value for the vertical displacement. This isn't possible within the context of the problem, so the vertical displacement at t = 5.05s is actually 0m

The problem hasn't stated whether the ground is frictionless or not. I'm going to assume it is frictionless and therefore the stone will keep moving across the ground after landing, so the horizontal displacement at t = 5.05s is 74.7m

Answer:

one ninth

Explanation:

d = 1 cm , v = v

D = 3d, V = ?

By the equation of continuity,

A V = a v

3.14 x D^2 / 4 x V = 3.14 x d^2 / 4 x v

9d^2 x V = d^2 x v

V = v / 9

Thus, the velocity becomes one ninth the initial velocity

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:

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.