What is the difference between Reynolds equation set and Navier Stokes equation?

Answer:

Option b

Explanation:

An object is said to fall freely when there is no force acting on the object other than the gravitational force. Thus the acceleration of the object is solely due to gravity and no other acceleration acts on the body.

Also the initial velocity of the body in free fall is zero and hence less than the final velocity.

As the body falls down closer to earth, it experiences more gravitational pull and the velocity increases as it falls down and the moment it touches the ground the period of free fall ends at that instant.

Thus the final velocity of an object in free fall is not zero because the final velocity is the velocity before coming in contact with the ground.

Answer:

v = 75  m/s

Explanation:

given data:

mass of gazelle is 25 kg

length of slope is 3 km

height of slope is 40 m

drag force is 20 N

jump height is 2 m

By work energy theorem we have

[tex]\frac{1}{2} mv^2 = F*L + mg(h_1 +h_2)[/tex]

[tex]\frac{1}{2}*25*v^2 = 20*3000 + 25 *9.8(40 + 2)[/tex]

solving for v

[tex]v = \sqrt{\frac{70290}{12.5}}[/tex]

v = 75  m/s

Answer:

The maximum height that the rocket reaches is 645.5 m.

Explanation:

Given that,

Mass = 10000 kg

Acceleration = 2.25 m/s²

Distance = 525 m

We need to calculate the velocity

Using equation of motion

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

Put the value in the equation

[tex]v^2=0+2\time2.25\times525[/tex]

[tex]v=\sqrt{2\times2.25\times525}[/tex]

[tex]v=48.60\ m/s[/tex]

We need to calculate the maximum height with initial velocity

Using equation of motion

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

[tex]h=\dfrac{v^2-u^2}{-2g}[/tex]

Put the value in the equation

[tex]h=\dfrac{0-(48.60)^2}{-2\times9.8}[/tex]

[tex]h=120.50\ m[/tex]

The total height  reached by the rocket is

[tex]h'=s+h[/tex]

[tex]h'=525+120.50[/tex]

[tex]h'=645.5\ m[/tex]

Hence, The maximum height that the rocket reaches is 645.5 m.

Answer:

t=12600ns

Explanation:

We use the relation between distance and velocity to solve this problem:

d=v*t

d=3.8km=3.8*10^3m

v=3*10^8 m/s^2         light's speed

we solve to find t:

t=d/v=(3.8*10^3)/(3*10^8)=1.26*10^(-5)s=12600*10^(-9)s=12600ns

Answer:

1.27 × 10⁴ ns

Explanation:

Given data

We can find the time (t) that it takes to the light to travel 3.80 km using the following expression.

v = d/t

t = d/v

t = (3.80 × 10³ m)/(3.00 × 10⁸ m/s)

t = 1.27 × 10⁻⁵ s

We know that 1 s = 10⁹ ns. Then,

1.27 × 10⁻⁵ s × (10⁹ ns/1s) = 1.27 × 10⁴ ns

Final answer:

To find the total increase in the length of each day over 43 centuries, we use the sum of an arithmetic series: S = n/2(2a1 + (n-1)d). With 4300 days in 43 centuries, an initial increase of 0 ms, and a daily increase of 0.002 ms, the total increase is 18471.9 milliseconds.

Explanation:

The length of the day increases because Earth's rotation is gradually slowing, primarily due to the friction of the tides. To calculate the total increase in the length of each day over 43 centuries, given that the length of a day at the end of a century is 1 ms longer than the day at the start of the century, we use the arithmetic progression formula where the total sum (S) is equal to n/2 times the first term (a1) plus the last term (an). In this case, the difference between each day (common difference, d) is constant at 0.002 ms.

The first term (a1) is 0 ms, as there is no increase on the first day of the first century, and the last term (an) after 43 centuries will be 43 ms, since we're told that each century contributes an additional 1 ms to the length of a day. So, for 43 centuries, we have n=4300 days (number of days in 43 centuries), a1=0 ms, an=43 ms, and d=0.002 ms/day.

Using the formula for the sum of an arithmetic series, S = n/2(2a1 + (n-1)d), we get the following:

S = 4300/2 [2(0) + (4300-1)(0.002 ms)]

S = 2150 [0 + 4299(0.002 ms)]

S = 2150 x 8.598 ms

S = 18471.9 ms

Therefore, over 43 centuries, the total increase in the length of each day adds up to 18471.9 milliseconds.

