A block slides down a frictionless incline with constant acceleration. After sliding 6.80 m down, it has a speed of 3.80 m/s. What was the speed when the block had slid 3.40 m down? O 1.90 m/s O 2.69 m/s O None of the above

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

[tex]m1*g*h-\frac{m1*V_B^2}{2}=0[/tex]  Solving for Vb:

[tex]V_B=\sqrt{2gh}=6.56658m/s[/tex]

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

[tex]V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s[/tex]

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

[tex]m1*g*h2-\frac{m1*V_{B'}^2}{2}=0[/tex] Solving for h2:

h2 = 0.092m

Answer:

[tex]q=+8.34*10^{-7}C}[/tex]

Explanation:

The potential V due to a charge q,  at a distance r, is:

[tex]V=k\frac{q}{r}[/tex]

k=8.99×109 N·m^2/C^2      :Coulomb constant

We replace the values in order to find q:

[tex]q=\frac{V*r}{k}=\frac{500*15}{8.99*10^{9}}=8.34*10^{-7}C[/tex]

Answer:

i apolagize im late but yeah bois 700 points

Explanation:

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

Answer:

1868.5 m

Explanation:

For AB :

u = 0 m/s

a = + 1.68 m/s^2

t = 14.2 s

Let the distance is s1 and the velocity at B is v.

Use first equation of motion

v = u + at

v = 0 + 1.68 x 14.2 = 23.856 m/s

Use third equation of motion

[tex]v^{2}=u^{2}+2as_{1}[/tex]

[tex]23.856^{2}=0^{2}+2\times1.68\times s_{1}[/tex]

s1 = 169.38 m

For BC:

Let the distance is s2.

s2 = v x t

s2 = 23.856 x 68 = 1622.21 m

For CD:

u = 23.856 m/s

a = - 3.7 m/s^2

v = 0

Let the distance is s3.

Use third equation of motion

[tex]v^{2}=u^{2}+2as_{3}[/tex]

[tex]0^{2}=23.856^{2}-2\times 3.7 \times s_{3}[/tex]

s3 = 76.91 m

The total distance traveled is

s = s1 + s2 + s3

s = 169.38 + 1622.21 + 76.91 = 1868.5 m

Thus, the total distance traveled is 1868.5 m.  

Answer:

average and instant

Explanation:

The average speed is the ratio of the total path traveled and the time it took to travel that path, that is why the first space must be average speed, this because it takes into account the total amount of distance, and the total amount of time.

Instant speed, is the speed an objet (in this case a car) has in a particular moment in time, for this speed it doesn't matter the distance or the time that the car has traveled, it only matters the speed in that moment, that is what the speedometer measures, thus the second blank space must be instant speed.

Answer:

Inverted

Real

Explanation:

u = Object distance =  30 cm

v = Image distance

f = Focal length = 10 cm

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{30}\\\Rightarrow \frac{1}{v}=\frac{1}{15}\\\Rightarrow v=15\ cm[/tex]

As, the image distance is positive the image is real and forms on the other side of the lens

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-15}{30}\\\Rightarrow m=-0.5[/tex]

As, the magnification is negative the image is inverted

By applying the lens equation, we calculate that the image is formed 15 cm behind the lens. This is a real image as it forms on the opposite side of the lens, and in the case of a converging lens, it will be inverted.

First, we use the lens equation, which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. In this case, the object distance 'do' is 30cm and the focal length 'f' is 10cm. Solving for 'di', we find that the image is located 15 cm behind the lens (i.e., on the opposite side from the object).

Since the image forms on the opposite side of the lens from where the object is, this indicates it's a real image. A positive image distance indicates a real image and a negative image distance indicates a virtual image.

For a converging lens, a real image is always inverted, and a virtual image is always upright. Therefore, in this case, the image would be inverted.

