As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it can be excluded from the calculations. For such a system, the potential energy is stored in the spring and is given by
U = 12k x 2
where k is the force constant of the spring and x is the distance from the equilibrium position. The kinetic energy of the system is, as always,
K = 12mv2
where m is the mass of the block and v is the speed of the block.
A) Find the total energy of the object at any point in its motion.
B) Find the amplitude of the motion.
C) Find the maximum speed attained by the object during its motion.

Answer:

P = 5.97 × 10^(5) Pa

Explanation:

We are given;

Mass of balloon;m_b = 4.3 kg

Radius;r = 1.54 m

Temperature;T = 289 K

Density;ρ = 1.19 kg/m³

We know that, density = mass/volume

So, mass = Volume x Density

We also know that Force = mg

Thus;

F = mg = Vρg

Where m = mass of balloon(m_b) + mass of helium (m_he)

So,

(m_b + m_he)g = Vρg

g will cancel out to give;

(m_b + m_he) = Vρ - - - eq1

Since a sphere shaped balloon, Volume(V) = (4/3)πr³

V = (4/3)π(1.54)³

V = 15.3 m³

Plugging relevant values into equation 1,we have;

(3 + m_he) = 15.3 × 1.19

m_he = 18.207 - 3

m_he = 15.207 kg = 15207 g

Molecular weight of helium gas is 4 g/mol

Thus, Number of moles of helium gas is ; no. of moles = 15207/4 ≈ 3802 moles

From ideal gas equation, we know that;

P = nRT/V

Where,

P is absolute pressure

n is number of moles

R is the gas constant and has a value lf 8.314 J/mol.k

T is temperature

V is volume

Plugging in the relevant values, we have;

P = (3802 × 8.314 × 289)/15.3

P = 597074.53 Pa

P = 5.97 × 10^(5) Pa

Answer:

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Explanation:

An electron has a mass of [tex]9.1 \times 10^{-31}\,kg[/tex] and a charge of [tex]-1.6 \times 10^{-19}\,C[/tex]. Based on the Principle of Charge Conservation, [tex]-2.10\times 10^{-6}\,C[/tex] in electrons must be removed in order to create a positive net charge. The amount of removed electrons is found after dividing remove charge by the charge of a electron:

[tex]n_{R} = \frac{-2.10\times 10^{-6}\,C}{-1.6 \times 10^{-19}\,C}[/tex]

[tex]n_{R} = 1.3125 \times 10^{13}\,electrons[/tex]

The number of atoms in 56 gram cooper ball is determined by the Avogadro's Law:

[tex]n_A = \frac{m_{ball}}{M_{Cu}}\cdot N_{A}[/tex]

Where:

[tex]m_{ball}[/tex] - Mass of the ball, measured in kilograms.

[tex]M_{Cu}[/tex] - Atomic mass of cooper, measured in grams per mole.

[tex]N_{A}[/tex] - Avogradro's Number, measured in atoms per mole.

If [tex]m_{ball} = 56\,g[/tex], [tex]M_{Cu} = 63.5\,\frac{g}{mol}[/tex] and [tex]N_{A} = 6.022\times 10^{23}\,\frac{atoms}{mol}[/tex], the number of atoms is:

[tex]n_{A} = \left(\frac{56\,g}{63.5\,\frac{g}{mol} } \right)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)[/tex]

[tex]n_{A} = 5.3107\times 10^{23}\,atoms[/tex]

As there are 29 protons per each atom of cooper, there are 29 electrons per atom. Hence, the number of electrons in cooper is:

[tex]n_{E} = \left(29\,\frac{electrons}{atom} \right)\cdot (5.3107\times 10^{23}\,atoms)[/tex]

[tex]n_{E} = 1.5401\times 10^{23}\,electrons[/tex]

The fraction of the cooper's electrons that is removed is the ratio of removed electrons to total amount of electrons when net charge is zero:

[tex]x = \frac{n_{R}}{n_{E}}[/tex]

[tex]x = \frac{1.3125\times 10^{13}\,electrons}{1.5401\times 10^{23}\,electrons}[/tex]