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:[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:

[tex] \Delta V=V_{2}-V_{1}=45.4V[/tex]

Explanation:

The energy, E, from a capacitor, with capacitance, C, and voltage V is:

[tex]E=\frac{1}{2} CV^{2}[/tex]

[tex]V=\sqrt{2E/C}[/tex]

If we increase the Voltage, the Energy increase also:

[tex]V_{1}=\sqrt{2E_{1}/C}[/tex]

[tex]V_{2}=\sqrt{2E_{2}/C}[/tex]

The voltage difference:

[tex]V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}[/tex]

[tex]V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V[/tex]

Answer:

E   = 5291.00 N/C

Explanation:

Expression for capacitance is

[tex]C = \frac{\epsilon  A}{d}[/tex]

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

[tex]C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}[/tex]

[tex]C = 24.06 \epsilon[/tex]

[tex]C = 24.06\times 8.854\times 10^{-12} F[/tex]

[tex]C =2.1\times 10^{-10} F[/tex]

We know that capacitrnce and charge is related as

[tex]V = \frac{Q}{C}[/tex]

 [tex]= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}[/tex]

v = 9.523 V

Electric field is given as

[tex]E = \frac{V}{d}[/tex]

   = [tex]\frac{9.52}{1.8*10^{-3}}[/tex]

E   = 5291.00 V/m

E   = 5291.00 N/C

Answer:

17.72° or 72.28°

Explanation:

u = 6.5 m/s

R = 2.5 m

Let the angle of projection is θ.

Use the formula for the horizontal range

[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]

[tex]2.5=\frac{6.5^{2}Sin2\theta }{9.8}[/tex]

Sin 2θ = 0.58

2θ = 35.5°

θ = 17.72°

As we know that the range is same for the two angles which are complementary to each other.

So, the other angle is 90° - 17.72° = 72.28°

Thus, the two angles of projection are 17.72° or 72.28°.

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:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, [tex]u_{x} = 10 m/s[/tex]

The distance between the buildings, [tex]d_{x} = 2.0 m[/tex]

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

Thus

From the second eqn of motion, we can write:

[tex]h = ut + \frac{1}{2}gt^{2}[/tex]

[tex]h = \frac{1}{2}gt^{2}[/tex]

[tex]2.5 = \frac{1}{2}\times 10\times t^{2}[/tex]

t = 0.707 s

Now,

When the policeman was chasing across:

[tex]d_{x} = u_{x}t + \frac{1}{2}gt^{2}[/tex]

[tex]d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m[/tex]

The distance they will meet at:

9.57 - 2.0 = 7.57 m

The policeman will fall a distance of 0.2 meters while jumping to the next building, calculated using the kinematic equation for vertical displacement under gravity.

The question relates to the topic of physics, specifically to the calculation of the distance fallen by an object under the influence of gravity, which is a fundamental kinematic concept. We want to find out how far the policeman will fall when jumping across to another building. We know that he's attempting to jump across a gap that is 2 meters wide, and the other building is 2.5 meters lower. Given that the acceleration due to gravity is approximated as 10 m/s² and assuming no air resistance, we can calculate the time it will take for the policeman to travel horizontally the 2 meters distance. Since the policeman's horizontal velocity is 10 m/s, the time to cross 2 meters is 2 meters / 10 m/s = 0.2 seconds.

Next, we need to calculate the vertical displacement during this time. Using the formula for vertical displacement s under constant acceleration a due to gravity, with initial vertical velocity u = 0 (since the policeman is only moving horizontally initially), s = u⋅t + 0.5⋅a⋅t². Substituting the values, we get s = 0⋅(0.2 s) + 0.5⋅(10 m/s²)⋅(0.2 s)². Simplifying this, we find s = 0.5⋅(10)⋅(0.04) = 0.2 meters.

Therefore, the policeman will fall a distance of 0.2 meters during the time it takes to jump across to the other rooftop.

Answer:

The rock thrown from the top of the cliff.

Explanation:

This is more of a conceptual question. The rock thrown from the top of the cliff will be accelerated downwards, that means, accelerated towards the bird nest, will the rock thrown from the bottom will be accelerated downwards too, but, in this case, this means that it will be accelerated against the direction of the bird nest.

The rock thrown downwards from the top of the cliff will hit the bird nest located at H/2 first, as it is continuously accelerated by gravity, unlike the upward-thrown rock which initially decelerates.

Which Rock Hits the Nest First?

When considering the rocks thrown by the two children - one upwards from the ground and one downwards from the top of the cliff - the rock that hits the bird nest located at H/2 first would be the one thrown downwards. This is based on the principles of kinematics in physics, which describe the motion of objects without considering the forces that cause the motion. The initial velocities of both rocks are the same in magnitude but opposite in direction; however, gravity only decelerates the upward-thrown rock and accelerates the downward-thrown rock, resulting in the latter reaching the nest quicker.

The rock thrown upwards will slow down as it reaches a height of H/2 and has to combat gravity's pull, which is not the case for the downward-throwing rock. Although both rocks are subjected to the same acceleration due to gravity, the downward-thrown rock will cover the distance to the nest faster due to its uninterrupted acceleration. Ignoring air resistance, we don't have to consider any opposing forces which might otherwise affect the time it takes for each rock to reach the nest.

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:

the force acting on the team mate is 1.19 kN.

Explanation:

given,

mass = 196 lbm

while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s

time taken for deceleration = 0.5 sec        

F = mass × acceleration

acceleration = [tex]\dfrac{0-6.7}{0.5}[/tex]              

                     = -13.4 m/s²                            

1 lbs  = 0.453 kg                      

196 lbs = 196 × 0.453  = 88.79 kg

F = 88.79 × 13.4                              

F = 1189.786 N = 1.19 kN                      

hence, the force acting on the team mate is 1.19 kN.

