A particular Bohr orbit in a hydrogen atom has a total energy of -0.28 eV . What is the kinetic energy of the electron in this orbit? What is the electric potential energy of the system?

Answer:

a) Reversible processes

b) 1222.73 J/K and 392.144 J/K

c) Because of breaking bonds that are a stable state, while accelerating molecules in the other is less difficult.

Explanation:

Since it is equality, it should be applied to reversible processes, and even better to isothermal processes, since the Temperature remains constant. The T is the absolute temperature (measured in °K) and dQ is the heat absorbed by the system, which DEPENDS on the process.

b) The heat absorbed in the fusion process depends on the latent heat of fusion, L, of the water, which is 334 J/g. It says t kg, but I assume it was a mistake in the typing, so the change in entropy is calculated for 1 kg of water melting as follows:

[tex]\Delta S=\frac{\Delta Q}{T}=\frac{mass*L}{T}=\frac{1kg*334J/g*(1000 g/kg)}{273.16 K} \\\Delta S= 1222.73 J/K[/tex]

Now, to proceed with the change of entropy for water heated from 273.16K to 300K we use the specific heat of water, which is 4184 J/kg°K as follows:

[tex]\Delta S=m_{mass}c_{water}*ln(\frac{T_f}{T_i} )=1kg*4184\frac{J}{Kg*K}*ln(\frac{300}{273.16})\\ \Delta S=392.144 J/K[/tex]

c) In the solid state, water molecules have different bonds with other water molecules creating the crystals. In the liquid state, each molecule moves freely with less interaction between molecules. So, it required more energy to break these bonds and alter this ordered state than just accelerating the molecules in the liquid state.

Answer:

a) The ball clears the crossbar by 10.6 m

b) The ball approaches the crossbar while falling

Explanation:

The position of the ball is described by the vector position r (see attached figure):

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

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle  

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s²)

The vector r is composed by rx and ry (see figure):

r = (rx ; ry)

a) Let´s find the time at which the ball flies a distance of  36.0 m. If at that time the vertical component of the vector r, ry, is equal or greater than 3.05 m, then the ball will clear the crossbar.

rx = x0 + v0 t cos α = 36.0 m

Since the origin of the system of reference is located where the kicker is, x0 = 0.

36.0 m = v0 t cos  α

36.0 m /(v0 cos  α) = t

36.0 m / (22.8 m/s * cos 51.0°) = t

t = 2.51 s

Now let´s calculate the height of the ball at that time:

ry = y0 + v0 t sin α + 1/2 g t²

Since the kicker is on the ground, y0 = 0

ry = 22.8 m/s * 2.51 s * sin 51.0° - 1/2 * 9.8 m/s² * (2.51 s)² = 13.6 m

Since the crossbar is 3.05 m high, the ball clears it by (13.6 m - 3.05 m) 10.6 m

b) Please see the figure to figure this out ;)

If the ball approaches the crossbar while still rising, the vertical component (vy) of the velocity vector will be positive. In change, if the ball approaches the crossbar while falling the vertical component of the velocity will be negative. See the figure.

The velocity vector is given by this equation:

v = (vx ; vy)

v = ( v0 cos α ; v0 sin α + g t)

Let´s see the vertical component at time t = 2.51

vy = v0 sin α + g t

vy = 22.8 m/s * sin 51.0° - 9.8 m/s² * 2.51 s

vy = -6.88 m/s

Then, the ball approaches the crossbar while falling.

Answer:

A. Howard Gardner

Explanation:

Howard Gardner was the educator who did not contribute to study or play.

- His parents fled Nazi Germany with their first son who died in accident before Howard was born.

-He was not allowed to play sports.

- He was an excellent pianist.

-He attended Harvard university.

-Influenced by Erickson

Among Howard Gardner, Jean Jacques Rousseau, Johann Amos Comenius, and Friedrich Froebel, Friedrich Froebel is the one most directly associated with the study of play for establishing the kindergarten system.

The question asks who among the listed educators did not contribute to the study or theory of play. Let's examine the contributions of each figure to determine the answer:

Considering the contributions of these educators, Friedrich Froebel is the one who stands out the most as being directly linked to the concept of play in education by virtue of establishing the kindergarten system.

