If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3000 m to an altitude of 420 m , where she will pull her ripcord?

Answer:

Distance between Karl and Joe is 38.467 m

Solution:

Let us assume that you are at origin

Now, as per the question:

Joe's tent is 19 m away from yours in the direction [tex]19.0^{\circ}[/tex] north of east.

Now,

Using vector notation for Joe's location, we get:

[tex]\vec{r_{J}} = 19cos(19.0^{\circ})\hat{i} + 19sin(19.0^{\circ})\hat{j}[/tex]

[tex]\vec{r_{J}} = 17.96\hat{i} + 6.185\hat{j} m[/tex]

Now,

Karl's tent is 45 m away from yours and is in the direction [tex]39.0^{\circ}[/tex]south of east, i.e.,  [tex]- 39.0^{\circ}[/tex] from the positive x-axis:

Again,  using vector notation for Karl's location, we get:

[tex]\vec{r_{K}} = 45cos(-319.0^{\circ})\hat{i} + 45sin(- 39.0^{\circ})\hat{j}[/tex]

[tex]\vec{r_{K}} = 34.97\hat{i} - 28.32\hat{j} m[/tex]

Now,  obtain the vector difference between [tex]\vec{r_{J}}[/tex] and [tex]\vec{r_{K}}[/tex]:

[tex]\vec{r_{K}} - \vec{r_{J}} = 34.97\hat{i} - 28.32\hat{j} - (17.96\hat{i} + 6.185\hat{j}) m[/tex]

[tex]\vec{d} = \vec{r_{K}} - \vec{r_{J}} = 17.01\hat{i} - 34.51\hat{j} m[/tex]

Now, the distance between Karl and Joe, d:

|\vec{d}| = |17.01\hat{i} - 34.51\hat{j}|

[tex]d = \sqrt{(17.01)^{2} + (34.51)^{2}} m[/tex]

d = 38.469 m

The distance between Karl's and Joe's tent is:

The distance between Joe's tent and Karl's tent is approximately 36.84 m.

To find the distance between Joe's tent and Karl's tent, we can use the concept of vector addition. We first need to break down the given distances and angles into their respective components:

Next, we can use the components to find the displacement from Joe's tent to Karl's tent:

Using the components, we can calculate the displacement from Joe's tent to Karl's tent:

Finally, we can use the Pythagorean theorem to find the magnitude of the displacement:

Magnitude = sqrt((-33.46 m)^2 + (16.49 m)^2) = 36.84 m

Therefore, the distance between Joe's tent and Karl's tent is approximately 36.84 m.

The answer is If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice, due to the time it takes light and gravitational force to travel from the Sun to Earth. The Sun is approximately 1.50×10¹¹ meters away, and light travels at about 300,000 kilometers per second. Therefore, we see the Sun as it was 8 minutes ago.

If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice any change. This is because the Sun is approximately 1.50×10¹¹ meters away, and light, which travels at the speed of about 300,000 kilometers per second, takes 8 minutes to travel from the Sun to Earth.

Therefore, for 8 minutes, we would continue to see the Sun as it was before the change or disappearance occurred.

Similarly, if there were an immediate change in the gravitational force due to the Sun’s disappearance, it would also take 8 minutes for Earth to experience the effect because gravitational information, like light, propagates at the speed of light.

Hence, The answer is If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice

Answer: Yes, he is exceeding the speed limit

Explanation:

Hi!

This is problem about unit conversion

1 mile = 1,609.344 m

Then the speed limit v is:

v = 75 mi/h = 120,700.8 m/h

1 hour = 60 min = 60*60 s = 3,600 s

v = (120,700.8/3,600) m/s = 33.52 m/s

38 m/s is higher than the speed limit v.

Answer:

[tex]t = \sqrt{L / ( sin(\theta) * (-1/2) * g * tg(\theta) )}[/tex]

Explanation:

The mass will have a weight, and since it is on a surface it will have a normal reaction.

