An arrow is shot straight up in the air with an initial speed of 250 ft/s. If on striking the ground, it embeds itself 4.00 in into the ground, find the magnitude of the acceleration (assumed constant) required to stop the arrow in units of ft/sec^2

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:

[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:

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:

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:

[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:

[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: [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: 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:

width of fringes are increased

Explanation:

The width of central maxima is given by the following expression

Width = 2 x Dλ / d

Due to increased width ,  position of a fringe  moves away from the centre.

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]

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.

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: The average speed is 27,24 mph (exactly 1008/37 mph)

Explanation:

This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:

1h--->18mi

X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h

1h--->56mi

X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h

Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.

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:

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:[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:

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:

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

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

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: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 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:

a) 231 yards

b) 55.2 yards

c) 0 yards to 260 yards

d) Infinite values

Explanation:

This situation can be described as a horizontal line that begins at point [tex]P_{1}=0 yards[/tex] (Meredith's house) and ends at point [tex]P_{2}=260 yards[/tex] (Bus stop). Where [tex]x[/tex] is the varying distance from her house, which can be calculated in the following way:

x=Final Position - Initial Position

or

[tex]x=x_{f} - x_{i}[/tex]

a) For the first case Meredith is at position [tex]x_{i}=29 y[/tex] and the bus stop at position [tex]x_{f}=260 y[/tex]. So the distance Meredith is from the bus stop is:

[tex]x=260 y - 29 y=231 y[/tex]

b) For the second case the initial position is [tex]x_{i}=204.8 y[/tex] and the final position [tex]x_{f}=260 y[/tex]. Hence:

[tex]x=260 y - 204.8  y=55.2  y[/tex]

c) If we take Meredith's initial position at her house  [tex]x_{i}=0 y[/tex] and her final position at the bus stop  [tex]x_{f}=260 y[/tex], the value of  [tex]x[/tex] varies from 0 yards to 260 yards.

d) As Meredith walks from her house to the bus stop, the variable [tex]x[/tex] assumes infinite values, since there are infinite position numbers from [tex]x=0 yards[/tex] to [tex]x=260 yards[/tex]

The answers to the possible distance covered by Meredith at the various distances from her house are;

A) distance = 231 yards

A) distance = 231 yardsB) distance = 55.2 yards

A) distance = 231 yardsB) distance = 55.2 yardsC) x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

A) We are told that meredith walks from her house to a bus stop that is 260 yards away.

After walking, she is now 29 yards from her house. This means that she has walked a total of 29 yards from her house.

Distance left to reach bus stop = 260 - 29 = 231 yards

B) We are told that Meredith is now 204.8 yards from her house. This means that she has walked a total of 204.8 yards from here house. Thus;

Distance left to reach bus stop = 260 - 204.8 = 55.2 yards.

C) This question is basically asking for all the possible values that Meredith could have walked from her house to the bus stop.

Since she starts from her house at 0m, then it means that if the bus stop is 260 m away, then if x is the possible distance, we can say that x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

Read more at; https://brainly.com/question/13242055

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:

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: 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:

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

Explanation:

Given that,

Initial sped of the runner, u = 0

Final speed of the runner, v = 30 ft/s

Acceleration of the runner, [tex]a=8\ ft/s^2[/tex]

Let t is the time taken by the runner. It can be calculated using first equation of motion as :

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{30-0}{8}[/tex]

t = 3.75 seconds

Let s is the distance covered by the runner. Using the second equation of motion as :

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

[tex]s=\dfrac{1}{2}\times 8\times (3.75)^2[/tex]

s = 56.25 feet

Hence, this is the required solution.

Final answer:

To reach a maximum speed of 30 f/s from rest with an acceleration of 8 f/s², it takes 3.75 seconds. During this time, the runner travels a distance of 56.25 feet.

Explanation:

The question involves calculating the time it takes for a runner to reach a maximum speed of 30 feet per second (f/s) from rest with an acceleration of 8 feet per second squared (f/s²), and then finding the distance traveled during this time. This can be solved using the basic kinematics equations.

Calculating Time to Reach Max Speed

To find the time, we use the equation v = at, where v is the final velocity (30 f/s), a is the acceleration (8 f/s²), and t is the time. Rearranging the equation to solve for t, we get t = v/a. Plugging in the values, t = 30 f/s / 8 f/s² = 3.75 seconds.

Calculating Distance Traveled

To find the distance traveled, we use the equation d = 0.5 * a * t², where d is the distance, a is the acceleration, and t is the time. Substituting the given values, d = 0.5 * 8 f/s² * (3.75 s)² = 56.25 feet.

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.