Answer:

All true except the second one.

Explanation:

The two charges act on one another no matter how far apart they are, all the way to infinity . True. Coulomb law doesn't have a distance limitation.

If charge "A" is attracted to "B", then charge "B" is repelled from "A" . False. This violates Newton 3rd Law (Action-Reaction)

Plus and plus repel . True . Charges of same sign repel and opposite sign atract.

If charge "A" is attracted to "B", then charge "B" is equally attracted to "A" . True. This is Newton 3rd Law (Action-Reaction)

Minus and minus repel . True . Charges of same sign repel and opposite sign atract.

Plus and minus attract . True . Charges of same sign repel and opposite sign atract.

Answer:

The magnitude of the current circulating through the coil is 1.68 mA.

Explanation:

Given that,

Number of turns = 100

Length = 5 cm

Resistance = 12.0 Ω

Rate of magnetic field = 0.081 T/s

We need to calculate the magnitude of the current circulating through the coil

Using formula of current

[tex]I=\dfrac{e}{R}[/tex]

[tex]I=\dfrac{-NA\dfrac{dB}{dt}}{R}[/tex]

Where, A = area

N = number of turns

R = resistance

Put the value into the formula

[tex]I=\dfrac{100\times(0.05)^2\times0.081}{12.0}[/tex]

[tex]I=0.0016875\ A[/tex]

[tex]I=1.68\times10^{-3}\ A[/tex]

[tex]I=1.68\ mA[/tex]

Hence, The magnitude of the current circulating through the coil is 1.68 mA.

The question involves Faraday's law, which relates changing magnetic fields to induced currents. Given the rate of change of the magnetic field and the resistance of the coil, the magnitude of the induced current is calculated to be 1.67 mA.

This question involves the concept of electromagnetic induction in physics. According to Faraday's law, a changing magnetic field will induce an electromotive force (EMF) in a circuit. In this case, we're given that the magnetic field is decreasing at a rate of 0.081 T/s. This rate of change can be used to calculate the induced EMF, which is equal to the rate of change of the magnetic flux.

The magnetic flux (Φ) through the coil is given by Φ = NBA, where N is the number of turns in the coil, B is the magnetic field, and A is the area of the coil. So, the rate of change of magnetic flux (dΦ/dt) is N*d(BA)/dt. Because the area of the coil is constant, this simplifies to NAdB/dt. A is 0.0025 m^2, N is 100, and dB/dt is -0.081T/s, so dΦ/dt is -0.02 Wb/s.

Faraday's law tells us that the induced EMF (ε) is equal to -dΦ/dt, so ε has magnitude 0.02 V. If we define positive current as current that would generate a magnetic field opposing the decrease in the existing field (Lenz's Law), the induced EMF 'pushes' positive current in this direction. The current (I) can then be calculated using Ohm's Law (V = IR), giving I = ε/R = 0.02V / 12 Ω = 0.00167 A, or 1.67 mA.

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Answer:

The difference between the velocity graph made walking at a steady rate means that its the same value in time, that means there's no slope on the graph, so its acceleration is 0

On the other hand, if the velocity is increasing with time, the slope of the graph becomes positive, which means that the acceleration of the particle is positive.

Answer:

book speed is 3.99 m/s

Explanation:

given data

mass m = 490 g = 0.490 kg

compressing x = 7.10 cm = 0.0710 m

spring constant k = 1550 N/m

to find out

book speed

solution

we know energy is conserve so

we can say

loss in spring energy is equal to gain in kinetic energy

so

[tex]\frac{1}{2}*k*x^2 = \frac{1}{2}*m*v^2[/tex]    ..................1

put here value

[tex]\frac{1}{2}*1550*0.071^2 = \frac{1}{2}*0.490*v^2[/tex]

v = 3.99 m/s

so book speed is 3.99 m/s

Answer:

It should fly 8° to west of south at 430km/h

Explanation:

According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.