The vertical component of the normal reaction will be equal and opposite to the weight.

w = g * m

Nv = N * sin(θ)

N is the normal reaction and Nh its vertical component

Nv = -w

N * sin(θ) = -g * m

The horizontal component of the normal will be

Nh = N * cos(θ)

N = Nh / cos(θ)

Then:

Nh / cos(θ) * sin(θ) = -g * m

sin/cos = tg

Nh * tg(θ) = -g * m

The horizontal component of the normal force will be the only force in the horizontal direction

It will cause an acceleration

Nh = ah * m

Then

ah * m * tg(θ) = -g * m

Simplifying the mass on each side

ah * tg(θ) = -g

ah = -g * tg(θ)

The mass will slide from a height related to the lenght of the ramp

L = D * sin(θ)

D = L / sin(θ) This is the distance it will slide

We set up a reference system with origin at the top of the ramp and the positive X axis pointing down the ramp in the direction of the slope.

In this reference system:

X(t) = X0 + V0*t + 1/2 * a * t^2

X0 = 0

V0 = 0

Then

X(t) = -1/2 * g * tg(θ) * t^2

It will move the distance D

L / sin(θ) = -1/2 * g * tg(θ) * t^2

t^2 = L / ( sin(θ) * (-1/2) * g * tg(θ) )

[tex]t = \sqrt{L / ( sin(\theta) * (-1/2) * g * tg(\theta) )}[/tex]

The negative sign will dissapear because gravity has a negative sign too.

Answer:

[tex]y_{max}=11m[/tex]

Explanation:

The maximum vertical displacement that the ball reaches can be calculate using the following formula:

[tex]v^{2}=v^{2} _{o}+2g(y-y_{o})[/tex]

At the highest point, its velocity becomes 0 because it stop going up and starts going down.

[tex]0=(29.4sin(30))^{2} -2(9.8)y[/tex]

Solving for y

[tex]y=\frac{(29.4sin(30))^{2}}{2(9.8)} =11m[/tex]

Answer:

The car overtakes the truck at a distance d = 3266.2ft from the starting point

Explanation:

Problem Analysis

When car catches truck:

dc = dt = d

dc: car displacement

dt: truck displacement

tc = tt = t

tc: car time

tt : truck time

car kinematics :

car moves with uniformly accelerated movement:

d = vi*t + (1/2)a*t²

vi = 0 : initial speed

d = (1/2)*a*t² Equation (1)

Truck kinematics:

Truck moves with constant speed:

d = v*t Equation (2)

Data

We know that the acceleration of the car is 3.00 ft / s² and the speed of the truck is 70.0 ft / s .

Development problem

Since the distance traveled by the car is equal to the distance traveled by the truck and the time elapsed is the same for both, then we equate equations (1 ) and (2)

Equation (1) = Equation (2)

(1/2)*a*t² = v*t

(1/2)*3*t² = 70*t  (We divide both sides by t)

1.5*t = 70

t = 70 ÷ 1.5

t = 46.66 s

We replace t = 46.66 s in equation (2) to calculate d:

d = 70*46.66 = 3266.2ft

d = 3266.2 ft

Answer: 20 pm=20*10^-12 m

Explanation: To solve this problem we have to use the relationship given by:

λmin=h*c/e*ΔV= 1240/60000 eV=20 pm

this expression is related with the bremsstrahlung radiation when a flux of energetic electrons are strongly stopped hitting to a catode. The electrons give their kinetic energy to the atoms of the catode.

Answer:

Explanation:

The mug is sliding towards the east but its velocity is going down . That means it has negative acceleration towards east   . In other words , it has positive acceleration towards west. Since it has positive acceleration towards

west , it must have positive force acting on it towards west.

Answer:

distance between object and screen is 2 m

Explanation:

given data

focal length = 0.18 m

magnification = 9.0 x

to find out

distance from the object

solution

we know that magnification is express as

magnification m = [tex]\frac{v}{u}[/tex]   .............1

here u is distance of object from lens

and v is distance of image from lens

so here

9 = [tex]\frac{v}{u}[/tex]

v = 9 u   ..................2

now we will apply here lens formula that is

[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]        ....................3

put here value f is focal length and v = 9 u

[tex]\frac{1}{0.18} = \frac{1}{u} + \frac{1}{9u}[/tex]  

solve it we get

u = 0.2

so v = 9 (0.2 )

v = 1.8

so here distance between object and screen is v +u

distance between object and screen = 1.8 + 0.2

distance between object and screen is 2 m

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

[tex]E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}[/tex]