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Answer:

electric flux is 280  Nm²/C  

so correct option is D 280  Nm²/C

Explanation:

radius r = 0.50 m

angle = 30 degree

field strength = 713 N/C

to find out

the electric flux through the surface

solution

we find here electric flux by given formula that is

electric flux = field strength × area× cos∅   .......1

here area = πr² = π(0.50)²

put here all value in equation  1

electric flux = field strength × area× cos∅  

electric flux = 713 × π(0.50)² × cos60

we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk

so we use cos60

electric flux = 280  Nm²/C

so correct option is D 280  Nm²/C

Answer:

The magnitude of the electric field is 129375 N/C toward south.

Explanation:

Given that,

Electric force [tex]F=2.07\times10^{-14}\ N[/tex]

(A). We need to calculate the magnitude of the electric field

Using formula of electric field

[tex]F = qE[/tex]

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

Where, q = charge of proton

E = electric field

[tex]E=\dfrac{2.07\times10^{-14}}{1.6\times10^{-19}}[/tex]

[tex]E=129375\ N/C[/tex]

(B). The direction of the electric field is toward the direction of the force.

So, The direction of the electric field is toward south

Hence, The magnitude of the electric field is 129375 N/C toward south.

Answer:

 % of water boils away= 12.64 %

Explanation:

given,

volume of block  = 50 cm³ removed from temperature of furnace = 800°C

mass of water = 200 mL = 200 g

temperature of water  = 20° C

the density of iron = 7.874 g/cm³ ,

so the mass of iron(m₁)  = density × volume = 7.874 × 50 g = 393.7 g

the specific heat of iron C₁ = 0.450 J/g⁰C

the specific heat of water Cw= 4.18 J/g⁰C

latent heat of vaporization of water is L_v = 2260 k J/kg = 2260 J/g

loss of heat from iron is equal to the gain of heat for the water

[tex]m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v[/tex]

[tex]393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260[/tex]

m₂ = 25.28 g

25.28 water will be vaporized

% of water boils away =[tex]\dfrac{25.28}{200}\times 100[/tex]

 % of water boils away= 12.64 %

The percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.

The heat transfer is the transfer of thermal energy due to the temperature difference.The heat flows from the higher temperature to the lower temperature.

The heat transfer of a closed system is the addition of change in internal energy and the total amount of work done by it.

As the initial volume of the iron block is 50 cm³ and the density of the iron is 7.874 g/cm³. Thus the mass of the iron block is,

[tex]m=50\times7.874\\m=393.7\rm g[/tex]

The temperature of the furnace is 800 degrees Celsius  and the specific heat of the iron block is 0.45 J/g-C.

As the boiling point of the water is 100 degree Celsius. Thus the heat loss by the block of iron is,

[tex]Q_L=393.7\times0.45\times(800-100)\\Q_L=124015.5[/tex]

The latent heat of the water is 2260 J/g. Thus the heat gain by vaporized water is,

[tex]Q_v=2260\times m_v\\[/tex]

Now the heat gain by the water is equal to the heat loss by the iron block.

As the specific heat of the water is 4.18 J/g-C and the temperature of the  water is 20 degrees and volume of water is 200 ml.

Thus heat gain by water can be given as,

[tex]Q_G=Q_L=200\times4.18(100-20)+2260m_v\\124015.5=200\times4.18(100-20)+2260m_w\\m_v=25.28\rm g[/tex]

Thus the total amount of the water boils away is 25.28 grams.

The percentage of the water boils away is,

[tex]p=\dfrac{25.25}{200}\times100\\p=12.64[/tex]

Thus the percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.

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Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

Explanation:

a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling  was 520 m.

b) As we are dealing with constant acceleration (due gravity) [tex]g=9.8 \frac{m}{s^{2}}[/tex] we can use the following formula:

[tex]y=y_{o}+V_{o} t-\frac{gt^{2}}{2}[/tex]   (1)

Where:

[tex]y_{o}=520 m[/tex]  is the initial height of the rocket (at the exact moment in which it stops due engines fail)

[tex]y=0[/tex]  is the final height of the rocket (when it finally hits the launch pad)

[tex]V_{o}=0[/tex] is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

[tex]t[/tex] is the time it takes to the rocket to hit the launch pad

Clearing [tex]t[/tex]:

[tex]0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}[/tex]   (2)

[tex]t^{2}=\frac{-520 m}{-4.9 m/s^{2}}[/tex]   (3)

