Suppose you made 5 measurements of the speed of a rocket:10.2 m/s, 11.0 m/s, 10.7 m/s, 11.0 m/s and 10.5 m/s. From these measurements you conclude the rocket is traveling at a constant speed. Calculate the mean, standard deviation, and error on the mean.

Answer:

The green car will catch the blue one after 10 s of the stop-light turning green.

Explanation:

The equation for the position of objects moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t

x0 = initial position

t = time

a = acceleration

v0 = initial velocity

For the blue car, this will be its position at time t

x = 0 m + 0 m/s · t + 1/2 · 0.5 m/s² · t²

x = 0.25 m/s² · t²

For the green car, that moves at constant speed, its position will be:

x = 0 m + v · t   (where v = velocity)

x = v · t

Now let´s find how much distance the blue car has traveled until the green car arrives at the stop-light:

x = 0.25 m/s² · (5s)² = 6.25 m

When the persecution begins (5 s after the stop-light turns green), the blue car is 6.25 ahead of the green car.

When the green car catches the blue one, its position is the same as the position of the blue car.

Then:

Position of the blue car = position of the green car

6.25 m + 0.25 m/s² · t² = v · t

0.25 m/s² · t² - v · t + 6.25 m = 0

Let´s use the formula for solving quadratic equations:

a = 0.25

b = -v

c = 6.25

(-b ± √(b²-4·a·c))/2·a

Let´s replace with the data:

(v ± √(v² - 4 · 0.25 m/s² · 6.25 m))/2· 6.25 m

(v ± √(v² - 6.25 m²/s²))/12.5 m

The minimum value of v that solves this equation is the value that makes the content inside the root to be null. A lower value than that and the content inside the root will be negative and have no solution (real solution).

Then:

v² - 6.25 m²/s² = 0

v² = 6.25 m²/s²

v = 2.5 m/s

Now the quadratic equation will be as follows:

0.25 m/s² · t² - 2.5 m/s · t + 6.25 m = 0

Solving this equation:

t = 5 s

Remember that this is the time that takes the green car to catch the blue one after the green car arrived at the stop-light (5 s after it turned green). If we count from when the light turns green, the elapsed time will be 10 s.

Answer:

36.13 m/s

Explanation:

initial speed, u = 56 km/h = 15.56 m/s

distance, s = 25 m

t = 2.11 s

Let the speed of the car at the impact is v and the acceleration is a.

Use second equation of motion

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

[tex]25=1.56 \times 2.11+\frac{1}{2}\times a \times 2.11^{2}[/tex]

a = 9.75 m/s^2

Use first equation of motion

v = u + at

v = 15.56 + 9.75 x 2.11

v = 36.13 m/s

Answer:

The maximum temperature is 90.06° C

Explanation:

Given that

t= 0.1 mm

Heat generation

[tex]q_g=0.3\ MW/m^3[/tex]

Heat transfer coefficient

[tex]h=500\ W/m^2K[/tex]

Here one side(left side) of the wall is insulated so the all heat will goes in to right side .

The maximum temperature will at the left side.

Lets take maximum temperature is T

Total heat flux ,q

[tex]q=q_g\times t[/tex]

[tex]q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2[/tex]

[tex]q=30\ W/m^2[/tex]

So the total thermal resistance per unit area

[tex]R=\dfrac{t}{K}+\dfrac{1}{h}[/tex]

[tex]R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}[/tex]

R=0.002 K/W

We know that

q=ΔT/R

30=(T-90)/0.002

T=90.06° C

The maximum temperature is 90.06° C

Answer: 7.66 8 m/s

Explanation:

Since the squirrel fell out and it is not thrown, we assume the initial velocity is zero ([tex]V_{o}=0[/tex]). On the other hand, the squirrel only experiences the acceleration due gravity, which is constant and in the downward direction [tex]g=-9.8 m/s^{2}[/tex]. So, the following equation will be useful to find the squirrel's final velocity [tex]V_{f}[/tex]:

[tex]{(V_{f})}^{2}={(V_{o})}^{2}-2gd[/tex]

Where  [tex]d=3 m[/tex] is the height from which the squirrel fell

As [tex]V_{o}=0[/tex]:

[tex]{(V_{f})}^{2}=-2gd[/tex]

Then:

[tex]{(V_{f})}^{2}=-2(-9.8 m/s^{2})(3 m)[/tex]