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

[tex]E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}[/tex]

E(0.89) = 306500 N/C

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

[tex]n = \frac{P_0}{P}[/tex]

[tex]P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}[/tex]

kinetic energy[tex] = \frac{1}{2} mv^2  =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2[/tex]

kinetic energy [tex]= 90590389.66 kg m^2/s^2[/tex]

gravitational energy [tex]= mgh = 3250*9.8*10000 = 315500000.00  kg m^2/s^2[/tex]

total energy [tex]= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2[/tex]

[tex]P_o =\frac{409091242.28}{750} = 545454.99 j/s[/tex]

[tex]effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49[/tex]

effeciency n = = 49%

To calculate the efficiency of an aircraft engine, we calculate the work done by the aircraft in climbing to its cruising altitude and reaching its cruising speed, then compare it to the total energy input from the engines. Using the aircraft's mass, altitude, speed, and power output in the efficiency formula allows us to determine its efficiency. Real-world factors like air resistance would normally be considered, but are omitted in this scenario.

Calculating Aircraft Engine Efficiency

The efficiency of an aircraft's engines can be determined by comparing the actual mechanical work done to the energy input as power from the engines. For the given aircraft scenario, we can calculate the work done by the aircraft in reaching both its cruising altitude and speed, and then determine efficiency using the average power output of the engines.

Work Done by the Aircraft

We start by calculating the work done against gravity to reach the cruising altitude (also known as potential energy, PE) and the kinetic energy (KE) gained by the aircraft to reach cruising speed:

PE = m * g * h

KE = 0.5 * m * v²

Where m is the mass of the aircraft, g is the acceleration due to gravity (9.81 m/s²), h is the altitude (10,000 meters), and v is the speed (converted to m/s).

Average Power and Efficiency

Next, we convert the aircraft's average power output from horsepower to watts:

1 horsepower = 745.7 watts

Average Power = 1500 hp * 745.7 W/hp

Now, we'll calculate efficiency:

Efficiency (η) = (Work done / Energy input) * 100%

The total work done is the sum of PE and KE. Energy input is the power multiplied by the time in seconds the power is delivered.

Let's apply these steps using the provided data:

Convert 12.5 minutes to seconds.

Calculate the work done based on mass, speed, and altitude.

Calculate the total energy input from the engines.

Finally, use these values to find the engine efficiency.

Please note, in a real-world scenario, additional factors like air resistance and variations in engine power output would affect these calculations.

Answer:

n = 3

Solution:

Since, the slit used is same and hence slit distance 'x' will also be same.

Also, the wavelengths coincide, [tex]\theta [/tex] will also be same.

Using Bragg's eqn for both the wavelengths:

[tex]xsin\theta = n\lambda[/tex]

[tex]xsin\theta = n\times 680.0\times 10^{- 9}[/tex]           (1)

[tex]xsin\theta = (n + 1)\lambda[/tex]

[tex]xsin\theta = (n + 1)\times 510.0\times 10^{- 9}[/tex]         (2)

equate eqn (1) and (2):

[tex] n\times 680.0\times 10^{- 9} = (n + 1)\times 510.0\times 10^{- 9}[/tex]

[tex]n = \frac{510.0\times 10^{- 9}}{680.0\times 10^{- 9} - 510.0\times 10^{- 9}}[/tex]

n = 3

Final answer:

Using the formula for the location of maxima in a double slit interference pattern and equating the equations for the two different wavelengths, we find that the value of n is 3 for this Young's two-slit experiment scenario.

Explanation:

To determine the value of n such that an nth-order maximum for a wavelength of 680.0 nm coincides with the (n+1)th maximum of light of wavelength 510.0 nm in a Young's two-slit experiment, we can use the formula for the location of maxima in a double slit interference pattern:

dsinθ = mλ, where d is the separation of the slits, sinθ is the sine of the angle of the maxima, m is the order of the maximum, and λ is the wavelength of the light.