[tex]x = 8.5222 \times 10^{-11}[/tex]

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Answer:

4.1 seconds

Explanation:

The height of the football is given by the equation:

[tex]H = -16t^2 + V*t + S[/tex]

Using the inicial position S = 4 and the inicial velocity V = 64, we can find the time when the football hits the ground (H = 0):

[tex]0 = -16t^2 + 64*t + 4[/tex]

[tex]4t^2 - 16t - 1 = 0[/tex]

Using Bhaskara's formula, we have:

[tex]\Delta = b^2 - 4ac = (-16)^2 - 4*4*(-1) = 272[/tex]

[tex]t_1 = (-b + \sqrt{\Delta})/2a[/tex]

[tex]t_1 = (16 + 16.49)/8 = 4.06\ seconds[/tex]

[tex]t_2 = (-b - \sqrt{\Delta})/2a[/tex]

[tex]t_2 = (16 - 16.49)/8 = -0.06\ seconds[/tex]

A negative time is not a valid result for this problem, so the amount of time the football is in the air before hitting the ground is 4.1 seconds.

The amount of time the football spent in air before it hits the ground is 4.1 s.

The given parameters;

To find:

Using the vertical model equation given as;

[tex]H = -16t^2 + Vt + S\\\\[/tex]

the final height when the ball hits the ground, H = 0

[tex]0 = -16t^2 + 64t + 4\\\\16t^2 - 64t - 4 = 0\\\\divide \ through \ by\ 4\\\\4t^2 - 16t - 1= 0\\\\solve \ the \ quadratic \ equation \ using \ the \ formula \ method;\\\\\\a = 4, \ b = -16, \ c = - 1\\\\t = \frac{-b \ \ + /- \ \ \ \sqrt{b^2 - 4ac} }{2a} \\\\[/tex]

[tex]t = \frac{-(-16) \ \ + /- \ \ \ \sqrt{(-16^2 )- 4(4\times -1)} }{2\times 4}\\\\t = \frac{16 \ \ + /- \ \ \sqrt{272} }{8} \\\\t = \frac{16 \ \ +/- \ \ 16.49}{8} \\\\t = \frac{16 - 16.49}{8} \ \ \ \ or \ \ \ \frac{16 + 16.49}{8} \\\\t = -0.61 \ s \ \ or \ \ \ 4.06 \ s\\\\t\approx 4.1 \ s[/tex]

Thus, the amount of time the football spent in air before it hits the ground is 4.1 s.

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Answer:

Option B:

The normal force

Explanation:

The normal force does no work as the box slides down the ramp.

Work can only be done when the force succeeds in moving the object in the direction of the force.

All the other forces involved have a component that is moving the box in their direction.

However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)

Answer:

v = 349.23 m/s

Explanation:

It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.

Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]

The orbital speed for a satellite is given by the formula as follows :

[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]

So, the orbital speed for a satellite is 349.23 m/s.

Answer:

0.15 m

Explanation:

First calculating the center of mass from the saw horse

[tex]\frac{3.15}{2} -1=0.575 m[/tex]

from the free body diagram we can write

Taking moment about the saw horse

55.9×9.81×y=14.5×0.575×9.81

y= 0.15 m

So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.

Answer:

the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]

Increase in volume of the glass =  10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]

[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]

[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Answer:

a) v₁ = 3.92 m / s , b)     ΔP =  = 9.0 10⁴ Pa, c)  t = 0.0297 s  

Explanation:

This is a fluid mechanics exercise

a) let's use the continuity equation

let's use index 1 for the hose and index 2 for the nozzle

        A₁ v₁ = A₂v₂

in area of ​​a circle is

       A = π r² = π d² / 4

we substitute in the continuity equation

        π d₁² / 4 v₁ = π d₂² / 4 v₂

        d₁² v₁ = d₂² v₂

the speed of the water in the hose is v1

       v₁ = v₂ d₂² / d₁²

       v₁ = 14 (1.8 / 3.4)²

        v₁ = 3.92 m / s

b) they ask us for the pressure difference, for this we use Bernoulli's equation

       P₁ + ½ ρ v₁² + m g y₁ = P₂ + ½ ρ v₂² + mg y2

as the hose is horizontal y₁ = y₂

       P₁ - P₂ = ½ ρ (v₂² - v₁²)

