A hot-air balloon is descending at a rate of 1.9 m/s when a passenger drops a camera. If the camera is 47 m above the ground when it is dropped, how much time does it take for the camera to reach the ground? If the camera is 47 m above the ground when it is dropped, what is its velocity just before it lands? Let upward be the positive direction for this problem.

Answer:

J = 14.4 kg*m^2

Explanation:

Assuming that the wheel is not moving anywhere, and the kinetic energy is only due to rotation:

Ek = 1/2 * J * w^2

J = 2 * Ek / (w^2)

We need the angular speed in rad / s

566 rev/min * (1 min/ 60 s) * (2π rad / rev) = 58.22 rad/s

Then:

J = 2 * 24400 / (58.22^2) = 14.4 kg*m^2

Answer:

a) 0.063 m/s

b)  243.83 m/s

Explanation:

given,

distance = 7900 km

time = 4 years

a) average velocity for the trip = [tex]\dfrac{7900}{4\times 365}[/tex]

                                                   = 5.41 km/day

                                                   = [tex]\dfrac{5.41\times 1000}{24\times 60\times 60}[/tex]

        average velocity for the trip = 0.063 m/s

b) average velocity  = [tex]\dfrac{7900\times 1000}{9\times 60\times 60}[/tex]

average velocity for the trip = 243.83 m/s

Answer:

"Magnitude of a vector can be zero only if all components of a vector are zero."

Explanation:

"The magnitude of a vector can be smaller than length of one of its components."

Wrong, the magnitude of a vector is at least equal to the length of a component. This is because of the Pythagoras theorem. It can never be smaller.

"Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction."

False. Magnitude of a vector is always positive.

"Magnitude of a vector can be zero if only one of components is zero."

Wrong. For the magnitude of a vector to be zero, all components must be zero.

"If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B."

Wrong. The magnitude of a vector depends on all components, not only the X component.

"Magnitude of a vector can be zero only if all components of a vector are zero."

True.

The correct statements are A and E.

A) True. The magnitude of a vector can indeed be smaller than the length of one of its components, especially when the vector has components in multiple directions.

B) False. The magnitude of a vector is always positive or zero, regardless of its direction. It is not negative if directed in the -x direction.

C) False. The magnitude of a vector can be zero only if all of its components are zero. Having only one component as zero does not guarantee a zero magnitude.

D) False. The magnitude of a vector depends on the vector's components in all directions, not just the component along the x direction. It cannot be immediately concluded that vector A has a bigger magnitude than vector B solely based on the x-component.

E) True. The magnitude of a vector can be zero only if all components of the vector are zero. This is a fundamental property of vectors; they have zero magnitude only if they have no components in any direction.

Complete question:

Which of the following statements about vectors is definitely true? (Section 3.3)

A) The magnitude of a vector can be smaller than the length of one of its components.

B) The magnitude of a vector is positive if it is directed in the +x direction and negative if it is directed in the -x direction.

C) The magnitude of a vector can be zero if only one of its components is zero.

D) If vector A has a bigger component along the x direction than vector B, it immediately means that vector A has a bigger magnitude than vector B.

E) The magnitude of a vector can be zero only if all components of the vector are zero.

Answer:0.558 m

Explanation:

Given

weight of Plank=228 N

weight of man=449 N

Length of plank=4.4 m

let [tex]R_a[/tex] and [tex]R_b[/tex] be the reactions at A & B

and Reaction at A becomes zero when plank is about to rotate at B

Taking moment about B at that instant so that plank is just about to rotate.