For two wavelengths to coincide, we equate the two equations:

nλ1 = (n+1)λ2

Substituting the given wavelengths:

n(680.0 nm) = (n+1)(510.0 nm)

Solving for n gives:

n = 510.0 / (680.0 - 510.0) = 3

Answer:

44.1 m

Explanation:

initial velocity of ball, u = 0

height of building, H = 106 m

g = 9.8 m/s^2

t = 3 second

Let the ball travels a distance of h in first 3 seconds.

Use second equation of motion

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

h = 0 + 0.5 x 9.8 x 3 x 3

h = 44.1 m

Thus, the distance traveled by the ball in first 3 seconds is 44.1 m.

Answer:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]

Explanation:

From the vertical movement, we know that initial speed is 0, and initial height is H, so:

[tex]Y_{f}=Y_{o}-g*\frac{t^{2}}{2}[/tex]

[tex]0=H-g*\frac{t^{2}}{2}[/tex]    solving for t:

[tex]t=\sqrt{\frac{2H}{g} }[/tex]

Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:

[tex]X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}[/tex]   Replacing values:

[tex]D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}[/tex]

Simplifying:

[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]

Answer:

Linear mass density,[tex]\lambda=2\times 10^{-4}\ kg/m[/tex]

Explanation:

Given that,

Mass of the string, m = 0.3 g = 0.0003 kg

Length of the string, l = 1.5 m

The linear mass density of a string is defined as the mass of the string per unit length. Mathematically, it is given by :

[tex]\lambda=\dfrac{m}{l}[/tex]

[tex]\lambda=\dfrac{0.0003\ kg}{1.5\ m}[/tex]

[tex]\lambda=0.0002\ kg/m[/tex]

or

[tex]\lambda=2\times 10^{-4}\ kg/m[/tex]

So, the linear mass density of a string is [tex]2\times 10^{-4}\ kg/m[/tex]. Hence, this is the required solution.

To calculate the linear mass density of the string, you divide the mass in kilograms by the length in meters. For a 0.3g and 1.5m string, this yields a linear mass density of 0.0002 kg/m.

The student asks how to calculate the linear mass density (μ) of a string. The linear mass density is defined as mass per unit length. To calculate it for a given string, you divide the mass of the string by its length. The provided string has a mass of 0.3g, which should be converted to kilograms (0.0003 kg), and a length of 1.5m.

The formula to calculate linear mass density is:
μ = mass/length.

Therefore, the linear mass density of the string is:
μ = 0.0003 kg / 1.5 m = 0.0002 kg/m.

Answer:

Second drop: 1.04 m

First drop: 1.66 m

Explanation:

Assuming the droplets are not affected by aerodynamic drag.

They are in free fall, affected only by gravity.

I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.

We can use the equation for position under constant acceleration.

X(t) = x0 + v0 * t + 1/2 * a *t^2

x0 = 0

a = 9.81 m/s^2

v0 = 0

Then:

X(t) = 4.9 * t^2

The drop will hit the floor when X(t) = 1.9

1.9 = 4.9 * t^2

t^2 = 1.9 / 4.9

[tex]t = \sqrt{0.388} = 0.62 s[/tex]

That is the moment when the 4th drop begins falling.

Assuming they fall at constant interval,

Δt = 0.62 / 3 = 0.2 s (approximately)

The second drop will be at:

X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m

And the third at:

X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m

The positions are:

1.9 - 0.86 = 1.04 m

1.9 - 0.24 = 1.66 m

above the floor

Answer:

Net charge,[tex]Q=7.84\times 10^{-6}\ C[/tex]

Explanation:

Number of protons, [tex]n_p=9\times 10^{13}[/tex]

Number of electrons, [tex]n_e=4.1\times 10^{13}[/tex]

Charge on electron, [tex]q_e=-1.6\times 10^{-19}\ C[/tex]

Charge on proton, [tex]q_p=1.6\times 10^{-19}\ C[/tex]

Net charge acting on the substance is :

[tex]Q=n_eq_e+n_pq_p[/tex]

[tex]Q=4.1\times 10^{13}\times (-1.6\times 10^{-19})+9\times 10^{13}\times 1.6\times 10^{-19}[/tex]

[tex]Q=0.00000784\ C[/tex]

or

[tex]Q=7.84\times 10^{-6}\ C[/tex]

So, the net charge on the substance is [tex]7.84\times 10^{-6}\ C[/tex]. Hence, this is the required solution.