[tex]V_{f}=\sqrt{58.8 m^{2}/s^{2}}[/tex]

Finally:

[tex]V_{f}=7.668 m/s[/tex]

Answer:[tex]16.234^{\circ}C[/tex]

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water([tex]m_w[/tex])=600 gm

temperature of water =15

specific heat of water [tex]=4.186 j/g-^{\circ}C[/tex]

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.)[tex]=mgh=6\times 9.81\times 10=588.6 J[/tex]

Heat supplied[tex]=m_wc(\Delta T)[/tex]

[tex]H.E.=600\times 4.186\times (T-16) [/tex]

[tex]588.6=2511.6\times (T-16)[/tex]

T-16=0.234

[tex]T=16.234^{\circ}C[/tex]

This is not an efficient way of heating water as there is only[tex] 0.234^{\circ}C [/tex]increase in temperature.

Answer:

Explanation:

Let's use the equation Δy = [tex]v(t)-\frac{1}{2}(g)(t)^{2}[/tex]

That would mean -h = [tex]v(\frac{9.8}{2})-\frac{1}{2}(9.80)(\frac{9.8}{2})^{2}.

Since the ball has stopped at t = 9.8 = g, then that would mean that the final velocity v = 0.

-h = [tex](0)(\frac{9.8}{2})-(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]

[tex]h = -(\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]

[tex]h = (\frac{1}{2})(9.80)(\frac{9.8}{2})^{2}[/tex]

[tex]h = (\frac{9.8}{2})(\frac{9.8}{2})(\frac{9.8}{2})[/tex]

[tex]h = \frac{9.8^{3}}{2^{3}}[/tex]

[tex]h = \frac{941.192}{8}[/tex]

[tex]h = 117.649[/tex]

The height of the cannonball at [tex]t = \frac{9.8}{2}[/tex] should be 117.649 m

Hope this helps!

Answer:

[tex]h = 67.5\,m[/tex]

Explanation:

The position of the cannonball is given by the following expressions:

Half position

[tex]h = H -\frac{1}{2}\cdot g \cdot \left(\frac{t_{g}}{2} \right)^{2}[/tex]

Final position

[tex]0 = H -\frac{1}{2}\cdot g \cdot t_{g}^{2}[/tex]

The instant when the cannonball hits the ground is:

[tex]t_{g} = \sqrt{\frac{2\cdot H}{g} }[/tex]

Lastly, this result is applied in the other equation, which simplified afterwards:

[tex]h = H -\frac{1}{8}\cdot g \cdot \left(\frac{2\cdot H}{g} \right)[/tex]

[tex]h = H -\frac{1}{4}\cdot H[/tex]

[tex]h = \frac{3}{4}H[/tex]

[tex]h = 67.5\,m[/tex]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v²   ........1

6.5 × 1.6 × [tex]10^{-19}[/tex] = 1/2 × 1.672 × [tex]10^{-27}[/tex] ×v²

v = 3.5 × [tex]10^{7}[/tex] m/s

so

radius will be

radius = [tex]\frac{m*v}{B*q}[/tex]   ........2

radius =  [tex]\frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}[/tex]  

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

The diameter of the largest orbit, just before the protons exit the cyclotron is equal to 0.614 meter.

Given the following data:

Scientific data:

To calculate the diameter of the largest orbit, just before the protons exit the cyclotron:

First of all, we would determine the velocity of protons.

According to the law of conservation of energy, the work done in accelerating the proton is equal to the kinetic energy of the proton possesses. Mathematically, this is given by this expression:

[tex]qV_d = \frac{1}{2} MV^2[/tex]

Where:

Making V the subject of formula, we have:

[tex]V=\sqrt{\frac{2qV_d}{M} }[/tex]

Substituting the given parameters into the formula, we have;

[tex]V=\sqrt{\frac{2 \times 1.6 \times 10^{-19}\times \;6.5 \times 10^6 }{1.67 \times 10^{-27}} }\\\\V=\sqrt{\frac{2.08 \times 10^{-12} }{1.67 \times 10^{-27}}}\\\\V=\sqrt{1.25 \times 10^{15}}\\\\V=3.53 \times 10^7\;m/s[/tex]

In a magnetic field, diameter is given by this formula:

[tex]Diameter = \frac{2MV}{Bq} \\\\Diameter = \frac{2 \times 1.67 \times 10^{-27}\times 3.53 \times 10^7}{1.2 \times 1.6 \times 10^{-19}} \\\\Diameter = \frac{1.179 \times 10^{-19}}{1.92 \times 10^{-19}}[/tex]

Diameter = 0.614 meter

Read more on magnetic field here: https://brainly.com/question/12757739

Answer:

B. it will decrease

Explanation:

If the comb is negatively charged, it's because electrons went from the hair to the comb. These electrons have a certain mass, that is why the mass of the hair will decrease.