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:

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:

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 velocity of car is 11.71 m/s.

Explanation:

Given that,

Mass of car = 2000 kg

Radius = 20 m

Coefficient of friction = 0.70

We need to calculate the velocity of car

Using relation centripetal force and frictional force

[tex]F= \dfrac{mv^2}{r}[/tex]...(I)

[tex]F=\mu mg[/tex]...(II)

From equation (I) and (II)

[tex]\dfrac{mv^2}{r}=\mu mg[/tex]

[tex]v=\sqrt{\mu\times r\times g}[/tex]

Put the value into the formula

[tex]v=\sqrt{0.70\times20\times9.8}[/tex]

[tex]v=11.71\ m/s[/tex]

Hence, The velocity of car is 11.71 m/s.

Answer:

car can move at  11 m/s without skidding.

Explanation:

given,

mass of car = 2000 kg

radius of turn = 20 m

μ = 0.7

using centripetal force

F = [tex]\dfrac{mV^2}{R}[/tex].....................(1)

and we know                                

F = μ N  = μ m g........................(2)

equating both the equation (1) and (2)                        

μ m g = [tex]\dfrac{mV^2}{R}[/tex]

v = [tex]\sqrt{\mu R g}[/tex]                                

v = [tex]\sqrt{0.7 \times 20 \times 9.81}[/tex]

v = 11.71 m/s                                            

hence, car can move at  11 m/s without skidding.

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:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average  force exerted by the ball on the glove. It is solved using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

F = mg

[tex]\dfrac{1}{2}mv^2=Fh[/tex]

[tex]F=\dfrac{mv^2}{2h}[/tex]

[tex]F=\dfrac{0.14\times (32)^2}{2\times 0.25}[/tex]

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

Answer:

0.98°

Explanation:

The Sun appears to move across various constellation because of the Earth's revolution around it. This apparent motion is essentially same as the actual motion of the Earth. The orbit of Earth around the Sun is almost circular.

Time taken to complete one revolution = 365 days

Degrees traveled in 365 Days = 360°

The motion per day is

[tex]=\frac{360}{365}[/tex]

= 0.98° per day

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:

a) 4.325*10^10 V/m

b) 1.689*10^10 V/m

c) 0

d) 6.384 C/m^3

Explanation:

Hello!

The electric field of a sphere of uniform charge is a piecewise function, let a be the raius of the sphere

For r<a:

  [tex]E = kQ\frac{r}{a^{3}}[/tex]

For r>a:

[tex]E=kQ/r^{2}[/tex]

Since a=0.26m and k= 8.987×10⁹ N·m²/C²

 a)  

[tex]E= 8.987\times10^{9}\frac{0.47C \times0.18m}{(0.26m)^{3}}[/tex]

     E=4.325*10^10 V/m

b)

[tex]E= 8.987\times10^{9}\frac{0.47C}{(0.5m)^{2}}[/tex]

   E=1.689*10^10 V/m

c)

Since r --> ∞         1/r^2 --> 0

 E(∞)=0

d)

The charge density may be obtained dividing the charge by the volume of the sphere:

[tex]\rho = \frac{Q}{V} =\frac{0.47C}{\frac{4}{3} \pi (0.26m)^{3}}=6.384 C/m^{3}[/tex]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy [tex]\mathrm{EPE}[/tex].

[tex]\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}[/tex],

where

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

Apply Coulomb's law:

Initial potential energy:

[tex]\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}[/tex].

Final potential energy:

[tex]\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}[/tex].

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

[tex]\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm  -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}[/tex].

Answer:

Answer is c

Explanation:

trust me.

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:

Explanation:

The motion of the pendulum will be be SHM with time period equal to

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

l = .75 m , g = 9.8

T = [tex]2\pi\sqrt{\frac{.75}{9.8} }[/tex]

T = 1.73 s .

Time to reach the point of maximum velocity or maximum kinetic energy

= T /4

= 1.73 /4

= 0.43 s

For notes on Guitar , The formula is

n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]

n is fequency of the note , T is tension of string , m is mass per unit length of string , l is length of string.

For fundamental note , l = .648 m and f = 147 Hz

147 = [tex]\frac{1}{2\times.648} \sqrt{\frac{T}{m} }[/tex]

For G note

110 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{m} }[/tex]

[tex]\frac{147}{110} =\frac{l}{.648}[/tex]

l = 866 mm

Distance from the end of string

866 - 648 = 218 mm or 245 mm

Option c ) is correct .

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:

8.953 s

Explanation:

Here a time of 8953 ms is given.

Some of the prefixes of the SI units are

mili = 10⁻³

micro = 10⁻⁶

nano = 10⁻⁹

kilo = 10³

The number is 8953.0

Here, the only solution where the number of significant figures is mili

1 milisecond = 1000 second

[tex]1\ second=\frac{1}{1000}\ milisecond[/tex]

[tex]\\\Rightarrow 8953\ milisecond=\frac{8953}{1000}\ second\\ =8.953\ second[/tex]

So 8953 ms = 8.953 s

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