Answer:[tex]h=160.84 W/m^2-K[/tex]

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied [tex]Q=mc\Delta T=0.3\times 2190\times (30)[/tex]

Heat due to convection =hA(100-30)

[tex]A=\pi d\cdot L[/tex]

[tex]A=\pi 0.05\times 10=1.571 m^2[/tex]

[tex]Q_{convection}=h\times 1.571\times (100-22)=122.538 h[/tex]

[tex]Q=Q_{convection}[/tex]

19,710=122.538 h

[tex]h=160.84 W/m^2-K[/tex]

The maximum theoretical efficiency for a heat engine operating between a high temperature of 300°C (573.15 K) and a low temperature of 27°C (300.15 K) is 47.63%, calculated using the Carnot efficiency formula.

To calculate the maximum theoretical efficiency for a heat engine operating between two temperatures, we can use the efficiency formula derived from the Carnot cycle, which is given by:

\(\eta = 1 - \frac{T_{cold}}{T_{hot}}\)

Where \(\eta\) is the efficiency, \(T_{cold}\) is the cold reservoir temperature, and \(T_{hot}\) is the hot reservoir temperature. Temperatures must be in Kelvin.

First, convert the temperatures from Celsius to Kelvin:

\(T_{cold} = 27 \degree C + 273.15 = 300.15 K\)

\(T_{hot} = 300 \degree C + 273.15 = 573.15 K\)

Now, substitute these values into the efficiency formula:

\(\eta = 1 - \frac{300.15}{573.15}\)

\(\eta = 1 - 0.5237\)

So, the maximum theoretical efficiency is:

\(\eta = 0.4763\)

Or in percentage:

\(\eta = 47.63\%\)

This calculation assumes an ideal Carnot engine, which is a theoretical limit and cannot be achieved in practical engines; the actual efficiency will be lower due to various inefficiencies.

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C   and q₂ = 11.0 n C

assume the distance be r from q₁  where the electric field is zero.

distance of point from q₂  be equal to 1.8 -r

now,

        E₁ = E₂

[tex]\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}[/tex]

[tex](\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}[/tex]

[tex]\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}[/tex]

1.8 = 6 r

r = 0.3 m

let us assume  q₁  be negative so, distance from  q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

   E₁ = E₂

[tex]\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}[/tex]

[tex](\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}[/tex]

[tex]\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}[/tex]

1.8 =4 r

r = 0.45 m

Answer:

19.642 cm

Explanation:

f₁ = Focal length of first lens = 11 cm

f₂ = Focal length of second lens = -25 cm

Combined focal length formula

[tex]\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\Rightarrow \frac{1}{f}=\frac{1}{11}+\frac{1}{-25}\\\Rightarrow \frac{1}{f}=\frac{14}{275}\\\Rightarrow f=\frac{275}{14}\\\Rightarrow f=19.642\ cm[/tex]

Combined focal length is 19.642 cm

The focal length of the combination of lenses is approximately 19.64 cm.

To find the focal length of the combination of lenses in this scenario, we need to use the lensmaker's formula for thin lenses in combination.

Given:

Focal length of lens 1, [tex]\( f_1 = +11 \)[/tex] cm

Focal length of lens 2, [tex]\( f_2 = -25 \)[/tex] cm

The formula for the focal length of two thin lenses in contact is given by:

[tex]\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \][/tex]

[tex]\[ \frac{1}{f} = \frac{1}{+11 \text{ cm}} + \frac{1}{-25 \text{ cm}} \][/tex]

[tex]\[ \frac{1}{f} = \frac{1}{11} - \frac{1}{25} \][/tex]

To subtract the fractions, find a common denominator, which is 275:

[tex]\[ \frac{1}{f} = \frac{25}{275} - \frac{11}{275} \][/tex]

[tex]\[ \frac{1}{f} = \frac{14}{275} \][/tex]

Now, invert both sides to solve for (f):

[tex]\[ f = \frac{275}{14} \][/tex]

[tex]\[ f \approx 19.64 \text{ cm} \][/tex]

Explanation:

Given that,

Frequency in the string, f = 110 Hz

Tension, T = 602 N

Tension, T' = 564 N

We know that frequency in a string is given by :

[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T}{m/L}}[/tex], T is the tension in the string

i.e.