[tex]t=\sqrt{106.12 s^{2}[/tex]   (4)

[tex]t=10.30 s[/tex]   (5)  This is the time

c) Now we need to find the final velocity [tex]V_{f}[/tex] for this rocket, and the following equation will be perfect to find it:

[tex]V_{f}=V_{o}-gt[/tex]  (6)

[tex]V_{f}=0-(9.8 m/s^{2})(10.30 s)[/tex]  (7)

[tex]V_{f}=-100.95 m/s[/tex]  (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = [tex]\frac{4\pi}{3}\times(0.0991^2-0.0882^2)[/tex]

or

volume of sphere = 0.0012 m³

also, volume of half sphere = [tex]\frac{\textup{Total volume}}{\textup{2}}[/tex]

or

volume of half sphere = [tex]\frac{\textup{0.0012}}{\textup{2}}[/tex]

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density =  [tex]\frac{\textup{Mass}}{\textup{Volume}}[/tex]

or

Density = [tex]\frac{\textup{0.568}}{\textup{0.0012}}[/tex]

or

Density = 474 kg/m³

Explanation:

Every rotating body experiences centrifugal force. Due to this force the body tends to bulge out around it mid point and gets flattened at the poles. Same is applicable to Earth as well. Since the Earth is rotating at a very high speed, its equator gets bulged out due to centrifugal force. Because of this bulged equator, Earth's pole to pole diameter and equatorial diameter has difference of around 42.76 km. It is flatter on the poles. This also proves that Earth is not a perfect sphere.

Answer and Explanation:

The reason for the not being perfectly spherical ad bulging out at the equator is that The centripetal force acting toward's the earth gravitational center tries to keep the Earth in perfect spherical shape.

Also the angular momentum of the orbiting planet influences the bulge,

The greater angular momentum results in more bulge while the lower value of it results in lesser bulge and more perfect spherical shape.

Also, a greater amount of force directed towards the center and acting on the object at the equator results in the bulges at the equator whereas at poles this force is not required and hence radius is lower in that region.

Answer:

The acceleration and the distance are 25200 mi/h² and 0.1008 mi.

Explanation:

Given that,

Initial speed = 71 mi/h

Final speed = 50 mi/h

Time = 3.0 s

(a). We need to calculate the acceleration

Using equation of motion

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

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

Put the value in the equation

[tex]a=\dfrac{(50-71)\times3600}{3}[/tex]

[tex]a=-25200\ mi/h^2[/tex]

Negative sign shows the deceleration.

(b). We need to calculate the distance

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex](50)^2=(71)^2+2\times(-25200)\times s[/tex]

[tex]s=\dfrac{(50)^2-(71)^2}{-25200}[/tex]

[tex]s=0.1008\ mi[/tex]

Hence, The acceleration and the distance are 25200 mi/h² and 0.1008 mi.

The distance traveled by the car when the car is constantly deaccelerating at a rate of 25200 miles/h² is 0.0504 miles.

Given to us

Initial Velocity of the car, u = 71 miles/h

Final Velocity of the car, v = 50 miles/h

Time = 3.0 s  [tex]=\dfrac{3}{3600}[/tex] hour

According to the first equation of motion, acceleration can be written as,

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

substituting the values we get,

[tex]a=\dfrac{50-71}{\dfrac{3}{3600}}[/tex]

[tex]a=-25,200\rm\ miles/h^2[/tex]

Thus, the acceleration of the car is -25,200 miles/h².

According to the third equation of motion,

[tex]v^2-u^2=2as[/tex]

Substituting the values we get,

[tex](50)^2-(71)^2=2(-25200)s[/tex]

[tex]s = 0.0504 \rm\ miles[/tex]

Thus, the distance car travel during the braking period is 0.0504 miles.

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Answer:

The strength of the field is 22.84 N/C.