[tex]V_{w}*cos(45) = V_{A}*sin(\alpha )[/tex]   Solving for α:

α = 8.03°

Answer:

Explanation:

Initial velocity u = 36 km/h = 10 m /s

v = 0 , accn = a , Time taken to stop = t , distance covered to stop = s

v² = u² - 2as

u² = 2as

a = u² / 2s

= 10 x 10 / 2 x 17

= 2.94 ms⁻²

Force applied = mass x acceleration

= 15000 / 9.8 x 2.94

= 4500 N

b )

v = u + at

0 = 10 - 2.94 t

t = 10 / 2.94

= 3.4 s

c )

from the relation

u² = 2as

it is clear that stopping distance is proportional to u², if acceleration a is constant .

If initial speed u is doubled , stopping distance d will become 4 times or 17 x 4 = 68 m .

d )

u = at

if a is constant time taken to stop will be proportional to initial velocity.

If initial velocity is doubled , time too will be doubled. Or time will become

3.4 x 2 = 6.8 s .

Answer:

u = 12962.11 m/s

Explanation:

Given that,

The radius of mercury, [tex]r=2440\ km=2440\times 10^3\ m[/tex]

Mass of Mercury, [tex]M=3\times 10^{24}\ kg[/tex]

Final speed of the object, v = 2000 m/s

Let u is its initial speed when the object is far from  Mercury. It can be calculated by applying the conservation of energy as :

Initial kinetic energy + gravitational potential energy = final kinetic energy

[tex]\dfrac{1}{2}mu^2+(-\dfrac{GmM}{r})=\dfrac{1}{2}mv^2[/tex]

[tex]\dfrac{1}{2}u^2+(-\dfrac{GM}{r})=\dfrac{1}{2}v^2[/tex]

[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+\dfrac{GM}{r}[/tex]

[tex]u^2=2\times (\dfrac{1}{2}v^2+\dfrac{GM}{r})[/tex]

[tex]u^2=2\times (\dfrac{1}{2}(2000)^2+\dfrac{6.67\times 10^{-11}\times 3\times 10^{24}}{2440\times 10^3})[/tex]

u = 12962.11 m/s

So, the initial speed of the object is 12962.11 kg. Hence, this is the required solution.

Explanation:

(a) We need to convert following angles in radians. The conversion formula from degrees to radian is given by :

[tex]Radians=(\dfrac{\pi}{180})\times degrees[/tex]

1. If angle is 16.8°

[tex]Radians=(\dfrac{\pi}{180})\times 16.8[/tex]

16.8 degrees = 0.29 radians

2. If angle is 53.7°

[tex]Radians=(\dfrac{\pi}{180})\times 53.7[/tex]

53.7 degrees = 0.937 radians

3. If angle is 94.2°

[tex]Radians=(\dfrac{\pi}{180})\times 94.2[/tex]

53.7 degrees = 1.64 radians

(b) We need to convert following angles to degrees. The conversion formula from radian degrees to is given by :

[tex]Degrees=(\dfrac{180}{\pi})\times radians[/tex]

1. If angle is 0.258 radian

[tex]Degrees=(\dfrac{180}{\pi})\times 0.258[/tex]

0.258 radian = 14.78 degrees

2.  If angle is 1.01 radian

[tex]Degrees=(\dfrac{180}{\pi})\times 1.01[/tex]

0.258 radian = 57.86 degrees

3.  If angle is 7.51 radian

[tex]Degrees=(\dfrac{180}{\pi})\times 7.51[/tex]

0.258 radian = 430.29 degrees

Hence, this is the required solution.

Answer:

Equivalent capacitance, [tex]C'=3.9\ \mu F[/tex]

Explanation:

Capacitance, [tex]C_1=9\ \mu F[/tex]

Capacitance, [tex]C_2=13\ \mu F[/tex]

Capacitance, [tex]C_3=16\ \mu F[/tex]

Let C' is the equivalent capacitance of the combination of capacitors. It is given by :

[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}[/tex]

[tex]\dfrac{1}{C'}=\dfrac{1}{9}+\dfrac{1}{13}+\dfrac{1}{16}[/tex]

[tex]C'=3.99\ \mu F[/tex]

or

[tex]C'=3.9\ \mu F[/tex]

So, their equivalent capacitance is [tex]3.9\ \mu F[/tex]. Hence, this is the required solution.

Final answer:

The equivalent capacitance of a 9.0 μF, 13 μF, and a 16 μF capacitors connected in series is approximately 4.0 μF when expressed with two significant figures.