[tex]f\propto\sqrt{T}[/tex]

[tex]\dfrac{f}{f'}=\sqrt{\dfrac{T}{T'}}[/tex], f' is the another frequency

[tex]{f'}=f\times \sqrt{\dfrac{T'}{T}}[/tex]

[tex]{f'}=110\times \sqrt{\dfrac{564}{602}}[/tex]

f' =106.47 Hz

We need to find the beat frequency when the hammer strikes the two strings simultaneously. The difference in frequency is called its beat frequency as :

[tex]f_b=|f-f'|[/tex]

[tex]f_b=|110-106.47|[/tex]

[tex]f_b=3.53\ beats/s[/tex]

So, the beat frequency when the hammer strikes the two strings simultaneously is 3.53 beats per second.

Final answer:

To calculate the beat frequency between two piano strings where one string's tension changes, it involves understanding sound production in instruments and the phenomenon of beats. However, without the length and mass of the strings, determining the beat frequency directly from the tension change is not straightforward.

Explanation:

The question involves calculating the beat frequency heard when a piano hammer strikes two strings tuned to the same note, where one string's tension has been altered. Beat frequency is the difference between the frequencies of two sounds. When two similar frequencies are played together, they produce beats that can be heard as a pulsation. However, to calculate the beat frequency from the given tensions, we first need to know the frequencies of the strings based on their tensions. Unfortunately, without specific information about the length and mass of the strings, calculating the exact frequencies and thus the beat frequency directly from the change in tension (602 N to 564 N) is not straightforward in this context. Normally, frequency can be related to tension in a string using the formula for the fundamental frequency of a vibrating string, which depends on the tension, length, and mass per unit length of the string. The question implies an understanding of the physical principles behind the production of sound in stringed instruments and the phenomenon of beats.

Answer:

1.  1.568 x 10^6 N m^2 / C

2. -  1.568 x 10^6 N m^2 / C

Explanation:

E = 4.9 x 10^4 N/C

Side of square, a = 8 m

Area, A = side x side = 8 x 8 = 64 m^2

Angle between lane of sheet and electric field = 30°

Angle between the normal of plane of sheet and electric field,

θ = 90°- 30° = 60°

The formula for the electric flux is given by

[tex]\phi = E A Cos\theta[/tex]

(1) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = 4.9 x 10^4 x 64 x Cos 60 = 1.568 x 10^6 N m^2 / C

(2) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = - 4.9 x 10^4 x 64 x Cos 60 = - 1.568 x 10^6 N m^2 / C

Answer: [tex]V_{0,x } = 297.7 \frac{m}{s}\\V_{0,y } = 637 \frac{m}{s}[/tex]

Explanation:

Hi!

We define the point (0,0) as the intial position of the ball. The initial velocity is [tex](V_{0,x}, V_{0,y})[/tex]

The motion of the ball in the horizontal direction (x) has constant velocity, because there is no force in that direction. :

[tex]x(t) = V_{0,x}t[/tex]

In the vertical direction (y), there is the downward acceleration g of gravity:

[tex]y(t) = -gt^2 + V_{0,y}t[/tex]

(note the minus sign of acceleration, because it points in the negative y-direction)

When the ball hits the ground, at t = 65s,  y(t = 65 s) = 0. We use this to find the value of the initial vertical velocity:

[tex]0 = t(-gt + V_{0,y})\\V_{0,y} = gt = 9.8 \frac{m}{s^2} 65 s = 637 \frac{m}{s}[/tex]

We used that g = 9.8 m/s²

To find the horizonttal component we use the angle:

[tex]\tan(65\º) = \frac{V_{y,0}}{V_{x,0}} = 2.14\\V_{x,0} = 297.7\frac{m}{s}[/tex]

Answer:

[tex]a=4m/s^{2}[/tex]

Explanation:

From the concept of average acceleration we know that

[tex]a=\frac{v_{2}-v_{1} }{t_{2}-t_{1}  }[/tex]

From the exercise we know that

[tex]v_{2}=30m/s\\v_{1}=10m/s\\t_{2}=5s\\t_{1}=0s[/tex]

So, the average acceleration of the car is:

[tex]a=\frac{30m/s-10m/s}{5s}=4m/s^{2}[/tex]

Answer:D

Explanation:

A secondary battery or rechargeable battery is a battery that can be discharged and recharged.