      ΔP = ½ 1000 (14² - 3.92²)

       ΔP = 90316.8 Pa = 9.0 10⁴ Pa

c) how long does a tub take to flat

the continuity equation is equal to the system flow

        Q = A₁v₁

        Q = V t

where V is the volume, let's equalize the equations

         V t = A₁ v₁

         t = A₁ v₁ / V

A₁ = π d₁² / 4

let's reduce it to SI units

         V = 120 l (1 m³ / 1000 l) = 0.120 m³

          d1 = 3.4 cm (1 m / 100cm) = 3.4 10⁻² m

let's substitute and calculate

         t = π d₁²/4   v1 / V

         t = π (3.4 10⁻²)²/4 3.92 / 0.120

         t = 0.0297 s

Answer:

The distance is  [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

        The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is  [tex]\frac{1}{10}[/tex] of  an arcsec

  This implies that  0.2 grid unit is  [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           [tex]d = \frac{1}{k}[/tex]

substituting values

           [tex]d = \frac{1}{0.02}[/tex]

           [tex]d = 50 \ parsec[/tex]

Note  [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So  [tex]d = 50 * 3.08 *10^{13}[/tex]

     [tex]d = 1.5 *10^{15} \ km[/tex]

Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264

Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]

The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]

To compare the power of the two cars, first find the Kinetic Energy each one has:

K.E. for Model A

[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]

K.E. for model B

[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]

[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]

Now, determine Power for each model:

Power for model A

[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]

Power for model B

[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]

Comparing power of model B to power of model A:

[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]

[tex]\frac{P_B}{P_A} =[/tex] 8.5264

Comparing power for each model, power for model B is 8.5264 better than model A.

Answer:

x = 46.54m

Explanation:

In order to find the length of the incline you use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]      (1)

vo: initial speed of the soccer ball = 0 m/s

t: time

a: acceleration

You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):

[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex]      (2)

Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:

[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex]       (3)

The length of the incline is 46.54 m

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

Answer:

88.29 N

Explanation:

mass of the ball = 4.5 kg

weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)

weight W = 4.5 x 9.81 = 44.145 N

centrifugal forces Tc act on the ball as it swings.

At the top point of the vertical swing,

Tension on the rope = Tc - W.

At the bottom point of the vertical swing,

Tension on the rope = Tc + W

therefore,

difference in tension between these two points will be;

Net tension = tension at bottom minus tension at the top

= Tc + W - (Tc - W) = Tc + W -Tc + W

= 2W

imputing the value of the weight W, we have

2W = 2 x 44.145 = 88.29 N

Answer:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

Answer:

The two value of the wavelength for the out of tune guitar is  

[tex]\lambda _2 = (6.48,6.52) \ cm[/tex]

Explanation:

From the question we are told that

     The wavelength of the note is [tex]\lambda = 6.50 \ cm = 0.065 \ m[/tex]

     The difference in beat frequency is [tex]\Delta f = 17.0 \ Hz[/tex]

Generally the frequency of the note played by the guitar that is in tune is  

        [tex]f_1 = \frac{v_s}{\lambda}[/tex]

Where [tex]v_s[/tex] is the speed of sound with a constant value [tex]v_s = 343 \ m/s[/tex]

       [tex]f_1 = \frac{343}{0.0065}[/tex]

      [tex]f_1 = 5276.9 \ Hz[/tex]

The difference in beat is mathematically represented as

       [tex]\Delta f = |f_1 - f_2|[/tex]

Where [tex]f_2[/tex] is the frequency of the sound from the out of tune guitar

     [tex]f_2 =f_1 \pm \Delta f[/tex]

substituting values

      [tex]f_2 =f_1 + \Delta f[/tex]

      [tex]f_2 = 5276.9 + 17.0[/tex]  