The mass of your hair will not change significantly when combing and the comb becomes negatively charged because the mass of electrons is extremely small. Small pieces of paper will be attracted to a negatively charged comb due to electrostatic forces.

When you comb your hair and the comb becomes negatively charged, it does so by gaining excess electrons from your hair. However, the mass of electrons is extremely small, so when considering the mass change of your hair due to the transfer of electrons, it's accurate to say that the mass will not change significantly enough to be noticeable or measurable with standard equipment. Therefore the correct answer is C. the mass will not change.

In the contexts of the related information provided, when you bring a negatively charged comb close to small pieces of paper, they will be attracted to the comb because of electrostatic forces. This is because the paper, being neutral, becomes polarized in the presence of the charged comb, leading to an attraction. Thus, the observed effect would be b. Pieces of paper are attracted to the comb.

The total work done on the crate by friction is -294.6 J for part (a) and -160.1 J for part (b).

The total work done on the crate by friction is determined by calculating the work done on each leg of the trip separately. For part (a), the crate is pushed 10.6 m south and then 10.6 m west. The work done on the crate by friction is the sum of the work done on each leg. The work done on the first leg is -219.1 J and the work done on the second leg is -75.5 J, so the total work done by friction is -294.6 J. For part (b), the crate is pushed along a straight-line path to the dock, traveling 15.0 m southwest. The work done on the crate by friction is -160.1 J.

https://brainly.com/question/30280752

#SPJ3

The total work done by friction on the crate when pushed south and then west (21.2 m total) is 1246.56 J, and when pushed directly southwest (15.0 m) is 882 J.

(a) Crate Pushed 10.6 m South and then 10.6 m West

In this case, the work done will be the sum of work done in pushing the crate 10.6 m south and in pushing the crate 10.6 m west.

Distance south: - 10.6 m = d₁

Distance west: - 10.6 m = d₂

force of friction = f = μ[tex]_k[/tex] N = 0.20 × m g = 0.20 × 30.0 kg × 9.8 m/s² = 58.8 N

∴ work done due south = W₁ = f d₁ = 58.8 N × 10.6 m = 623.28 N

and, work done due west = W₂ = f d₂ = 58.8 N × 10.6 m = 623.28 N

so total work done = W₁ + W₂ = 623.28 N + 623.28 N = 1246.56 N

(b) Crate Pushed Along a Straight-Line Path of 15.0 m Southwest

Let us analyze this problem with the vector approach:

Distance south: - 10.6 m

Distance west: - 10.6 m

The total displacement of the crate can be calculated using the Pythagoras Theorem:

d² = (10.6 m)² + (10.6 m)² = 224.72 m²

d = 14.99 m ≈ 15 m

the angle between the resultant vector and the negative x-axis can be given as :

θ = tan⁻¹ [tex]|\frac{10.6}{10.6} |[/tex] = 45°, the direction of the resultant vector be towards south-west (graphically in the [tex]3^r^d[/tex] quadrant)

The worker is applying a horizontal force, which means that the force of friction (f) will also be horizontal. So,

f = μ[tex]_k[/tex] N = 0.20 × m g = 0.20 × 30.0 kg × 9.8 m/s² = 58.8 N

then the work done by the force of friction will be,

W = f d = 58.8 N × 15 m

W = 882 J

Answer:

Acceleration, [tex]a=8.57\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the Cheetah, u = 5 m/s

Distance covered, s = 35 m

Final speed of the cheetah, v = 25 m/s

We need to find the acceleration required for the cheetah to reach its final speed. The formula is as follows :

[tex]a=\dfrac{v^2-u^2}{2s}[/tex]

[tex]a=\dfrac{(25)^2-(5)^2}{2\times 35}[/tex]

[tex]a=8.57\ m/s^2[/tex]

So, the acceleration of the cheetah is [tex]8.57\ m/s^2[/tex]. Hence, this is the required solution.