[tex]228\times 1.1=449\times x[/tex]

Where x is the maximum distance that man can walk

x=0.558 m

Answer:

Sound Intensity at microphone's position is [tex]9.417\times 10^{- 4} W/m^{2}[/tex]

The amount of energy impinging on the microphone is [tex]9.417\times 10^{- 8} W/m^{2}[/tex]

Solution:

As per the question:

Emitted Sound Power, [tex]P_{E} = 32.0 W[/tex]

Area of the microphone, [tex]A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex]

Distance of microphone from the speaker, d = 52.0 m

Now, the intensity of sound, [tex]I_{s}[/tex] at a distance away from the souce of sound follows law of inverse square and is given as:

[tex]I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}[/tex]

[tex]I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}[/tex]

Now, the amount of sound energy impinging on the microphone is calculated as:

If [tex]I_{s}[/tex] be the Incident Energy/[tex]m^{2}/s[/tex]

Then

The amount of energy incident per 1.00 [tex]cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex] is:

[tex]I_{s}(1.00\times 10^{- 4}) = 9.417\times 10^{- 4}\times 1.00\times 10^{- 4} = 9.417\times 10^{- 8} J[/tex]

Final answer:

The sound intensity at the position of the microphone is calculated using the formula Intensity = Power/Area. The amount of sound energy impinging on the microphone each second is found using the formula Energy = Power * Time.

Explanation:

To find the sound intensity at the position of the microphone, we can use the formula:

Intensity = Power/Area

Given that the sound power is 32.0W and the microphone has an area of 1.00cm^2 (converted to m^2 by dividing by 10000), we can calculate:

Intensity = 32.0W / (1.00cm^2 / 10000)

Next, to find the amount of sound energy impinging on the microphone each second, we can use the formula:

Energy = Power * Time

Since the time is 1 second, we have:

Energy = 32.0W * 1s

Therefore, the sound intensity at the position of the microphone is the calculated value, and the amount of sound energy impinging on the microphone each second is 32.0 joules.

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s[/tex]

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m[/tex]

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

Answer:

39.8 m

Explanation:

Let h be the required height and t be the time to reach the egg which was thrown later on. The egg thrown earlier took ( t +2.2) s , with initial velocity zero to cover distance of h.,

So

h = 1/2 g( t +2.2)²

For the egg thrown with velocity of 58 m/s

h = 58 t + 1/2 g t²

Equating these two equations

1/2 g( t +2.2)²  =  58 t + 1/2 g t²

1/2 g ( t² + 4.4 t + 4.84 ) = 58 t + 1/2 g t²

58 t = 21.56 t + 23.716

t = .65 s

h = 1/2 x 9.8 x (2.2+.65)²

h = 39.8 m

Answer:

It means at a diverging boundary when magma spreads to both sides it is almost identical in both the side.

Explanation:

At a diverging boundary when magma spreads to both sides it is almost identical in both the side.

If you take a picture of one side of the boundry and the spreading of sea floor and place it before a mirror you can see the image is identical to the picture of other side.

Therefore, the meaning of saying, each side of the sea floor away from the mid ocean ridge is a mirror image of the other side is explained above.

The statement that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side refers to the symmetric nature of seafloor spreading. Molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust, creating symmetric magnetic striping patterns. This is akin to the bilaterally symmetric structure observed in certain organisms.

When we say that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side, we are referencing the symmetric nature of seafloor spreading. Just as a mirror image reflects an object exactly, so does the seafloor on one side of the mid-ocean ridge reflect the seafloor on the opposite side. This is because molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust. This new crust then moves away from the ridge due to tectonic forces, and the process repeats, creating a pattern of symmetrical magnetic striping on the seafloor.

This 'mirroring' effect is similar to the bilaterally symmetric structure seen in certain organisms, where a plane cut from the front to back of the organism produces distinct left and right sides that are mirror images of each other. You can see this symmetry in the images of the Moon provided – two different sides, yet mirroring similar physical traits.