It is an elcetrochemical cell which involve redox reaction i.e. oxidation and reduction. Oxidation is a process of loosing the electrons while reduction involves gaining of electron.

During discharging battery act as galvanic cell in which Chemical energy is converted into Electrical energy.

During Charging battery act as Electrolytic cell in which Electrical energy is converted in to chemical energy.

Answer:

The snail travel at the end of 5 s with a velocity of 12 m/s and the distance of the snail is 22.5 m.

Explanation:

Given that, the initial velocity of the snail is,

[tex]u=2m/s[/tex]

And the acceleration of the snail is,

[tex]a=1m/s^{2}[/tex]

And the time taken by the snail is,

[tex]t=5 sec[/tex]

Now according to first equation of motion,

[tex]v=u+at[/tex]

Here, u is the initial velocity, t is the time, v is the final velocity and a is the acceleration.

Now substitute all the variables

[tex]v=2m/s+ 1 \times 5 sec\\v=7m/s[/tex]

Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.

Now according to third equation of motion.

[tex]v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a} \\[/tex]

Here, u is the initial velocity, a is the acceleration, s is the displacement, v is the final velocity.

Substitute all the variables in above equation.

[tex]s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m[/tex]

Therefore the distance of the snail is 22.5 m.

Answer:

The answer is [tex] -9.239 \times 10^{-8}\ N = [/tex]

Explanation:

The definition of electric force between two puntual charges is

[tex]F_e = \frac{K q_1 q_2}{d^2}[/tex]

where

[tex]K = 9 \times 10^9\ Nm^2/C^2[/tex].

In this case,

[tex]q_1 = e = 1.602\times 10^{-19}\ C[/tex],

[tex]q_2 = -e = -1.602\times 10^{-19}\ C[/tex]

and

[tex]d = 5 \times 10^{-11}\ m[/tex].

So the force is

[tex]F_e = -9.239 \times 10^{-8}\ N [/tex]

where the negative sign implies force of attraction.

Answer:

In second time period will be [tex]1.4\times 10^{-6}sec[/tex]

Explanation:

We have given the time period  of wave [tex]T=1.4microsecond[/tex]

We have to change this time period in unit of second

We know that 1 micro sec [tex]10^{-6}sec[/tex]

We have to change 1.4 micro second

To change the time period from micro second to second we have to multiply with [tex]10^{-6}[/tex]

So [tex]1.4microsecond =1.4\times 10^{-6}sec[/tex]

Answer:

the critical angle of the flint glass is 37.04⁰

Explanation:

to calculate the critical angle for total internal reflection.

given,

wavelength of the flint glass =  λ  = 580.0 nm

                                                       = 580 × 10⁻⁹ m        

critical angle  = sin^{-1}(\dfrac{\mu_a}{\mu_g})

at the wavelength of 580.0 nm the refractive index of the glass is 1.66

refractive index of air = 1                        

critical angle  = sin^{-1}(\dfrac{1}{1.66})

                      = 37.04⁰              

hence, the critical angle of the flint glass is 37.04⁰

Answer:

Explanation:

Resistance in series is given by the sum of all the resistor in series

value of Total Resistance is given by

[tex]R_{th}=R_1+R_2+R_3+R_4+..............R_n[/tex]

Where [tex]R_{th}[/tex] is the total resistance

[tex]R_1,R_2[/tex] are the resistance in series

Current in series remains same while potential drop is different for different resistor

The value of net resistor is always greater than the value of individual resistor.

If a there is a defect in a single resistor then it affects the whole circuit in series.