Explanation:

Given that,

Speed [tex]v= 7.5\times10^{5}\ m/s[/tex]

Distance = 7.0 cm

We need to calculate the acceleration

Using equation of motion

[tex]v^2-u^2=2as[/tex]

Put the value in the equation

[tex]0-(7.5\times10^{5})^2=2\times a\times7.0\times10^{-2}[/tex]

[tex]a =-\dfrac{(7.5\times10^{5})^2}{2\times7.0\times10^{-2}}[/tex]

[tex]a =-4.017\times10^{12}\ m/s^2[/tex]

We need to calculate the strength of the field

Using newton's second law and electric force

[tex]F = ma = qE[/tex]

[tex]-qE=-ma[/tex]

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

Put the value into the formula

[tex]E=\dfrac{9.1\times10^{-31}\times4.017\times10^{12}}{1.6\times10^{-19}}[/tex]

[tex]E=22.84\ N/C[/tex]

Hence, The strength of the field is 22.84 N/C.

Final answer:

The strength of the electric field that can stop an electron moving at 7.5 x 10^5 m/s within a distance of 7 cm can be calculated using the work-energy principle, where the work done by the electric field is equal to the change in kinetic energy of the electron.

Explanation:

The strength of the electric field required to bring an electron moving to the right at 7.5 x 105 m/s to rest in a distance of 7.0 cm can be found using the work-energy principle and knowing the force exerted by an electric field on a charge. The work done by the electric field is equal to the change in kinetic energy of the electron, which is initially ½ mv2 and becomes zero when the electron is at rest.

To calculate the strength of the field, we can use:

Work (W) = Electric field (E) x Charge (e) x Distance (d)

½ mv2 = E * e * d

where m is the mass of the electron, v is its initial velocity, e is the elementary charge (approximately 1.6 x 10−12 C), and d is the distance (7.0 cm or 0.07 m). We solve for E to find the strength of the electric field

Answer:

3.06 seconds time passes before the watermelon has the same velocity

watermelon going at speed 59.9 m/s

watermelon traveling when it hits the ground at speed is 79.19 m/s

Explanation:

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at    ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

so 3.06 seconds time passes before the watermelon has the same velocity

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t²    .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so  by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

so watermelon going at speed 59.9 m/s

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as      ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

so watermelon traveling when it hits the ground at speed is 79.19 m/s

Answer:

The wave equation is [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]

a sinusoidal wave can be u = Acos( ax + bt) + B*sin(ax + bt)

where A, a, B and b are real constants. (here you also can add a phase to the arguments of the sin and cosine)

then [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]b^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))

and [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]= [tex]ac^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))

then if a*c = b, this is a solution of the wave equation.

Answer:

a) You should position the cannon at 981 m from the wall.

b) You could position the cannon either at 975 m or 7.8 m (not recomended).

Explanation:

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

In a parabolic motion, the position of the flying object is given by the vector position:

r =( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)

where:

r = position vector

x0 = initial horizontal position

v0 = module of the initial velocity vector

α = angle of lanching

y0 = initial vertical position

t = time

g = gravity acceleration (-9.8 m/s²)

The vector "r" can be expressed as a sum of vectors:

r = rx + ry

where

rx = ( x0 + v0 t cos α ; 0)

ry = (0 ; y0 + v0 t sin α + 1/2 g t²)

rx and ry are the x-component and the y-component of "r" respectively (see figure).

a) We have to find the module of r1 in the figure. Note that the y-component of r1 is null.

r1 = ( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)

Knowing the the y-component is 0, we can obtain the time of flight of the cannon ball.

0 = y0 + v0 t sin α + 1/2 g t²

If the origin of the reference system is located where the cannon is, the y0 and x0 = 0.

0 = v0 t sin α + 1/2 g t²

0 = t (v0 sin α + 1/2 g t)         (we discard the solution t = 0)    

0 = v0 sin α + 1/2 g t

t = -2v0 sin α / g

t = -2 * 100 m/s * sin 53° / (-9.8 m/s²) = 16.3 s  

Now, we can obtain the x-component of r1 and its module will be the distance from the wall at which the cannon sould be placed:

x = x0 + v0 t cos α

x = 0 m + 100m/s * 16.3 s * cos 53

x = 981 m

The vector r1 can be written as:

r1 = (981 m ; 0)

The module of r1 will be: [tex]x = \sqrt{(981 m)^{2} + (0 m)^{2}}[/tex]