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Answer:

(a): [tex]F_e = 8.202\times 10^{-8}\ \rm N.[/tex]

(b): [tex]F_g = 3.6125\times 10^{-47}\ \rm N.[/tex]

(c): [tex]\dfrac{F_e}{F_g}=2.27\times 10^{39}.[/tex]

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053[tex]\times 10^{-9}[/tex] m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges [tex]q_1[/tex] and [tex]q_2[/tex] respectively is given by

[tex]F_e = \dfrac{k|q_1||q_2|}{r^2}[/tex]

where,

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, [tex]q_1 = +1.6\times 10^{-19}\ C.[/tex]

The charge on the electron, [tex]q_2 = -1.6\times 10^{-19}\ C.[/tex]

These two are separated by the distance, [tex]r = 0.053\times 10^{-9}\ m.[/tex]

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

[tex]F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.[/tex]

Part (b):

The gravitational force of attraction between two objects of masses [tex]m_1[/tex] and [tex]m_1[/tex] respectively is given by

[tex]F_g = \dfrac{Gm_1m_2}{r^2}.[/tex]

where,

For the given system,

The mass of proton, [tex]m_1 = 1.67\times 10^{-27}\ kg.[/tex]

The mass of the electron, [tex]m_2 = 9.11\times 10^{-31}\ kg.[/tex]

Distance between the two, [tex]r = 0.053\times 10^{-9}\ m.[/tex]

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

[tex]F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.[/tex]

The ratio [tex]\dfrac{F_e}{F_g}[/tex]:

[tex]\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.[/tex]

Answer: A) we can write a wave packet like Y = ∑aₙ*cos(w*n*t)

So you have lots of different wavelengths here, and you think that a wave packet has a unified frequency, but not wavelength.

B) mathematically speaking you will write a wave packet like i did up there, has a sum of many waves, where for each one you can have an intensity aₙ and a different frequency, the only thing you must see is that all the waves you are suming are moving in the same direction.

Answer:

if Y is the position and X the time: in the first one you will see a crescent function that starts sharp and starts to curve down as the time pases. as the cart is slowing down, you will need more time to move the same as before.

Y (position)

I

sensor-------------------------------------------------------------------

I                                                    o

I                                     o

I                           o

I                   o

I            o

I       o

I   o

I------------------------------------------------------------------------------------- X (time)

in the second case the cart starts close to the sensor and starts getting sharper and sharper as the time pases. This is because the velocity is increasing, so for each second that pases, you will travel more distance that the second before it.

Y (position)

I

sensor ----------------------

I       o

I                 o

I                          o

I                                 o

I                                       o

I                                            o

I                                              o

I------------------------------------------------------------------------------------- X (time)

i hope you can understand it, kinda hard to do graphs here.

Answer:

2.2 µm

Explanation:

For constructive interference, the expression is:

[tex]d\times sin\theta=m\times \lambda[/tex]

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle, [tex]\theta[/tex]  = 15.8°

First bright fringe means , m = 1

So,

[tex]d\times sin\ 15.8^0=1\times \597\ nm[/tex]

[tex]d\times 0.2723=1\times \597\ nm[/tex]

[tex]d=2192.43481\ nm[/tex]

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

Distance between slits ≅ 2.2 µm

Answer:

The separation between the slit is [tex]2.19\mu m[/tex]

Solution:

As per the question:

Wavelength of light, [tex]\lambda = 597 nm = 597\times 10^{-9} m[/tex]

[tex]\theta = 15.8^{\circ}[/tex]

Now, by Young's double slit experiment:

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

here,

n = 1

x = slit width

Therefore,

[tex]x = \frac{597\times 10^{-9}}{sin15.8^{\circ}} = 2.19\times 10^{- 6} m[/tex]

[tex]x = 2.19\mu m[/tex]

Final answer:

The average speed for the trip to the hospital, which covers a distance of 6.0 miles in 15 minutes, is 24 miles per hour (mph).

Explanation:

To calculate the average speed of the trip to the hospital, you take the total distance traveled and divide it by the total time taken. The student has specified that it takes 15 minutes to drive 6.0 miles in total. Therefore, the average speed can be calculated as follows:

Speed = Total Distance / Total Time
Speed = 6.0 miles / 15 minutes
Speed = 0.4 miles per minute

To convert this to miles per hour (mph), multiply by 60 (since there are 60 minutes in an hour):

Speed = 0.4 miles/minute × 60 minutes/hour
Speed = 24 mph.