Explanation:

The question concerns the calculation of the equivalent capacitance when multiple capacitors are connected in series. To find the equivalent capacitance of the 9.0 μF, 13 μF, and 16 μF capacitors connected in series, you use the formula for capacitors in series:

1/Ceq = 1/C1 + 1/C2 + 1/C3

Substituting the given values:

1/Ceq = 1/9.0 μF + 1/13 μF + 1/16 μF

Calculating the reciprocals and summing them gives:
1/Ceq = 0.111 + 0.077 + 0.063

1/Ceq = 0.251

Now, the equivalent series capacitance, Ceq, can be calculated by taking the reciprocal of the result:
Ceq = 1 / 0.251

Ceq is approximately 3.98 μF.

To express this answer using two significant figures, the equivalent capacitance is 4.0 μF.

Answer:

water leak is 650 gallons

time required to full drain is 80 hrs

Explanation:

given data

volume V = 3200 gallon

rate = V(t) = 80 - t

to find out

how much water leak between 10 and 20 hour and  drain complete

solution

we know here rate is 80 - t

so here rate will be

[tex]\frac{dV(t)}{dt}[/tex] = 80 - t

and for time 10 and 20 hour

take integrate between 10 and 20

so water leak = [tex]\int\limits^ {20}_ {10} {(80-t)} \, dt[/tex]   .....................1

water leak = ( 80t - [tex]\frac{t^{2} }{2} )^{20}_{10}[/tex]

water leak = 650

so water leak is 650 gallons

and

we know here for full tank drain condition

water leak full = 80 t - [tex]\frac{t^{2} }{2} [/tex]

3200  = 80 t - [tex]\frac{t^{2} }{2} [/tex]

6400 = t² - 160 t

t = 80

so time required to full drain is 80 hrs

Answer:

The tension is 14 N

Explanation:

For this problem we have to use newton's law, so:

[tex]F=m*a[/tex]

The second string is connected to the mass of 8 kg, but the mass of 8 kg is connected to the mass of 20kg, so we can say that the second string is handling the two masses. so:

[tex]F=28kg*0.5\frac{m}{s^2}=14N[/tex]

Answer:

Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]

Explanation:

Tortoise travels an additional distance 580 m .

Since there is no acceleration, the distance traveled by any object is d = v t , where v is the total distance and t is the total time taken.

Speed of hare = [tex]v_{h}[/tex] = 9 m/s

Speed of tortoise = [tex]v_{t}[/tex] =5 m/s

Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]

Distance traveled by tortoise = d +580 m = 5 [tex]t_{t}[/tex]

Final answer:

The distance that the hare travels from the starting line after time t is represented by the expression 9t meters.

Explanation:

To write an expression that represents the hare's distance from the starting line in meters as a function of time, we use the formula for distance traveled at constant speed d = vt, where d is distance, v is speed, and t is time. Since the hare starts running at 9 meters per second, the distance the hare travels from the starting line after t seconds can be represented by the expression 9t meters. Note that we do not include the 580-meter head start given to the tortoise in the hare's expression since we are only considering the hare's distance from the starting line and not its position relative to the tortoise.

Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

m = 10 kg

Problem development

We replace the known data in formula (1)

1200 = 10*a

a = 1200/10

a = 120 m/s²

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by [tex]\frac{1}{2} mv^2[/tex]

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = [tex]\frac{1}{2} (4m)(v^2) = 2mv^2[/tex]

B) KE =[tex]\frac{1}{2} (3m)(2v)^2 = 6mv^2[/tex]

C) KE = [tex]\frac{1}{2} (2m)(3v)^2 = 9mv^2[/tex]

D) KE = [tex]\frac{1}{2} (3)(4v)^2 = 8mv^2[/tex]

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

Answer:

The position-time graph and the velocity-time graph has been shown in the figure attached.

Explanation:

Given:

For t = 0 s to t = 20 s,

The car moves with a constant velocity whose position-time and the velocity time graph has been shown with a red line.

For t = 20 s to t = 30 s,

The car moves with a constant acceleration whose position-time and the velocity time graph has been shown with a green line.