Answer:

a) is more precise

b) is more accurate

Explanation:

(a) 7.72, 7.74. 7.73. 7.75, 7.74

For this set the average is: (7.72+7.74+7.73+7.75+7.74)/5=7.736

(b) 7.86, 7.90, 7.78, 7.93, 7.83

For this set the average is: (7.86+7.90+7.78+7.93+7.83)/5=7.86

Accuracy refers to the closeness of a measured set of measures to a known value.In our case that the book value is 7.86g/cm3

We can see that the set (b) is more accurate.

Precision correspond to the closeness of the measurements to each other.

We can see that the set of measures a) is more precise, they are more close one of the each other than the set b)

Answer:

(a) is more precise

(b) is more accurate

Explanation:

When a given set of values are precise, this implies that all the values are close to each other but may not be accurate. But a set of values are accurate when they are compared and it can be observed that they are close to a given measured value. So we can infer that the values in option (a) are precise, while that in (b) are accurate when compared with the book value given.

Answer:

Car travel a distance of 60.06 m in 6 sec

Explanation:

We have given initial velocity v = 20 m/sec

Time = 6 sec

As the car stops finally so final velocity v = 0

From the first equation of motion

v = u+at (as the car velocity is slows down means it is a case of deceleration)

So v = u-at

[tex]0=20-a\times 6[/tex]

[tex]a=3.33m/sec^2[/tex]

Now from second equation of motion [tex]s=ut-\frac{1}{2}at^2=20\times 6-\frac{1}{2}\times 3.33\times 6^2=60.06m[/tex]

Answer:

E = 7.334 KeV

Explanation:

given,

initial energy = 60 keV

Δ x = 30 cm

[tex]E = E_0e^{-\mu \Delta x}[/tex]

μ(for soft tissue) = 0.7 cm⁻¹ (taken from table)

[tex]E = E_0e^{-0.7\times 20}[/tex]

         E = 60 × 0.1224

         E = 7.334 KeV

energy of x-ray photon after travelling through 30 cm soft tissue in body is E = 7.334 KeV

Answer:

1.5×10^4 N

Explanation:

mass of earth M= 6×10^(24) kg

mass of object m= 6.5×10^2 kg

distance between them R = 4.15×10^6 m

we know from the newtons law of gravitation

[tex]F=G\frac{Mm}{R^2}[/tex]

putting values in the above equation we get

[tex]F=6.67\times10^{-11}\frac{6\times10^{24}6.5\times10^2}{4.15^2\times10^{12}}[/tex]

F= 1.5×10^4 N

the gravitational force between earth and the object

= 1.5×10^4 N

The gravitational force exerted on a 6.50 x 10^2 kg object that is 4.15 x 10^6 m above the Earth's surface is approximately 3573 N, as calculated using Newton's Universal Law of Gravitation.

The question asks us to calculate the gravitational force acting on a 6.50 x 10^2 kg mass that is 4.15 x 10^6 m above the Earth's surface. This calculation can be done using Newton's Universal Law of Gravitation, specifically the formula F = G * (M1*M2) / d². In this formula, F is the force, G is the gravitational constant, M1 and M2 are the masses of the two objects, and d is the distance between their centers.

Given the mass of the Earth as 5.98 x 10^24 kg and the radius of the Earth as 6.371 x 10^6 m, we can calculate the gravitational force. The distance from the center of the Earth to the location of our object is the radius of the Earth plus the height of the object above the Earth, which is 6.371 x 10^6 m + 4.15 x 10^6 m = 1.0521 x 10^7 m. Plugging these values into the equation, we get F = (6.67 x 10^-11 N(m²/kg²) * 6.50 x 10^2 kg * 5.98 x 10^24 kg) / (1.0521 x 10^7 m)² = 3573 N, approximately.

https://brainly.com/question/32609171

#SPJ3

Answer: Ok, this problems gives the next info:

Initial velocity = 30m/s

initial position = 15 m

So the only force in our problem is the gravitational, ence the acceleration will be:

a(t) = -9.8 [tex]\frac{m}{s^{2} }[/tex] constant.

for the velocity we must integrate the acceleration over time, and add the integration constant, in this case the initial velocity.

we get v(t) = -9.8 [tex]\frac{m}{s^{2} }[/tex]*t + 30m/s

for the position we integrate over time again, this time the integration constant will be the initial position.

x(t) = [tex]\frac{9.8}{2}[/tex] [tex]\frac{m}{s^{2} }[/tex]*[tex]t^{2}[/tex]+ 30m/s*t + 1.5m

and start doing some resolutions.