Therefore, the average speed for the entire trip is 24 mph.

Answer:

p = 727.273 kg/m3,  y = 7134.545 N/m^3, SG = 0.7273

Explanation:

Density is simply the amount of mass of a substance per unit of volume. It can be found by dividing the mass in kg by the volume im m^3:

[tex]p = \frac{m}{v}  = \frac{40kg}{55lt*\frac{1 m^3}{1000 lt}} = 727.273 kg/m^3[/tex]

Specific weight is the weight of the substance per unit of volume. The weight is the mass of the material times the gravity, and it represents the force that the earth exerts on an object. Another way of calculate this value, its multiplying the density of the fuel times the gravity. Then:

[tex]y =  p*g = 727.273 kg/m^3 * 9.81 m/s^2 = 7134.545 N/m^3[/tex]

Specific Gravity is the ratio of the density of the substance to the density of a reference substance. For liquids, the reference substance is water at 4°C, which has a density of about 1000 kg/m^3.

[tex]SG =\frac{ p_{fuel}}{p_{water}}  = \frac{727.273 kg/m^3}{1000 kg/m^3} = 0.7273[/tex]

Answer:169.1 N

Explanation:

Given

bird weigh 38 pounds

and we know

1 Pound is equal to 0.453592 kg

Thus 38 pounds is equal to 17.236 kg

Thus the weight of bird is [tex]17.236\times 9.8=169.09 N \approx 169.1 N[/tex]

Answer:

The weight in newtons is 169.1.

Explanation:

This question can be solved as a simple rule of three problem.

We have that each pound has 4.45 newtons. So, how many newtons are there in 38 pounds?

So

1 pound - 4.45N

38 pounds - xN

[tex]x = 38*4.45[/tex]

[tex]x = 169.1N[/tex]

The weight in newtons is 169.1.

Answer:

Explanation:

length L = 5±.1

percentage error

= 0.1/5 x 100 = 2%

breadth b = 3.8 ±.1

percentage error

= .1 / 3.8 x 100

= 2.6 %

Maximum reading of length = 5.1 m

maximum reading of breadth = 3.9m

maximum area possible = 5.1 x 3.9

= 19.9 m².

maximum possible error in the measurement of area in percentage terms

= 2+2.6 = 4.6 %

error in maximum area

19.9 x 4.6 /100

= .9 m²

Minimum possible measurement of area

4.9 x 3.7 = 18.1 m²

possible error

18.1 x 4.6 / 100

= .8 m²

Answer:

20.88 mN

Explanation:

given data:

mass of object = 7.2 gm

acceleration of object = 2.9 m/s2

we know that force is given as

F = ma

where m is mass of object and a is acceleration of moving object.

putting all value to get required force

    = 7.2*10^{-3}\ kg *2.9 m/s2

   = 20.88*10^{-3} N

force in milli newton is

  = 20.88*10^{-3} * 1000 = 20.88 mN

Answer:

The speed of the ball when it hits the ground is 83.4 m/s

Explanation:

Please see the attached figure for a description of the problem.

The vector velocity at time "t" can be written as follows:

[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]

where:

v0 : module of the initial velocity vector.

α: launching angle.

g: acceleration due to gravity.

t: time

To calculate the impact speed, we can use this equation once we know the time at which the ball hits the ground. For this, we can use the equation for position:

[tex]r = ( x0 + v0*t*cos\alpha ; y0 + v0*t*sin\alpha + 1/2 g*t^{2})[/tex]

where:

r= vector position

x0 = horizontal initial position

y0 = vertical initial position

The problem gives us information about the vertical displacement. If we take the base of the wall as the center of the reference system, at the time at which the ball hits the ground, the module of the vertical component of the vector position, r, is 0 m (see figure). The initial vertical position is  22 m.

if ry is the vertical component of the vector r at final time:

[tex]ry = (0; y0 + v0*t*sin\alpha +1/2 g*t^{2})[/tex]

then, the module of ry is (see figure):

module ry =[tex]0 = 22m + v0*t*sin\alpha + 1/2*g*t^{2}[/tex]

Let´s replace with the given data:

[tex]0 = 22m + 39.3 m/s*t -4.9 m/s^{2}*t^{2}[/tex]

Solving the quadratic equation:

t = -0.5 and t = 8.5 s

At 8.5 s after firing, the ball hits the ground.