Then, the cannon should be placed 981 m from the wall.

b) The procedure is the same as in point a) only that now the y-component of the vector r2 ( see figure) is not null:

r2y = (0 ; y0 + v0 t sin α + 1/2 g t² )

The module of this vector is 10 m, then, we can obtain the time and with that time we can calculate at which distance the cannon should be placed as in point a).

module of r2y = 10 m

10 m = v0 t sin α + 1/2 g t²

0 = 1/2 g t² + v0 t sin α - 10 m

Let´s replace with the data:

0 = 1/2 (-9.8 m/s² ) t² + 100 m/s * sin 53 * t - 10 m

0= -4.9 m/s² * t² + 79.9 m/s * t - 10 m

Solving the quadratic equation we obtain two values of "t"

t = 0.13 s and t = 16.2 s

Now, we can calculate the module of the vector r2x at each time:

r2x = ( x0 + v0 t cos α ; 0)

r2x = (0 m + 100m/s * 16.2 s * cos 53 ; 0)

r2x = (975 m; 0)

Module of r2x = 975 m

at t = 0.13 s

r2x = ( 0 m + 100m/s * 0.13 s * cos 53 ; 0)

r2x = (7.8 m ; 0)

module r2x = 7.8 m

You can place the cannon either at 975 m or at 7.8 m (see the red trajectory in the figure) although it could be dangerous to place it too close to the enemy fortress!

Explanation:

The momentum of an object is given by :

[tex]p=m\times v[/tex]............(1)

m is the mas of the object

v is the speed of the object

According to question, arrow's mass is doubled and the speed is halved. So,

m' = 2m

v' = v/2

The new momentum becomes :

[tex]p'=2m\times \dfrac{v}{2}[/tex]

p' = mv

p' = p

So, the momentum remains the same. The momentum is changed by a factor of 1. Hence, this is the required solution.

Answer:

The answer is very close to [tex]N_e=2.989\times10^{26}[/tex], where [tex]N_e[/tex] is the number of electrons.

Explanation:

First we take into account that the block weighs 1Kg, and that the number of protons and electrons is the same. As the electron mass is tiny even compared to that of the proton and neutrons we can neglect it in our considerations.

Let's start by equating the mass of all protons and neutrons to the mass of the  of the object:

[tex]N_p m_p+N_n m_n=1[/tex]

Where [tex]N_p[/tex] and [tex]N_n[/tex] is the number of protons and neutrons respectively. [tex]m_p[/tex] and [tex]m_n[/tex] is the mass of an proton and a neutron respectively. Because the number of protons and neutrons is equal we can say the following [tex]N_p=N_n=N[/tex], thus we have:

[tex]N_p m_p+N_n m_n=N(m_e+m_p)=1Kg \implies N=\frac{1}{m_p+m_n} [/tex]

On the other hand we have that the sum of all charges is less than the absolute value of 1C, we can express this by the following:

[tex]-1<N_p\cdot e-N_e\cdot e<1[/tex]

[tex]\implies -1<N\cdot e-N_e\cdot e<1[/tex]

Where [tex]e[/tex] is the proton charge (same as for the electron). We continue with the inequality:

[tex]-N\cdot e-1<N_p\cdot e-N_e\cdot e<-N\cdot e+1[/tex]

[tex]\implies \frac{N\cdot e+1}{e}>N_e>\frac{N\cdot e-1}{e}[/tex]

[tex]\implies \frac{(m_e+m_p)^{-1}\cdot e+1}{e}>N_e>\frac{(m_e+m_p)^{-1}\cdot e-1}{e}[/tex]

We have the estimated number of electrons bound. Because

[tex](m_e+m_p)^{-1}\cdot e>>1[/tex] We can neglect the ones on the rightmost and leftmost parts of the inequality. We then have

[tex]N_e\approx\frac{(m_p+m_n)^{-1}\cdot e}{e}[/tex]

Using the table values of the mass of the proton, mass of the neutron and the electron charge e we get

[tex]N_e\approx\frac{(1.672\times 10^{-27}+1.674\times 10^{-27})^{-1}\cdot e}{e}=(1.672\times 10^{-27}+1.674\times 10^{-27})^{-1}[/tex]

[tex]\, =2.989\times 10^{26}[/tex] electrons

Answer:

(a) - 165.032 m/s

(b) 238.37 m/s

Explanation:

initial horizontal velocity, ux = 172 m/s

height, h = 1390 m

g = 9.8 m/s^2

Let it strikes the ground after time t.