(a) how high does the ball go

in this problem you need to obtain the time where the ball stops goin up and starts going down, and put that time in the position equation.

For this, we see v(t0) = 0 so t0 = 30/9.8 = 3.06s

then x(3.06) = 50 meters.

(b) how much time does it take for the ball to reach its maximum height.

Well, we already obtained it it, is 3.06 seconds.

(c) what is the total time the ball is in the air before striking the ground?

here you must see when x(t1) = 0, because if the position is zero, then it means that the ball striking the ground.

As the position is a quadratic function of the time, we must use the bashkara equation so t = [tex]\frac{-30 +- \sqrt{30^{2} +4*4.9*1.5}  }{-2*4.9}[/tex]

this gives us two times, we only took the positive one, because is the one that makes physical sense.

then t = 6.172 seconds.

Answer:

12 nC

Explanation:

Capacity of the parallel plate capacitor

C = ε₀ A/d

ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate

Area of plate = π r²

= 3.14 x (0.8x 10⁻²)²

= 2 x 10⁻⁴

C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³

= 7.08 x 10⁻¹³

Potential difference between plate = field strength x distance between plate

= 6 x 10⁶ x 2.8 x 10⁻³

= 16.8 x 10³ V

Charge on plate = CV

=7.08 x 10⁻¹³ X 16.8 X 10³

11.9 X 10⁻⁹ C

12 nC .

The charge on each electrode of the parallel-plate capacitor is 534 nC.

The charge on each electrode of a parallel-plate capacitor can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.

First, we need to find the capacitance of the capacitor using the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the electrodes, and d is the separation between the electrodes.

Substituting the given values (diameter = 1.6 cm, separation = 2.8 mm) into the formula, we have:

C = (8.85 × 10⁻¹² F/m)(π(0.08 m)²)/0.0028 m = 8.90 × 10⁻¹⁰ F

Finally, we can calculate the charge on each electrode using Q = CV:

Q = (8.90 × 10⁻¹⁰ F)(6.0 × 10⁶ N/C) = 5.34 × 10⁻⁴ C = 534 nC

Answer:

total work is 99.138 kJ

Explanation:

given data

diameter = 5 cm

depth = 75 m

density = 1830 kg/m³

to find out

the total work

solution

we know mass of volume is

volume = [tex]\frac{\pi}{4} d^2 dx[/tex]

volume = [tex]\frac{\pi}{4} d^2 1830 dx[/tex]

so

work required to rise the mass to the height of x m

dw = [tex]\frac{\pi}{4} d^2 1830[/tex] gx dx

so total work is integrate it with 0 to 75

w = [tex]\int\limits^{75}_{0} {\frac{\pi}{4} d^2 1830 gx dx}[/tex]

w = [tex]\frac{\pi}{4}[/tex] × 0.05² × 1830× 9.81× [tex](\frac{x^2}{2})^{75}_0[/tex]

w = 99138.53 J

so total work is 99.138 kJ

Explanation:

It is given that,

Initial state of electron, [tex]n_i=1[/tex]

Final state of electron, [tex]n_f=3[/tex]

The wavelength of the excited electron is given by :

[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})[/tex]

Where

R is Rydberg's constant

[tex]\dfrac{1}{\lambda}=1.097\times 10^{7}\ J\times (\dfrac{1}{3^2}-\dfrac{1}{1^2})[/tex]

[tex]\lambda=-1.02\times 10^{-7}\ m[/tex]

or

[tex]\lambda=102\ nm[/tex]

Let f is the frequency of the observed photon. It is given by :

[tex]f=\dfrac{c}{\lambda}[/tex]

[tex]f=\dfrac{3\times 10^8\ m/s}{1.02\times 10^{-7}\ m}[/tex]

[tex]f=2.94\times 10^{15}\ Hz[/tex]

Hence, this is the required solution.

Answer:

1.032 x 10^-7 m, 2.9 x 10^15 Hz

Explanation:

n = 1 to n = 3

Rydberg's constant, R = 1.09 × 10^7 per metre

Use the formula for the wavelength

[tex]\frac{1}{\lambda }=R\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )[/tex]

here, n1 = 1 and n2 = 3

[tex]\frac{1}{\lambda }=1.09\times 10^{7}\left ( \frac{1}{1^{2}}-\frac{1}{3^{2}} \right )[/tex]

[tex]\frac{1}{\lambda }=1.09\times 10^{7}\times \frac{8}{9}[/tex]

λ = 1.032 x 10^-7 m

Let the frequency is f.