Now, we can find the module of the velocity vector when the ball hits the ground:

[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]

at time t = 8.5s

[tex]v = (81.0m/s * cos\alpha ; 81.0m/s * sin\alpha - 9.8 m/s^{2} *8.5s)[/tex]

v = (70.8 m/s; -44.0 m/s)

module of v = [tex]\sqrt{(70.8m/s)^{2} + (-44.0m/s)^{2}}[/tex] = 83.4 m/s

Answer:

4 %

2 ) 3.42 %

Explanation:

Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .

Given , 4 milligram is the maximum error possible .

Measured weight = 100 milligram

So percentage maximum potential error

= (4 / 100)  x 100

4 %

2 )

As per measurement

weight of 6 milliliters of water

= 48.540 - 42.745 = 5.795 gram

6 milliliters of water should measure 6 grams

Deviation = 6 - 5.795 = - 0.205 gram.

Percentage of error =(.205 / 6 )x 100

= 3.42 %

Answer:

H = 171.90 m

Explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H = [tex]\frac{1}{2} gt^2[/tex]    ......................1

here t is time and a is acceleration

so

we find t first

we know during time (t -1) s , it fall distance (H - 53.2) m

so equation of distance

[tex]H - 53.2 = \frac{1}{2} g (t-1)^2[/tex]

[tex]H - 53.2 = \frac{1}{2} g (t^2-2t+1)[/tex]

[tex]H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g[/tex]     ................2

now subtract equation 2 from equation 1 so we get

[tex]H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)[/tex]

53.2 = gt - [tex]\frac{1}{2} g [/tex]

53.2 = 9.81 t - [tex]\frac{1}{2} 9.8 [/tex]

t = 5.92 s

so from equation 1

H = [tex]\frac{1}{2} (9.81)5.92^2[/tex]

H = 171.90 m

Answer:

news will reach to the listener who are 42 km apart

Explanation:

given,

distance of the radios from the station = 42 Km

speed of the sound in the air = 343 m/s

distance of the people = 9.26 m

[tex]time =\dfrac{distance}{speed}[/tex]

time taken by the signal to reach to the radio

speed of the electromagnetic wave to reach to the people

speed of electromagnetic wave = 3 × 10⁸ m/s

[tex]t =\dfrac{42000}{3 \times 10^8}[/tex]

t = 1.4 μs

time taken to reach to the people

[tex]time =\dfrac{distance}{speed}[/tex]

[tex]t =\dfrac{9.26}{343}[/tex]

t = 27 ms

time taken by the station to radio is less.

hence, news will reach to the listener who are 42 km apart

Explanation:

Given that,

Separation between the plates, d = 3.1 mm = 0.0031 m

Capacitance of the capacitor, C = 23 pF

We need to find the area of the plate. The capacitance of a parallel plate capacitor is given by :

[tex]C=\dfrac{A\epsilon_0}{d}[/tex]

[tex]A=\dfrac{Cd}{\epsilon_0}[/tex]

[tex]A=\dfrac{23\times 10^{-12}\times 0.0031}{8.85\times 10^{-12}}[/tex]

[tex]A=0.0080\ m^2[/tex]

or

[tex]A=80\ cm^2[/tex]

So, the area of the plate is 80 square centimetres. Hence, this is the required solution.

The question is asking to calculate the required minimum deceleration of a speeding car to fit the speed limit when it crosses the end of a measured stretch of the road. The deceleration required can be calculated using the motion equations of physics by first calculating the time it takes to travel the distance between the two strips and then using this time to calculate the required deceleration.