Use second equation of motion in vertical direction

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

-1390 = 0 - 0.5 x 9.8 x t^2

t = 16.84 second

(a) Let vy be the vertical component of velocity as it strikes the ground

Use first equation of motion in vertical direction

vy = uy - gt

vy = 0 - 9.8 x 16.84

vy = - 165.032 m/s

Thus, the vertical component of velocity as it strikes the ground is 165.032 m/s downward direction.

(b)

The horizontal component of velocity remains constant throughout the motion.

vx = 172 m/s

vy = - 165.032 m/s

The resultant velocity is v.

[tex]v=\sqrt{172^{2}+165.032^{2}}[/tex]

v = 238.37 m/s

Thus, teh velocity with which it hits the ground is 238.37 m/s.

Answer:

Explanation:

Energy stored in a capacitor

= 1/2 CV²

C is capacitance and V is potential of the capacitor .

When capacitor is charged to 24 V ,

E₁ = 1/2 x 2.4 x 24 x24 = 691.2 J

When it is charged to 12 volt

E₂ = 1/2 CV²

.5 X 2.4 X 12 X12

= 172.8 J

Answer:

Explanation:

The car is moving with uniform velocity . Hence there is no acceleration in the car .It indicates that net force on the car is zero . Since force in backward direction is exerted by the wind,  to make net force zero , the forward push by the car must be equal to backward force by the wind. In other words

Forward force by car = backward force by wind = 5000 N.

Along the three Cartesian coordinates. The Miller indices of the plane are [tex](1/2,1/4,1/3)[/tex].

A system known as Miller indices is used to explain how crystal planes and directions inside a crystalline substance are oriented. They serve as a tool to depict the crystal lattice's three-dimensional configuration of atoms or lattice points. In order to describe the surfaces and axes of crystals, Miller indices are frequently used in crystallography.

Miller indices are a crucial tool that crystallographers use to convey crystallographic data and comprehend the geometric arrangement of atoms in crystals. They support the characterization of crystal structures, the prediction of material characteristics, and the comprehension of the microscopic behavior of materials.

Given:

Intercepts  =  [tex]9.66\ A, 19.32\ A, 14.49\ A[/tex]

Lattice constant, [tex]a = 4.83\ A[/tex]

The reciprocal of intercepts is given as:

[tex]r = (1/9.66),1/19.32,1/14.49)[/tex]

The Miller indices are given as:

[tex]M = r/(1/a)\\M = ((1/9.66),1/19.32,1/14.49))/(1/4.83)\\M = (1/2,1/4,1/3)[/tex]

Hence, along the three Cartesian coordinates. The Miller indices of the plane are [tex](1/2,1/4,1/3)[/tex].

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The Miller indices for a plane intercepting a body-centered cubic lattice with intercepts of 9.66 Å, 19.32 Å, and 14.49 Å are (6,3,4). These are found by taking reciprocals of the intercepts, then multiplying by a common factor to get the smallest set of integers.

The student's question is about finding the Miller indices of a plane in a body-centered cubic lattice with given lattice constants and plane intercepts. Miller indices describe the orientation of a plane or set of planes in a crystal lattice.

To find these, we first take the reciprocals of the intercepts, which in this case gives us 1/2, 1/4, and 1/3. We then need to multiply these by a common factor to eliminate any fractions and get the smallest set of integers. Multiplying by 12 gives us the Miller indices of (6,3,4).

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Answer:

[tex] v_f = 15 \frac{m}{s}  [/tex]

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum [tex]\vec{L}[/tex] is

[tex]\vec{L}  = \vec{r} \times \vec{p}[/tex]

where [tex]\vec{r}[/tex] is the position and [tex]\vec{p}[/tex] the linear momentum.