For t = 30 s to t = 50 s,

The car moves with a constant velocity whose position-time and the velocity time graph has been shown with a blue line.

For t = 50 s to t = 60 s,

The car moves with a constant deceleration whose position-time and the velocity time graph has been shown with a black line.

For a constant velocity, the velocity-time graph of the particle is a straight line parallel to a time axis and the position-time graph is a straight line inclined at some positive angle with the time axis.

For a constant acceleration, the velocity-time graph is a straight line incline at a positive angle with the time axis and the position-time graph is a parabolic curved line having upward concavity.  

For a constant deceleration, the velocity-time graph is a straight line incline at a negative angle with the time axis and the position-time graph is a parabolic curved line having downward concavity.

Answer:

Magnitude of the resultant force (Fn₁) on q₁

Fn₁ = 0.142N (directed toward the center of the square)

Explanation:

Theory of electrical forces

Because the particle q₁ is close to three other electrically charged particles, it will experience three electrical forces and the solution of the problem is of a vector nature.

Graphic attached

The directions of the individual forces exerted by q₂, q₃ and q₄ on q₁ are shown in the attached figure.

The force (F₁₄) of q₄ on q₁ is repulsive because the charges have equal signs and the forces (F₁₂) and (F₁₃) of q₂ and q₃ on q₁ are attractive because the charges have opposite signs.

Calculation of the forces exerted on the charge q₁

To calculate the magnitudes of the forces exerted by the charges q₂, q₃, and q₄ on the charge q₁ we apply Coulomb's law:

[tex]F_{12} = \frac{k*q_1*q_2}{r_{12}^2}[/tex]: Magnitude of the electrical force of q₂ over q₁. Equation((1)

[tex]F_{13} = \frac{k*q_1*q_3}{r_{13}^2}[/tex]: Magnitude of the electrical force of q₃ over q₁. Equation (2)

[tex]F_{14} = \frac{k*q_1*q_4}{r_{14}^2}[/tex]: Magnitude of the electrical force of q₄ over q₁. Equation (3)

Equivalences

1µC= 10⁻⁶ C

Known data

q₁=q₄= 1.96 µC = 1.96*10⁻⁶C

q₂=q₃= -1.96 µC = -1.96*10⁻⁶C

r₁₂= r₁₃ = 0.47m: distance between q₁ and q₂ and q₁ and q₄

[tex]r_{14} = \sqrt{0.47^2+ 0.47^2}=0.664m[/tex]

k=8.99x10⁹N*m²/C² : Coulomb constant

F₁₂ calculation

We replace data in the equation (1):

[tex]F_{12} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]

F₁₂ = 0.156 N Direction of the positive x axis (+x)

F₁₃ calculation

We replace data in the equation (2):

[tex]F_{13} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]

F₁₃ = 0.156 N Direction of the negative y axis (-y)

Magnitude of the net electrostatic force between F₁₃ and F₁₂

[tex]F_{n23}= \sqrt{0.156^2+0.156^2} = 0.22N[/tex] (directed toward the center of the square)

F₁₄ calculation

We replace F₁₄ data in the equation (3):

[tex]F_{14} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.664^2}[/tex]

F₁₄ = 0.078 N (In the opposite direction to Fn₂₃)  

Calculation of the resulting force on q₁: Fn₁

Fn₁ = Fn₂₃ - F₁₄ = 0.22 - 0.078 = 0.142 N

Final answer:

The net electrostatic force on a charge can be found using Coulomb's law, considering the effects of both attraction and repulsion due to the charges' arrangement at the corners of a square. Calculations account for diagonal opposites and adjacent corners, taking advantage of the square's symmetry.

Explanation:

To calculate the magnitude of the net electrostatic force on a charge in a configuration where two positive charges of 1.96 µC and two negative charges of the same magnitude are placed at the corners of a 0.47-m square, we can use Coulomb's law. Coulomb's law states that the electrostatic force (F) between two point charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. It is given by F = k * |q1*q2| / r², where k is Coulomb's constant (k ≈ 8.99 x 10⁹ N·m²/C²).