Use the relation

v = f x λ

[tex]3 \times 10^8=f\times 1.032 \times 10^{-7}[/tex]

f = 2.9 x 10^15 Hz

Answer:

surface charge density = 5.91  µC/m²

Explanation:

given data

diameter = 1 μm

radius = 0.5 × [tex]10^{-6}[/tex] m

speed = 3.5 m/s

density 900 kg/m³

distance = 2 mm = 2 × [tex]10^{-3}[/tex] m

to find out

surface charge density of the plane

solution

volume is express as

volume = [tex]\frac{4}{3} \pi r^3[/tex]

volume = [tex]\frac{4}{3} \pi (0.5*10^{-6})^3[/tex]

volume = 5.23 × [tex]10^{-19}[/tex] m³

and

mass = density × volume

mass = 900 ×5.23× [tex]10^{-19}[/tex]

mass = 4.712 × [tex]10^{-16}[/tex] kg

and

from motion of equation

v² -u² = 2×a×s

here v is speed 3.5 and u initial is 0 and a is acceleration and s is distance

3.5²  = 2×a×2 × [tex]10^{-3}[/tex]

acceleration = 3062.5 m/s²

so from newton second law

Force = mass × acceleration

force = 4.712 × [tex]10^{-16}[/tex] ×3062.5

force = 1.44305 × [tex]10^{-12}[/tex] N

so

surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]  

here q is charge with addition 27 extra electron and f is force and ∈ = 8.85× [tex]10^{-12}[/tex]

surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]  

surface charge density = [tex]\frac{2*8.85*10^{-12}*1.443*10^{-12}}{27*1.60*10^{-19}}[/tex]  

surface charge density = 5.91 × [tex]10^{-6}[/tex] C/m²

so surface charge density = 5.91  µC/m²

Answer:

Time period, [tex]T=3.05\times 10^{-5}\ s[/tex]

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

[tex]T=\dfrac{1}{f}[/tex]

T is the time period of the crystal's motion.

Time period is given by :

[tex]T=\dfrac{1}{32768}[/tex]

[tex]T=3.05\times 10^{-5}\ s[/tex]

So, the time period of the crystal's motion is [tex]3.05\times 10^{-5}\ s[/tex]. Hence, this is the required solution.

Answer:

Acceleration of the car, [tex]a=3.47\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the car, u = 0 (at rest)

Final speed of the car, v = 100 km/h = 27.77 m/s

Time, t = 8 s

We need to find the acceleration of the car. Mathematically, it is given by :

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

[tex]a=\dfrac{27.77\ m/s}{8\ s}[/tex]

[tex]a=3.47\ m/s^2[/tex]

So, the acceleration of the car is [tex]3.47\ m/s^2[/tex]. Hence, this is the required solution.

Answer:

acceleration is 3.47 m/s²

Explanation:

given data

speed = 100 km/hr = 27.78 m/s

time = 8 s

to find out

acceleration?

solution

we will apply here first equation of motion

v = u +at  ..............1

here v is final speed and u is initial speed that is 0 and t is time that is 8 sec

so put all value in equation  1

v = u +at

27.78 = 0 +a(8)

a = [tex]\frac{27.78}{8}[/tex]

a = 3.47 m/s²

so acceleration is 3.47 m/s²

Final answer:

To convert 72.4 miles per hour to meters per second, multiply by 1609.34 to convert miles to meters, then divide by 3600 to convert hours to seconds. The result is approximately 32.34 meters per second.

Explanation:

To convert a speed of 72.4 miles per hour to its equivalent in meters per second, you can use the conversion factors that 1 mile is equal to 1609.34 meters and 1 hour is equal to 3600 seconds. Since the conversion from kilometers per hour to meters per second involves dividing by 3.6, and knowing from the reference material that 60 miles per hour is approximately equal to 27.8 meters per second, we can calculate the speed in meters per second as follows:

Doing the math:

Check if the answer is reasonable: Since 72.4 mi/h is a little more than 60 mi/h, and we know that 60 mi/h is approximately 27.8 m/s, it seems reasonable that 72.4 mi/h would be slightly over 27.8 m/s, so our answer of 32.34 m/s seems plausible within the context.

Answer:

The necessary condition is that the force that is applied should not be able to produce any torque about the axis of rotation which is achieved only if the force passes through the axis of rotation.