     [tex]f_2 = 5293.9 \ Hz[/tex]

The wavelength for this frequency is

      [tex]\lambda_2 = \frac{343 }{5293.9}[/tex]

     [tex]\lambda_2 = 0.0648 \ m[/tex]

    [tex]\lambda_2 = 6.48 \ cm[/tex]

For the second value of the second frequency

     [tex]f_2 = f_1 - \Delta f[/tex]

     [tex]f_2 = 5276.9 -17[/tex]

      [tex]f_2 = 5259.9 Hz[/tex]

The wavelength for this frequency is

   [tex]\lambda _2 = \frac{343}{5259.9}[/tex]

   [tex]\lambda _2 = 0.0652 \ m[/tex]

   [tex]\lambda _2 = 6.52 \ cm[/tex]

This question involves the concepts of beat frequency and wavelength.

The possible wavelengths of the out-of-tune guitar are "6.48 cm" and "6.52 cm".

The beat frequency is given by the following formula:

[tex]f_b=|f_1-f_2|\\\\[/tex]

f₂ = [tex]f_b[/tex] ± f₁

where,

f₂ = frequency of the out-of-tune guitar = ?

[tex]f_b[/tex] = beat frequency = 17 Hz

f₁ = frequency of in-tune guitar = [tex]\frac{speed\ of\ sound\ in\ air}{\lambda_1}=\frac{343\ m/s}{0.065\ m}=5276.9\ Hz[/tex]

Therefore,

f₂ = 5276.9 Hz ± 17 HZ

f₂ = 5293.9 Hz (OR) 5259.9 Hz

Now, calculating the possible wavelengths:

[tex]\lambda_2=\frac{speed\ of\ sound}{f_2}\\\\\lambda_2 = \frac{343\ m/s}{5293.9\ Hz}\ (OR)\ \frac{343\ m/s}{5259.9\ Hz}\\\\[/tex]

λ₂ = 6.48 cm (OR) 6.52 cm

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Answer:

Explanation:

Given that:

The mass of the cell is 9.0 x 10^-14 kg

The charges of the cell is -2.5pC and the other -3.30 pC

[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]

Radius is  3.75 × 10-6 m

The final distance is twice the radius

i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]

The formula for the velocity of the cell is

[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]

[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]

[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]

The maximum acceleration of the cells as they move toward each other and just barely touch is

[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]

[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]

[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]

The answers obtained are;

1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].

2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].

The answer is explained as shown below.

We have, the mass of the red blood cell;

Also, the charges of the cells are;

The distance between the charges when they barely touch will be two times the radius of each charge.

1. As both the cells are negatively charged they will repel each other.

Now, substituting all the known values in the equation, we get;

[tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]

2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.

Substituting the known values, we get;

[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]

[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]

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Answer:

46648 J

Explanation:

mass m= 85 Kg

velocity v = 56 m/s

distance covered s =40 m

According to Question,

frictional drag force to him that matched his weight

[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]

Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J

W=Q= F_d×s

=833×56 = 46648 J

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Answer:

[tex]v_B=3.78\times 10^5\ m/s[/tex]

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]

Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]

Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]

Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]

So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Answer:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

[tex]v^2=v_o^2+2ax[/tex]         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]

Next, you use the second Newton law to calculate the force:

[tex]F=ma[/tex]

m: mass of the electron = 9.11*10^-31kg

[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]

The quotient between the weight of the electron and the force F is:

[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]

The force F is 2.59*10^11 times the weight of the electron

Answer:

The magnitude of the electric force on the particle in terms of the variables given is, F = qE

The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)

Explanation:

Given;

a charged particle, q

magnitude of electric field, E

magnitude of magnetic field, B

The magnitude of the electric force on the particle in terms of the variables given;

F = qE

The magnitude of the magnetic force on the particle in terms of the variables given;

F = q (v x B)

where;

v is the constant velocity of the charged particle

Answer:

The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|

The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|

Explanation:

The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).

Electric force = qE

The magnitude of the electric force = |qE|

That is, magnitude of the product of the charge and the electric field vector.