Since the charges are at the corners of a square, the opposing charges will attract, and like charges will repel. Each charge experiences forces due to its interaction with the other three charges. The net force on any charge is the vector sum of the forces exerted by the other three charges. This sum can be simplified as the square has symmetry so that the forces along one diagonal cancel and the forces along the sides of the square add together to pull the charge towards the center.

The diagonal distance (r) is √2 times the side of the square (a), which is r = √2 * 0.47 m. For simplicity, let's calculate the force between one charge and its diagonal opposite, and then double it for symmetry reasons (each charge has two diagonally opposite charges).

The force due to one diagonal is: F_diagonal = (k * (1.96 x 10⁻⁶)^2) / ((√2 * 0.47)²)
Now we need to calculate this force and then double it to account for both diagonals.

The final step will be to add the forces due to charges on adjacent corners, which are of the same polarity and hence exert repulsion. Their vector sum will be directed towards the square's center due to symmetry.

Answer:

Explanation:

Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds

s = 1/2at²

= .5 x a x 100

=50a

Velocity after 10 s

u = at = 10a

Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.

s = ut + 1/2 gt²

50a = -10ax20 + .5 x 10x400

250a = 2000

a = 8 m s⁻² .

Velocity after 10 s

= at = 80 m/s

further height reached with this speed under free fall

h = v² / 2g

= 80 x 80 / 2 x 10

= 320 m

height achieved under acceleration

= 50a

= 50 x 8 = 400m

Total height

= 320 + 400= 720 m

velocity after falling from 720 m

v² = 2gh = 2 x10 x 720

v = 120 m/s

Answer:

a) The initial acceleration is 7.84 m/s²

b) The impact speed is 117.6 m/s

c) The height is 705.6 m

Explanation:

a) The speed from A to B is:

v = u + at

Where

u = initial speed = 0

t = 10 s

Replacing:

v = 10t (eq. 1)

The vertical distance between A to B is:

[tex]h=\frac{1}{2} at^{2} +ut=\frac{1}{2} a*(10)^{2}+0=50a[/tex] (eq. 2)

From B to C, the time it take is equal to 20 s, then:

[tex]h=vt+\frac{1}{2} at^{2}[/tex]

Replacing eq. 1 and 2:

[tex]-50a=(10a*20)-\frac{1}{2} *g*20^{2} \\-250a=-\frac{1}{2} *9.8*20^{2} \\a=7.84m/s^{2}[/tex]

b) The impact speed is equal:

[tex]v_{i} ^{2} =v^{2} +2gs[/tex]

Where

s = h = -50a

[tex]v^{2} _{i} =(10a)^{2} +2*(-9.8)*(-50a)\\v=\sqrt{100a^{2}+980a } \\v=\sqrt{(100*7.84^{2})+(980*7.84) } =117.6m/s[/tex]

c) The height is:

[tex]v_{i} ^{2} =v^{2} +2gs\\0=(10a)^{2} -2gs\\(10a)^{2} =2gs\\s=\frac{(10a)^{2} }{2g} \\s=\frac{(10*7.84)^{2} }{2*9.8} =313.6m[/tex]

htotal = 313.6 + 50a = 313.6 + (50*7.84) = 705.6 m

Answer:

The speed of the object in the last instant prior to hitting the ground is -24.1 m/s

Explanation:

The equation for the position and velocity of the object will be:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

v0 = initial velocity

y0 = initial height

g = acceleration due to gravity

t = time

v = velocity at time t

We know that at its maximum height, the velocity of the object is 0. We can obtain the time it takes the object to reach the maximum height and with that time we can calculate the maximum height:

v = v0 + g · t

0 = 18.5 m/s - 9.8 m/s² · t

-18.5 m/s / -9.8 m/s² = t

t = 1.89 s

Now,let´s find the max-height:

y = y0 + v0 · t + 1/2 · g · t²

y = 12.2 m + 18.5 m/s · 1.89 s + 1/2 ·(-9.8 m/s²) · (1.89 s)²

y = 29.7 m

Now, let´s see how much it takes the object to hit the ground:

In that instant, y = 0.

y = y0 + v0 · t + 1/2 · g · t²

0 = 29.7 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²   (notice that v0 = 0 because the object starts from its maximum height, where v = 0)

