In attempting to
pass thepuck to a teammate, a hockeyplayer
gives it an initialspeedof 1.7
m/s. However, this speed is inadequate tocompensate for
the kinetic frictionbetween the puck and theice. As
a result, the puck travels only half the
distancebetweenthe players before slidingto a
halt. What minimum initial speed should the puck
havebeengiven so that it reached
theteammate, assuming that the same force of
kineticfrictionacted on the puck
everywherebetween the two players?

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: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:

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:

Part a)

[tex]v = 21.9 m/s[/tex]

Part b)

[tex]y = 4.17 m[/tex]

So he will not able to block the goal

Part c)

[tex]y = 0.98 m[/tex]

yes he can stop the goal

Explanation:

As we know by the equation of trajectory of the ball

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 3 m[/tex]

[tex]x = 45.7 m[/tex]

[tex]\theta = 45 degree[/tex]

now from above equation we have

[tex]y = 45.7 tan 45 - \frac{9.81(45.7)^2}{2v^2cos^245}[/tex]

[tex]3 = 45.7 - \frac{20488}{v^2}[/tex]

[tex]v^2 = 479.81[/tex]

[tex]v = 21.9 m/s[/tex]

Part b)

If lineman is 4.6 m from the football

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 4.6tan45 - \frac{9.81(4.6^2)}{2(21.9^2)cos^245}[/tex]

[tex]y = 4.17 m[/tex]

So he will not able to block the goal

Part c)

If lineman is 1 m from the football

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 1tan45 - \frac{9.81(1^2)}{2(21.9^2)cos^245}[/tex]

[tex]y = 0.98 m[/tex]

yes he can stop the goal

Answer:

u(intial velocty)= 10.688 m/s apprx

Explanation:

given data:

height of apartment = 16 m

time of hiitng = 1.02 s

Using equation of motion we have

h = ut + 0.5gt2 putting all value to get inital velocity value

16 = u(1.02)+ 0.5(9.8)(1.02)^2

SOLVING FOR u

16 - 5.09 = U1.02

u(intial velocty)= 10.688 m/s apprx

Direction will be towards ground

To find the ball's initial velocity, use the equation of motion for free fall and solve for v0. The initial velocity of the ball is approximately 10.78 m/s.

To find the ball's initial velocity, we can use the equation of motion for free fall: d = v0t + 0.5gt2, where d is the distance, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity. In this case, d = 16.0 m, t = 1.02 s, and g = 9.8 m/s2. Plugging in these values, we can solve for v0.

16.0 = v0(1.02) + 0.5(9.8)(1.02)2

Simplifying the equation gives us:

16.0 = 1.02v0 + 5.01

Subtracting 5.01 from both sides gives:

10.99 = 1.02v0

Finally, dividing both sides by 1.02 gives:

v0 ≈ 10.78 m/s

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

[tex]D = \sqrt{(d_{m})^2+(d_{w})^2}[/tex]

Where, [tex]d_{m}[/tex] =distance of Machmer Hall

[tex]d_{w}[/tex] =distance of Witless

Put the value into the formula

[tex]D = \sqrt{(400)^2+(180)^2}[/tex]

[tex]D=438.63\ m[/tex]

Hence, The distance from Witless to Machmer is 438.63 m.

Answer:

The volume of the slab is 2527.84 cm

Explanation:

Only the first two decimals are significant digits because so are the data that the problem gives us.

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:

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:[tex]\theta =32.08 ^{\circ}[/tex]

Explanation:

Given

Ball launcher is mounted on car at angle of [tex]39^{\circ}[/tex]

launching velocity is u=9.7 m/s

Speed of car =2.2 m/s

So in horizontal component of balloon speed of car will be added

Thus [tex]u_x=9.7cos39+2.2=7.53+2.2=9.73 m/s[/tex]

[tex]u_y=9.7sin39=6.10 m/s[/tex]

therefore Appeared trajectory angle is

[tex]tan\theta =\frac{6.10}{9.73}=0.627[/tex]

[tex]\theta =32.08 ^{\circ}[/tex]

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:

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:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × [tex]10^{-8}[/tex] s

speed = 0.8 c = 0.8 × 3 × [tex]10^{8}[/tex]

to find out

expected distance over

solution

we know c is speed of light in air is 3 × [tex]10^{8}[/tex] m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × [tex]10^{-8}[/tex] ×  0.8 × 3 × [tex]10^{8}[/tex]

expected distance = 4.32

so the expected distance is 4.32 m

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.

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:

a) 215 cm

b) 1290 mm

c) 2.81 m³

d) 265 cm

e) 1590 mm

f) 5.26 m³

Explanation:

Lower value

1 cubit = 43 cm

Length of cylinder = 5 cubits

So,

a) Length of cylinder = 5×43 = 215 cm

Diameter of cylinder = 3 cubit

1 cubit = 43 cm = 430 mm

b) Diameter of cylinder = 3×430 = 1290 mm

Radius of cylinder = 129/2 = 64.5 cm

Volume of cylinder

[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.645^2\times 2.15\\\Rightarrow v=2.81 m^3[/tex]

c) Volume of cylinder = 2.81 m³

Upper value = 53 cm

Length of cylinder = 5 cubits

So,

d) Length of cylinder = 5×53 = 265 cm

Diameter of cylinder = 3 cubit

1 cubit = 53 cm = 530 mm

e) Diameter of cylinder = 3×530 = 1590 mm

Radius of cylinder = 159/2 = 79.5 cm

Volume of cylinder

[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.795^2\times 2.65\\\Rightarrow v=5.26 m^3[/tex]

f) Volume of cylinder = 5.26 m³

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: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

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:

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.

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:

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]

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

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:

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:

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]

Answer:

Speed= 6cm/s and velocity= 6cm/s in the negative direction

Explanation:

the change in position is from 45cm to 27 cm (moving towards the negative x direction)

[tex] \Delta x = 45 cm - 27 cm = 18 cm[/tex]

And the change in time:

[tex] \Delta t= 3 s[/tex]

Now we must define the difference between speed and velocity:

Speed is a scalar quantity, which means that it is a number. Velocity ​​is also a number but you must also indicate the direction of the movement.

Thus, the speed is:

[tex] speed= \Delta x/ \Delta t = 18cm/3s=6cm/s[/tex]

An the velocity is:

6cm/s in the negative direction

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:

Ball A

Explanation:

Let the initial speed of the balls be u .

Angle of projection for ball A = 20°

Angle of projection for ball B = 75°

As we know that at highest point, the ball has only horizontal speed which always remains constant throughout the motion because the acceleration in horizontal direction is zero.

Speed of ball A at highest point = u Cos 20° = 0.94 u

Speed of ball B at highest point = u Cos 75° = 0.26 u

So, the ball A has bigger speed than B.

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.