The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).

It is given mathematically as (qv × B)

The magnitude of the magnetic force is the magnitude of the vector product obtained.

Magnitude of the magnetic force = |qv × B|

Hope this Helps!!!

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, [tex]V_s[/tex] = 450 V

potential at radial distance, [tex]V_r[/tex] = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]

[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]

Therefore, the radius of the sphere is 4.05 m

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

Answer:

a) The distance from the cornea vertex to the retina is 2.37 cm

b) This distance is shorter than for the normal eye.

Explanation:

a) Let refractive index of air,

n(air) = x = 1

Let refractive index of lens,

n(lens) = y = 1.4

Object distance, s = 36 cm

Radius of curvature, R = 0.65 cm

The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.

Image distance, s' = ?

(x/s) + (y/s') = (y-x)/R

(1/36) + (1.4/s') = (1.4 - 1)/0.65

1.4/s' = 0.62 - 0.028

1.4/s' = 0.592

s' = 1.4/0.592

s' = 2.37 cm

Distance from the cornea vertex to the retina is 2.37 cm

(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.

Answer:

F₂ = 925.92 N

Explanation:

In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,

σ₁ = σ₂

F₁/A₁ = F₂/A₂

F₂ = F₁ A₂/A₁

where,

F₂ = Initial force that must be applied to narrow arm = ?

F₁ = Load on Wider Arm to be raised = 12000 N

A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²

A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²

Therefore,

F₂ = (12000 N)(25π cm²)/(324π cm²)

F₂ = 925.92 N

Answer:

d) Both blocks have had equal losses of energy to friction

Explanation:

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces

Moreover, the block A is twice as fast than block B at the time of crossing the finish line

So based on the above information,  it contains the losses of identical friction

And we also know that

Friction energy loss is

[tex]= \mu \times m \times g \times D[/tex]

It would be the same for both the blocks

hence, the option d is correct

The correct answer will be both blocks have had equal losses of energy to friction.

Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.

Moreover, the block A is twice as fast as block B at the time of crossing the finish line.

So based on the above information,  it contains the losses of identical friction.

And we also know that

Friction energy loss is

[tex]E_f=\mu m g D[/tex]

It would be the same for both the blocks

Hence both blocks have had equal losses of energy to friction.

To know more about friction, follow

https://brainly.com/question/24386803

Answer:

(b) electrons will not be ejected

Explanation:

Determine the number of photons ejected by 10 W red laser pointer.

The wavelength (λ) of red light is  700 nm = 700 x 10⁻⁹ m

Energy of a photon is given as;

[tex]E = \frac{hc}{\lambda}[/tex]

where;

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

c is speed of light, = 3 x 10⁸ m/s

[tex]E = \frac{6.626*10^{-34} *3*10^8}{700 X 10^{-9}} \\\\E = 2.8397 *10^{-19} \ J/photon[/tex]

The number of photons emitted by 10 W red laser pointer

10 W = 10 J/s

[tex]Number \ of \ photons = 10(\frac{ J}{s}) * \frac{1}{2.8397*10^{-19}} (\frac{photon}{J} ) = 3.522 *10^{19} \ photons/s[/tex]

Determine the number of photons ejected by 5 W red green pointer

The wavelength (λ) of green light is  500 nm = 500 x 10⁻⁹ m

[tex]E = \frac{hc}{\lambda} = \frac{6.626*10^{-34} *3*10^8}{500*10^{-9}} = 3.9756 *10^{-19} \ J/photon[/tex]

The number of photons emitted by 5 W green laser pointer

5 W = 5 J/s

[tex]Number \ of \ photons = \frac{5J}{s} *\frac{photon}{3.9756*10^{-19}J} = 1.258 *10^{19} \ Photons/s[/tex]

The number of photons emitted by 10 W red laser pointer is greater than the number of photons emitted by 5 W green laser pointer.

Thus, 5 W green laser pointer will not be able to eject electron from the same metal.

The correct option is "(b) electrons will not be ejected"

Answer:

v = 4.08m/s₂

Explanation:

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s