-29.7 m = -4.9 m/s² · t²

t² = -29.7 m / -4.9 m/s²

t = 2.46 s

Now, we can calculate the speed at t= 2.46 s, the instant prior to hitting the ground.

v = v0 + g · t

v = g · t

v = -9.8 m/s² · 2.46 s

v = -24.1 m/s

Answer:264 m

Explanation:

Given

acceleration of rocket([tex]a_1[/tex])= 5 m/s^2[/tex]

velocity after 10 s

v=u+at

[tex]v=0+5\times 10[/tex]

v=50 m/s

after first stage rocket is ejected

acceleration of second stage=[tex]8 m/s^2[/tex]

distance between first and second part after 4 sec

[tex]s=u_1t+\frac{1}{2}at^2 [/tex]

here [tex]u_1=50 m/s[/tex]

[tex]s=50\times 4+\frac{1}{2}\times 8\times 4^2[/tex]

s=200+64=264 m  

Answer:

i)0.1213m

ii)0.2426 m

Explanation:

The formula to apply here are;

For a closed conduit at one end the fundamental frequency produced by a pipe of length L is ;

f=v/4L where v is speed of sound and L is length of conduit

707=343/4L

707*4L=343

2828L=343

L=343/2828

L=0.1213m

For an open conduit, the fundamental frequency produced by pipe of length L is given by;

f=v/2L where v is speed of sound and L is length of conduit

707=343/2L

707*2L=343

1414L=343

L=343/1414

L=0.2426 m

Answer:

Thickness = 103.38 nm

Explanation:

Given that:

The refractive index of the thin soap bubble = 1.33

The wavelength of the light = 550 nm

The minimum thickness that produces a bright fringe can be calculated by using the formula shown below as:

[tex]Thickness=\frac {\lambda}{4\times n}[/tex]

Where, n is the refractive index of the thin soap bubble = 1.33

[tex]{\lambda}[/tex] is the wavelength

So, thickness is:

[tex]Thickness=\frac {550\ nm}{4\times 1.33}[/tex]

Thickness = 103.38 nm

The object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s with a constant acceleration of 5 m/s^2.

To find the distance the object will move before it achieves a velocity of 6.4 m/s, we can use the equation v^2 = v0^2 + 2ax, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and x is the distance. Rearranging the equation, we get x = (v^2 - v0^2) / (2a). Plugging in the values, we have x = (6.4^2 - 4.3^2) / (2 * 5) = 10.12 m. Therefore, the object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s.

Final answer:

To find how far an object with constant acceleration a = 5 [tex]m/s^2[/tex] and initial velocity of 4.3 m/s moves before reaching a velocity of 6.4 m/s, use the kinematic equation [tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0). After calculation, the object will move 5.0 m before reaching 6.4 m/s.

Explanation:

We need to calculate how far the object moves before it achieves a velocity of v = 6.4 m/s given constant acceleration a = 5 [tex]m/s^2[/tex], initial velocity v0 = 4.3 m/s, and starting position x0 = 2.7 m. We can use the kinematic equation which relates final velocity, initial velocity, acceleration, and displacement:

[tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0)

Rearranging for displacement (x - x0):

(x - x0) = ([tex]v^2[/tex] - [tex]v0^2[/tex]) / (2a)

Substituting the given values:

(x - 2.7 m) = [tex](6.4 m/s)^2[/tex] - [tex](4.3 m/s)^2[/tex] / [tex](2 * 5 m/s^2)[/tex]

(x - 2.7 m) = (40.96 [tex]m^2/s^2[/tex] - 18.49 [tex]m^2/s^2[/tex]) / (10 [tex]m/s^2[/tex])

(x - 2.7 m) = 22.47 [tex]m^2/s^2[/tex] / (10 [tex]m/s^2[/tex])

(x - 2.7 m) = 2.247 m

x = 2.7 m + 2.247 m

x = 4.947 m

To the nearest 0.1 m, the object will move 5.0 m before it achieves a velocity of 6.4 m/s.

Answer:

The concentration downstream reduces to 0.03463mg/L

Explanation:

Initially let us calculate the time it is required to fill the lake, since for that period of time the pollutant shall remain in the lake before being flushed out.