We also know that the torque is

[tex]\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )[/tex]

[tex]\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p} [/tex]

[tex]\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F} [/tex]

but, as the linear momentum is [tex]\vec{p} = m \vec{v}[/tex] this means that is parallel to the velocity, and the first term must equal zero

[tex]\vec{v} \times \vec{p}=0[/tex]

so

[tex]\vec{\tau} =   \vec{r} \times \vec{F} [/tex]

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

[tex]\vec{\tau}_{rod} =   0 [/tex]

this means, for the angular momentum measure from the rod:

[tex]\frac{d\vec{L}_{rod}}{dt} =   0 [/tex]

that means :

[tex]\vec{L}_{rod} = constant[/tex]

So, the magnitude of initial angular momentum is :

[tex]| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)[/tex]

but the angle is 90°, so:

[tex]| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| [/tex]

[tex]| \vec{L}_{rod_i} | = r_i * m * v_i[/tex]

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

[tex]| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s} [/tex]

[tex]| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s} [/tex]

For our final angular momentum we have:

[tex]| \vec{L}_{rod_f} | = r_f * m * v_f[/tex]

and the radius is 0.250 m and the mass is 2.00 kg

[tex]| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f [/tex]

but, as the angular momentum is constant, this must be equal to the initial angular momentum

[tex] 7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f [/tex]

[tex] v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg} [/tex]

[tex] v_f = 15 \frac{m}{s}  [/tex]

Answer:

15 m/s

Explanation:

L = mvr

Li = (2.00 kg)(0.750 m)(5m/s) = 7.5 kgm^2/s

conservation of angular momentum --> Li=Lf

Lf = 7.5 kgm^2/s

7.5 kgm^2/s = (2.00 kg)(0.250 m)(vf)

vf = 15 m/s

Answer:

110.27 m/s or 396.972 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.7

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.16 m²

v = Terminal velocity

F = ma

[tex]F=\frac{1}{2}\rho CAv^2\\\Rightarrow ma=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{ma}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{84\times 9.81}{1.21\times 0.7\times 0.16}}\\\Rightarrow v=110.27\ m/s[/tex]

Converting to km/h

[tex]\frac{110.27}{1000}\times 3600=396.972\ km/h[/tex]

Terminal velocity of the skydiver is 110.27 m/s or 396.972 km/h

Answer:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

t=1.07

Vo=0

we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

To get a sharp image of the Sun, the paper should be placed at the focal length of the lens, which is 10 centimeters away from the lens. The image is in focus at this point because the light rays from the Sun are effectively parallel when they reach the lens, which then focuses these rays at its focal point.

Given the sun is so far away, the light it emits is nearly parallel by the time it reaches Earth. When using a lens to cast an image of the Sun, the point where the image is in focus, that is, the focal point, is also the focal length of the lens.

In this case, the student uses a lens with f = 10 cm, meaning the focal length of the lens is 10 centimeters. To get a sharp image, the paper on which the image is being projected should be placed 10 cm away from the lens, or at the focal length of the lens. This is because the light is in sharp focus at this distance, creating a clear image on the paper.

An important concept here is that the Sun is an astronomical unit away, so the light rays from the Sun are essentially parallel when they reach the lens. The lens then focuses these parallel rays to its focal point, forming a sharp image at a distance equal to its focal length.

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Answer:

Part (a) 10.15 m

Part (b) Rising

Explanation:

Given,

part (a)

Let 't' be the time taken to reach the ball to the cross bar,

In x-direction,

[tex]\therefore r\ =\ u_xt\\\Rightarrow t\ =\ \dfrac{r}{u_x}\ =\ \dfrac{r}{ucos\theta}\\\Rightarrow t\ =\ \dfrac{36.0}{23.6cos45^o}\\\Rightarrow t\ =\ 2.15\ sec[/tex]

Let y be the height attained by the ball at time t = 2.15 sec,

[tex]y\ =\ u_yt\ \ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ usin\theta t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\  23.6\times sin45^o\times 2.15\ -\ 0.5\times 9.81\ 2.15^2\\\Rightarrow y\ =\ 13.205\ m[/tex]

Now Let H be the height by which the ball is clear the crossbar.