Explanation:

Mathematically the angular momentum is related to force as

[tex]\overrightarrow{F}\times \overrightarrow{r}=\frac{d\overrightarrow{l}}{dt}[/tex]

where

'r' is arm of the force 'F'

For angular momentum to be conserved the right hand side of the above equation should be zero which can happen only if:

1) F = 0

2) r = 0

3) F and r are parallel.

Thus fulfilling any of the above condition shall conserve the angular momentum of a particle.

Answer:

[tex]a=1.33 m/s^{2}[/tex]

A. [tex]F=2.67N[/tex]

B. [tex]x=37.50m[/tex]

Explanation:

From the exercise we know that the swallow:

[tex]m_{total} =2kg[/tex]

[tex]v_{o}=0\\v_{f}=10m/s\\t=7.5s[/tex]

To find its acceleration we must calculate:

[tex]a=\frac{v_{f}-v_{o}  }{t_{f}-t_{o}  }=\frac{(10-0)m/s}{(7.5-0)s}=1.33m/s^{2}[/tex]

A. Now, from Newton's second law we know that force is:

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

[tex]F=(2kg)*(1.33m/s^{2})=2.67N[/tex]

B. To find the distance which the bird travels we need to find how long does it take

[tex]v_{f}=v_{o}+at[/tex]

Solving for t

[tex]t=\frac{v_{f} }{a}=\frac{10m/s}{1.33m/s^{2} }=7.51s[/tex]

Now, from the equation of position we know that

[tex]x=x_{o}+v_{o}t+\frac{1}{2}at^{2}[/tex]

[tex]x=\frac{1}{2}(1.33m/s^2)(7.51s)^2=37.50m[/tex]

Answer:

Explanation:

u = 85 km / h = 23.61 m /s

Acceleration a = - 0.47 m /s²

Distance travelled in n th sec

S_n = u + ( 2t -1 )a/2

S_n is distance travelled in n th seond

Distance traveled in 1 st second

= 23.61 - .5 x .47

= 23.37  m

=23  m

Distance travelled when t = 5 th

= 23.61 - (2x5 -1)/2 x .47

= 23.61 - 4.5 x .47

23.61 -2.115

=21.495

21 m  is distance travelled in  5 th second .

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

[tex]a=\frac{v^2}{r}[/tex]

r is the radius of the loop

on substituting the respective values, we get

[tex]8\times9.81=\frac{340^2}{r}[/tex]

or

r = 1472.98 m

Answer:

138.46 ft

Explanation:

When the ball is dropped until the moment it hits the water, the ball moves in a uniform acceleration motion. Therefore, the equation that describes the movement of the ball is:

[tex]X = \frac{1}{2}*g*t^{2}  + V_{0} *t + x_0[/tex]

Where X is the distance that the ball has fallen at a time t. [tex]V_0[/tex] is the initial velocity, which is 0 ft/s as the ball was simply dropped. [tex]x_0[/tex] is the initial position, we will say that this value is 0 in the position where the ball was dropped for simplicity, and it increases as the ball is falling. Now, we replace x with 16 feets and solves for t:

[tex]16 ft = \frac{1}{2} * 32.2 \frac{ft}{s^{2}} *t^{2} +0 \frac{ft}{s} * t + 0 ft[/tex]

[tex]t = \sqrt{2* \frac{16 ft}{32.2 \frac{ft}{s^2}}} = 1 s[/tex]

The velocity that the ball will have at the moment the ball that the ball hits the water will be:

[tex]V = V_o+g*t=0\frac{ft}{s}+32.2\frac{ft}{s^2}*1s =32.2\frac{ft}{s}[/tex]

The time that will take the ball to reach the bottom from the top of the lake will be t = 5.3s - 1s = 4.3s. And as the ball will travel with constant velocity equal to 32.2 ft/s^2, the depth of the lake will be:

[tex]d = v*t = 32.2\frac{ft}{s}  * 4.3s = 138.46 ft[/tex]

The depth of the lake is approximately 153.6 feet.

Step 1: Understanding the Problem

We have a lead ball that is dropped from a height of 16.0 ft. After being dropped, it hits the water and sinks to the bottom with a constant velocity. We are told that it reaches the bottom of the lake 5.30 seconds after it is dropped. Our goal is to calculate the depth of the lake.