Thus the detention period is calculated as

[tex]t=\frac{V}{Q}\\\\t=\frac{9000000}{7}=1.285\times 10^{6}seconds\\\\\therefore t = 14.872days[/tex]

Now the concentration of the pollutant after 14.872 days is calculated as

[tex]N_{t}=N_{0}e^{-kt}[/tex]

where

[tex]N_{o}[/tex] is the initial concentration

't' is the time elapsed after which the remaining concentration is calculated

k is the dissociation constant.

Applying values we get

[tex]N_{t}=3\times e^{-0.3\times 14.872}[/tex]

[tex]N_{t}=0.03463mg/L[/tex]

Final answer:

The force that the second child must exert to keep the door from moving is 10.5 N.

Explanation:

To find the force that the second child must exert to keep the door from moving, we need to consider the torque applied by each child.

First, we calculate the torque applied by the first child:

T1 = F1 * r1 * sin(θ)

T1 = (17.5 N) * (0.600 m) * sin(90°) = 10.5 N·m

The torque applied by the second child is:

T2 = F2 * r2

T2 = F2 * (0.450 m)

Since the door is not moving, the net torque must be zero. Therefore:

T1 + T2 = 0

Substituting the values:

10.5 N·m + F2 * (0.450 m) = 0

Simplifying, we find that F2 = -10.5 N * (0.450 m) / (0.450 m) = -10.5 N

Therefore, the force that the second child must exert to keep the door from moving is 10.5 N.

Answer:

Explanation:

The electric Force and gravitational Force are the same:

[tex]m*g=E*q[/tex]

[tex]E=m*g/q=1.0*10^{-2}*9.81/(2.9*10^{-6})=3.38*10^{4}N/C[/tex]

If the Electric field is doubled, the Electric Force is doubled, so its new value is twice the weight of the object. If we add this new electric force (upwards), with the weight (downwards), we have a resulting force upwards = one time the weight. in conclusion the acceleration will have a value equal to gravity but pointing upwards.

Answer:

Focal length: 44.16cm, Power: 2.2645 diopters

Explanation:

The object and image locations are the distance from the point to the lens.

object location =  25 - 2 = 23cm

image location = 50 - 2 = -48cm

the image location is negative since the image is virtual (the light rays do not pass through the image)

We can then use the lens equation to find the focal length.

1/object location + 1/image location = 1/focal length

(1/23)-(1/48)=1/f

Focal length = 44.16cm

To find the power we use the equation

D=1/f where f is the focal length in meters

44.16cm = 0.4416m

power = 2.2645 diopters

The prescription for glasses needed to correct the person's farsightedness should have lenses with a power of approximately 4.35 diopters.

To correct the vision of a person with a near point of 50.0 cm, we must find a lens that will allow them to clearly see objects at the normal near point of 25.0 cm. First, we calculate the lens's focal length (f), which is the distance at which it would bring parallel rays to focus. The formula for lens power (P) in diopters (D) is given by P = 1/f (in meters). The needed focal length is the new near point (25.0 cm, or 0.25 m) minus the distance the glasses are from the eye (2.00 cm, or 0.02 m). Thus, the focal length is f = 0.25 m - 0.02 m = 0.23 m. After conversion to meters, calculating the power gives P = 1/0.23 m ≈ 4.35 D.

The prescription for the glasses should therefore have lenses with a power of approximately 4.35 diopters to correct the vision to a normal near point.

Answer:

In order to avoid collision, the ferry must stop at a distance of 12.5 m from the dock.

Solution:

The initial velocity of the taxi, [tex]v_{o} = 5 m/s[/tex]

The minimum value of acceleration , [tex]a_{min} = - 1 m/s^{2}[/tex]

The maximum value of acceleration , [tex]a_{max} = 1 m/s^{2}[/tex]

Now,

When the deceleration starts the ferry slows down and at minimum deceleration of [tex]- 1 m/s^{2}[/tex], the ferry stops.

Thus, inthis case, the final velocity, v' is 0.

Now, to calculate the distance covered, 'd' in decelerated motion is given by the third eqn of motion:

[tex]v'^{2} = v_{o}^{2} + 2ad[/tex]

[tex]0^{2} = 5^{2} + 2\times (- 1)d[/tex]

[tex]d = 12.5 m[/tex]