[tex]\therefore H\ =\ y\ -\ h\ =\ 13.205\ -\ 3.05\ =\ 10.15\ m[/tex]

part (b)

At the maximum height the vertical velocity of the ball becomes zero.

i,e, [tex]v_y\ =\ 0[/tex]

Let h be the maximum height attained by the ball.

[tex]\therefore v_y^2\ =\ u_y^2\ -\ 2gh\\\Rightarrow 0\ =\ (usin\theta)^2\ -\ 2gh\\\Rightarrow h\ =\ \dfrac{(usin\theta)^2}{2g}\\\Rightarrow h\ =\ \dfrac{23.6\times sin45.0^o)^2}{2\times 9.81}\\\Rightarrow h\ =\ 14.19\ m[/tex]

Hence at the cross bar the ball attains the height 13.205 m but the maximum height is 14.19 m. Therefore the ball is rising when it reaches at the crossbar.

Answer:

a) [tex]F_a=0.152 \mu N[/tex]

b) [tex]F_b=203.182 \mu N[/tex]

Explanation:

The center of mass of an homogeneous sphere is its center, therefore you can use Newton's universal law of gravitation to find both questions.

[tex]F_g=G\frac{m_1m_2}{d}[/tex]

[tex]G=6.674*10^{-11} NmKg^{-2}[/tex]

a) d = 19m

[tex]F_a = G\frac{8.4*10^{4}*9.8}{19^2}[/tex]

[tex]F_a=0.152 \mu N[/tex]

b) d = 0.52

[tex]F_b = G\frac{8.4*10^{4}*9.8}{0.52^2}[/tex]

[tex]F_b=203.182 \mu N[/tex]

Answer:

(a) GF = 1.522 x (10 ^ -7)  N

(b) GF = 2.032 x (10 ^ -4)  N

Explanation:

The magnitude of the gravitational force follows this equation :

GF = (G x m1 x m2) / (d ^ 2)

Where G is the gravitational constant universal.

G = 6.674 x (10 ^ -11).{[N.(m^ 2)] / (Kg ^ 2)}

m1 is the mass from the first body

m2 is the mass from the second body

And d is the distance between each center of mass

m2 is a particle so m2 it is a center of mass itself

The center of mass from the sphere is in it center because the sphere has uniform density

For (a) d = 19 m

GF = {6.674 x (10 ^ -11).{[N.(m ^ 2)] / (Kg ^ 2)} x 8.4 x (10 ^ 4) Kg x 9.8 Kg} / [(19 m)^ 2]

GF = 1.522 x (10 ^ -7)  N

For (b) d = 0.52 m

GF = 2.032 x (10 ^ -4)  N

Notice that we have got all the data in congruent units

Also notice that the force in (b) is bigger than the force in (a) because the distance is shorter

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

[tex]V=E\times d[/tex]

[tex]V=3000\ N/C\times 0.7\ m[/tex]

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

Answer:

[tex]g'=3.71\ m/s^2[/tex]

Explanation:

Given that,

Time period of a pendulum on the earth's surface, T₁ = 1.2 s

Time period of the same pendulum on Mercury, T₂ = 1.95 s

The time period of the pendulum is given by :

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

On earth :

[tex]T_1=2\pi \sqrt{\dfrac{l}{g}}[/tex]

[tex]1.2=2\pi \sqrt{\dfrac{l}{9.8}}[/tex].............(1)

Let g' is the acceleration due to gravity on Mercury. So,

[tex]1.95=2\pi \sqrt{\dfrac{l}{g'}}[/tex]............(2)

From equation (1) and (2) :

[tex]\dfrac{1.2}{1.95}=\sqrt{\dfrac{g'}{9.8}}[/tex]

[tex]g'=(\dfrac{1.2}{1.95})^2\times 9.8[/tex]

[tex]g'=3.71\ m/s^2[/tex]

So, the acceleration due to gravity on the mercury is [tex]3.71\ m/s^2[/tex]. Hence, this is the required solution.