Step 2: Analyzing the Time

The total time from the moment the ball is dropped until it hits the bottom of the lake is 5.30 seconds. This time includes:

Step 3: Calculate the Fall Time

We can first determine the time it takes for the ball to drop 16.0 ft. The formula for the distance fallen under gravity is:

[tex]d = \frac{1}{2} g t^2[/tex]

where:

Rearranging the formula for time:

[tex]t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \times 16.0\text{ ft}}{32 \text{ ft/s}^2}} = \sqrt{1} = 1 \text{ second}[/tex]

Step 4: Calculate Time to Sink

If the total time from the drop to reaching the bottom is 5.30 seconds, the time taken to sink in the water is:

[tex]\text{Time to sink} = 5.30 \text{ s} - 1 \text{ s} = 4.30 \text{ s}[/tex]

Step 5: Calculate Depth of the Lake

Since the ball sinks at a constant velocity, we need to find the velocity of the ball once it is in the water. The velocity of the ball just before entering the water can be found using the formula:

[tex]v = g t = 32 \text{ ft/s}^2 \times 1 \text{ s} = 32 \text{ ft/s}[/tex]

This means the ball will sink at a velocity of 32 ft/s. Now, we can calculate the depth of the lake (d):

[tex]d = \text{velocity} \times \text{time} = 32 \text{ ft/s} \times 4.30 \text{ s} = 137.6 \text{ ft}[/tex]

Step 6: Add the Initial Drop

Finally, to find the total depth of the lake, we need to add the initial drop of 16.0 ft:

[tex]\text{Total depth} = 137.6 \text{ ft} + 16.0 \text{ ft} = 153.6 \text{ ft}[/tex]

Answer:

Option a) conductors

Explanation:

Conductors are the solids which allows the movement of electrons from atom to atom.

Answer:

a) Conductors.  

Explanation:

Conductors are solids that have free electrons. Some of the examples of solids that are conductors are Steel, Iron, Silver, Copper etc. How best a metallic solid can conduct depends on the number of conduction electrons present in it.These electrons are called conduction electrons.

Not all solids are conductors. For example, wood is a solid, but it is not a conductor. It is an insulator.

Insulators are those materials that do not have free electrons.

Semiconductors are the materials that behave as conductors under certain conditions

Answer:

Explanation:

A capacitor stores energy in the form of an electric field formed between the plates of the capacitor. To establish this electric field ( or to move against it) the charges must do work, and this work its stored in the capacitor.

A resistor can’t store energy. There is a loss of energy in a current passing a resistor, but this is thanks to the heat dissipated by the resistor, and can’t be recovered.

Final answer:

A capacitor stores energy in its electric field between its plates, whereas a resistor dissipates energy as heat.

Explanation:

A capacitor stores energy in the electric field between its plates. When a voltage is applied across the capacitor, it charges up and stores electrical potential energy.

The energy stored in a capacitor can be calculated using the formula:

E = ½CV²

Where E is the energy stored in joules, C is the capacitance of the capacitor in farads, and V is the voltage across the capacitor in volts.

On the other hand, a resistor dissipates energy as heat when a current passes through it and cannot store energy like a capacitor.

Answer with Explanation:

The problem can be simplified as follows

The number of possible outcomes after tossing a coin N times is [tex]2^N[/tex] since for each toss 2 outcomes are possible

Since we need equal heads and equal tails the no of cases amont the [tex]2^n[/tex] cases are

[tex]\binom{N}{N/2}=\frac{N!}{(N-N/2)!(N/2)!}=\frac{n!^2}{(n/2)!^2}[/tex]

Thus the required probability is

[tex]P(E)=\frac{\frac{n!}{(n/2)!^2}}{2^n}=\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]

Part 2)

For the limit as N approaches infinity we have

[tex]P(E')=\lim_{n\rightarrow \infty }\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]

Using Stirling's approximation and solving we get

[tex]\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{n}{2\pi i(n-i)}}\times \frac{n^n}{i^i(n-i)^i}\\\\\lim_{n\rightarrow \infty }\binom{n}{n/2}\approx \sqrt{\frac{2n}{\pi n^2}}\times \frac{n^n}{(n/2)^{n/2}\cdot (n/2)^{n/2}}\\\\\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{2n}{\pi n^2}}\times 2^n\\\\P(E')=\frac{\sqrt{\frac{2n}{\pi n^2}}\times 2^n}{2^n}=\sqrt{\frac{2}{\pi N}}[/tex]