The subject area of this question is kinematics, which is a branch of physics that deals with the concepts of distance, displacement, speed, velocity, and acceleration. The problem is asking for the minimum deceleration the car needs to have in order to not exceed the speed limit when it reaches the second strip. To solve this, you would use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is acceleration (in this case deceleration, so it will be negative), and t is time. But first, we need to figure out the time. The time it takes to travel between the two strips can be calculated by t = d/v. The distance d is given as 110m, and the average speed v we want is 21 m/s (the speed limit).

Once we have the time, we can substitute the values into the first equation and solve for a. The values are v = 21 m/s (the speed we want to have in the end), v0 = 33 m/s (the initial speed), and we've calculated t from the previous step. For deceleration, as the speed is decreasing, a will be a negative number.

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Answer:

[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]

Part c)

[tex]E = \frac{2.07 \times 10^5}{D}[/tex]

Explanation:

Part a)

As per Coulomb's law we know that force on a charge placed in electrostatic field is given as

[tex]F = QE[/tex]

now acceleration of charge is given as

[tex]a = \frac{QE}{M}[/tex]

now if charge moved through the distance D in electric field and its speed changes from vo to fraction f of speed of light c

then we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex](fc)^2 - v_o^2 = 2(\frac{QE}{M})D[/tex]

so we have

[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]

Part b)

Now using work energy theorem we can say that total work done by electric force on moving charge will convert into kinetic energy

So we will have

[tex]QED = \frac{1}{2}M(cf)^2 - \frac{1}{2}Mv_o^2[/tex]

so we have

[tex]E = \frac{M(c^2f^2 - v_o^2)}{2QD}[/tex]

Part c)

Now if an electron is accelerated using this field

then we have

[tex]M = 9.11 \times 10^{-31} kg[/tex]

[tex]Q = 1.6 \times 10^{-19} C[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

so we have

[tex]E = \frac{(9.1 \times 10^{-31})(0.9^2 - 0.001^2)\times 9 \times 10^{16}}{2(1.6 \times 10^{-19})D}[/tex]

[tex]E = \frac{2.07 \times 10^5}{D}[/tex]

Answer:

They have a difference in energy of 35 eV.

Explanation:

The energy at rest of a particle is given by:

[tex]E_{R} = m_{0}c^2[/tex]   (1)

Where [tex]m_{0}[/tex] is the mass of the particle at rest and c is the speed of light.

Beta particles are high energy and high velocity electrons or positrons ejected from the nucleus of an atom as a consequence of a radioactive decay. Either if the beta particle is an electron¹ or a positron² it will have the same mass.

Hence, the mass of the beta particle at rest in equation (1) will be equal to the mass of an electron:

[tex]m_{e} = 9.1095x10^{-31} Kg[/tex]

Replacing the values of [tex]m_{e}[/tex] and c in equation (1) it is gotten:

[tex]E_{R} = (9.1095x10^{-31} Kg)(3.00x10^{8} m/s)^{2}[/tex]

[tex]E_{R} = 8.19x10^{-14} Kg.m^{2}/s^{2}[/tex]

But [tex]1 J = Kg.m^{2}/s^{2}[/tex], therefore:

[tex]E_{R} = 8.19x10^{-14} J[/tex]

It is better to express the rest energy in electronvolts (eV):

[tex]1eV = 1.60x10^{-19} J[/tex]

[tex]8.19x10^{-14} J . \frac{1 eV}{1.60x10^{-19} J}[/tex] ⇒ [tex]511.875 eV[/tex]

[tex]E_{R} = 511.875 eV[/tex]

So the energy of the beta particle at rest is 511.875 eV.

Case for the one traveling at 0.35c:

Since it is traveling at 35% of the speed of light it is necessary to express equation (1) in a relativistic way, that can be done adding the Lorentz factor to it:

[tex]E = \frac{m_{0}c^{2}}{sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]   (2)

Where v is the velocity of the particle (for this case 0.35c).

[tex]E = \frac{511.875 eV}{sqrt{1-\frac{(0.35c)^{2}}{c^{2}}}}[/tex]

[tex]E = \frac{511.875 eV}{sqrt{1-\frac{0.1225c^{2}}{c^{2}}}}[/tex]

[tex]E = \frac{511.875 eV}{sqrt{1-0.1225}}[/tex]

[tex]E = \over{511.875 eV}{sqrt{0.8775}}[/tex]

[tex]E = \over{511.875 eV}{0.936}[/tex]

[tex]E = 546.875 eV[/tex]

The difference in energy between the two particles can be determined using the relativistic form of the kinetic energy:

[tex]K = E – E_{R}[/tex]  (3)

Where E is the energy of the particle traveling at 0.35c and [tex]E_{R}[/tex] is the energy of the beta particle at rest.

[tex]K = 546.875 eV – 511.875 eV[/tex]

[tex]K = 35 eV[/tex]

They have a difference in energy of 35 eV.

Key terms:

¹Electron: Fundamental particle of negative electric charge.

²Positron: Is an electron with positive electric charge (similar to an electron in all its properties except in electric charge and magnetic moment).

Answer:

D. Equalize voltage

Explanation:

Of the following lead-acid  battery the battery with voltage value is Equalize voltage. EQUALIZING lead acid battery is process of de-sulphating the battery by carrying out a controlled overcharge. When battery plates acquire sulphate coating over time, their efficiency reduces, by overcharging this sulpahte coating is blown-off and the battery is rejuvenated.

Answer:

a) The coin´s maximum height is 9.79 m above the ground.

b) The coin is 2.17 s in the air.

c) The speed is 13.82 m/s when the coin hits the ground

Explanation:

The equations for the position and velocity of the coin are the following:

y = y0 + v0 · t + 1/2 · g · t²

v =  v0 + g · t

Where

y = height at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

v = velocity at time t

a) At its max-height, the velocity of the coin is 0. Using the equation of velocity, we can obtain the time at which the velocity is 0.

v =  v0 + g · t

0 = 7.4 m/s - 9.8 m/s² · t

- 7.4 m/s / - 9.8 m/s² = t

t = 0.76 s

Now calculating the height of the coin at t = 0.76 s, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 7.4 m/s · 0.76 s - 1/2 · 9.8 m/s² · (0.76 s)²

y = 2.79 m

The coin´s maximum height above the ground is 7 m + 2.79 m = 9.79 m

b) After the coin reaches its maximum height, it falls to the ground. The initial position will be the max-height (2.8 m) and the final position is the ground (-7 m). The initial velocity, v0, will be 0, because the coin is at the max-height. Then, using the equation of position we can calculate the time the coin is falling:

y = y0 + v0 · t + 1/2 · g · t²

-7 m = 2.79 m - 1/2 · 9.8 m/s² · t²

-2 ·(-7 m - 2.79 m)/ 9.8 m/s² = t²

t = 1.41 s

The coin is (1.41 s + 0.76 s) 2.17 s in the air

c) Using the equation of velocity, we can calculate the speed at time 1.41 s, when the coin hits the ground.

v =  v0 + g · t

v = 0 m/s - 9.8 m/s² · (1.41 s)

v = -13.82 m/s

The speed is 13.82 m/s when the coin hits the ground.

Answer:14.72 m/s

Explanation:

Given

Initial velocity (u)=16.6 m/s

[tex]\theta =40.9^{\circ}[/tex]

Horizontal velocity component ([tex]u_x[/tex])=16.6cos40.9=12.54 m/s

As the ball comes down so its vertical displacement is zero except 3 m elevation

Thus [tex]v_y=\sqrt{\left ( 16.6sin40.9\right )^2+2\left ( -9.81\right )\left ( 3\right )}[/tex]

[tex]v_y=\sqrt{10.868^2-58.86}[/tex]

[tex]v_y=\sqrt{59.253}[/tex]

[tex]v_y=7.69 m/s[/tex]

there will be no change is horizontal velocity as there is no acceleration

Therefore Final Velocity

[tex]v=\sqrt{u_x^2+v_y^2}[/tex]

[tex]v=\sqrt{12.54^2+7.69^2}[/tex]

v=14.72 m/s

Answer:

[tex]4.50*10^8\frac{N}{C}[/tex]

Explanation:

The electric field is generated by a charge and represents the force exerted on a test charge, that is, the electric force per unit of charge. Therefore the equation for the electric field can be obtained from Coulomb's law.

[tex]E=\frac{F}{q}\\E=\frac{kq}{r^2}\\E=\frac{5C*8.99*10^9\frac{Nm^2}{C^2}}{(10 m)^2}=4.50*10^8\frac{N}{C}[/tex]

The magnitude of the electric field created by a larger charge Q at the location of a test charge is calculated by the formula E = k*Q/r² where k is Coulomb's constant, Q is the charge creating the field, and r is the distance between the charges. Substituting the given values into the formula, we find the electric field magnitude to be 4.495 x 10^8 N/C.

The magnitude of the electric field created by a point charge Q can be calculated using Coulomb's law as follows:

The equation for calculating the electric field E in relation to the force F imparted on a small test charge q is defined as E = F/q.

However, the Coulomb’s law gives the force F between two charges as F = k*Q*q/r², where, k is the Coulomb's constant (8.99 × 10^9 N.m²/C²), Q is the charge creating the field, q is the test charge, and r is the distance between them.

In the case where we want to find the electric field created by a larger charge Q at the location of a small charge q, we consider the force on the small charge exerted by Q, and therefore, we rewrite the equation as E = k*Q/r².

Therefore for your question, the magnitude of the electric field (E) at the location of the test charge is calculated as E = (8.99 x 10^9 N.m²/C² * 5.0 C) / (10m)² = 4.495 x 10^8 N/C.

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Answer:

[tex]n_l = 1.97[/tex]

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

[tex]\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

[tex]\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

[tex]\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

focal length of lens in liquid is

[tex]\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

                [tex]=\frac{n_{g} -n_{l}}{n_{l}}  [\frac{1}{(n_{g} - 1) f}[/tex]

rearrange fro[tex] n_l[/tex]

[tex]n_l = \frac{n_g f_l}{f_l+f(n_g-1)}[/tex]

[tex]n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}[/tex]

[tex]n_l = 1.97[/tex]

Final answer:

To calculate the refractive index of a liquid in which a convex lens acts as a diverging lens, the lensmaker's formula and given focal lengths are used. The formula is manipulated to solve for the refractive index of the liquid, providing an approximation of 0.2394 for the unknown index.

Explanation:

To find the refractive index of the liquid in which a convex lens becomes a diverging lens, we need to apply the lensmaker's formula, considering refractive indices of the lens material and the surrounding medium. The lens has a focal length of 30 cm in air, and its refractive index is 1.50. When the lens is immersed in the liquid, its focal length changes to -188 cm, indicating that it now diverges light rays.

We use the formula for the focal length of a lens in a medium:

1 / f = (n_lens / n_medium - 1) * (1 / R1 - 1 / R2)

which is derived from the lensmaker's formula, where f is the focal length, n_lens is the refractive index of the lens, n_medium is the refractive index of the surrounding medium, and R1 and R2 are the radii of curvature for the lens surfaces. For a thin lens, this can be simplified to:

f in medium = (n_medium / n_lens) * f in air.

Rearranging for n_medium, we get:

n_medium = (f in air / f in medium) * n_lens.

Plugging in the values, we calculate:

n_medium = (30 cm / -188 cm) * 1.50 = -0.2394 (approximately).

The negative sign indicates that we need to take the absolute value, so the refractive index of the liquid is approximately 0.2394.