Does percent difference give indication of accuracy or precision? Discuss

Answer:

14 N

Explanation:

The tension in the second string is puling both the masses of 20 kg and 8 kg with acceleration of 0.5 m s⁻²

So tension in the second string = total mass x acceleration

= 28 x .5 = 14 N . Ans..

Answer:

a) 1200 kN/m²

b) 1,200,000 kg/ms²

c) 1.2 × 10⁹ kg/km.s²

Explanation:

Given:

Pressure = 1200 kPa

a) 1 Pa = 1 N/m²

thus,

1000 N = 1 kN

1200 kPa = 1200 kN/m²

b)  1 Pa = 1 N/m² =  1 kg/ms²

Thus,

1200 kPa = 1200000 Pa

or

1200000 Pa = 1200000 × 1 kg/ms²

or

= 1,200,000 kg/ms²

c) 1 km = 1000 m

or

1 m = 0.001 Km

thus,

1,200,000 kg/ms² = [tex]\frac{1,200,000}{0.001}\frac{kg}{km.s^2}[/tex]

or

= 1.2 × 10⁹ kg/km.s²

(a) In kN and m units, the pressure is [tex]$\boxed{1200 \text{ kPa}}$[/tex].  (b) In kg, m, and s units, the pressure is [tex]$\boxed{1200 \text{ kg/(m·s}^2\text{)}}.[/tex] (c) In kg, km, and s units, the pressure is [tex]\boxed{1200 \text{ kg/(km·s}^2\text{)}}[/tex].

Explanation and logic of the

To convert pressure from kPa to the required units, we need to understand the relationship between pressure, force, and area. Pressure (P) is defined as the force (F) applied per unit area (A), given by P = F/A.

(a) In the International System of Units (SI), pressure is measured in pascals (Pa), where 1 Pa = 1 N/m². Since 1 kPa = 1000 Pa, and 1 kN = 1000 N, the pressure in kN/m² is numerically the same as in kPa. Therefore, the pressure in the tank is 1200 kN/m².

(b) To express the pressure in base SI units of kg, m, and s, we need to consider that 1 N = 1 kg·m/s². Since 1 kPa = 1 kN/m², and 1 kN = 1000 N, we have:

[tex]\[ 1200 \text{ kPa} = 1200 \times \frac{1000 \text{ kg\·m/s}^2}{1 \text{ m}^2} = 1200 \times 1000 \text{ kg/(m\s}^2\text{)} \][/tex]

(c) To express the pressure in kg, km, and s, we convert the area from m² to km²:

[tex]\[ 1 \text{ m}^2 = \frac{1}{1000 \times 1000} \text{ km}^2 = \frac{1}{1000000} \text{ km}^2 \][/tex]

Thus, the pressure in kg/(km.s²) is:

[tex]\[ 1200 \text{ kPa} = 1200 \times \frac{1000 \text{ kg\·m/s}^2}{\frac{1}{1000000} \text{ km}^2} = 1200 \times 1000000 \text{ kg/(km\·s}^2\text{)} \][/tex]

However, since 1 kPa is equivalent to 1 kN/m², and 1 kN = 1000 N, we can simplify the expression by recognizing that the conversion from m² to km² results in a factor of 1/1000000, which cancels out the factor of 1000 from the conversion of kN to N, leaving us with:

[tex]\[ 1200 \text{ kPa} = 1200 \text{ kg/(km\s}^2\text{)} \][/tex]

Therefore, the pressure in the tank, when expressed in kg/(km·s²), is 1200 kg/(km·s²).

Answer:

frequency wren hear when approaches is 470.29 Hz

frequency wren hear after hawk pass is 415.75 Hz

Explanation:

given data

hawk  velocity Vs= 38 m/s

frequency f = 440 Hz

wren velocity Vo= 17 m/s

speed of sound s = 343 m/s

to find out

What frequency wren hear and What frequency wren hear after hawk pass

solution

we apply here frequency formula that is

[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex]    .........1

here f1 is frequency hear by observer

put here all value as Vo and Vs negative because it approaches

[tex]f1=f ( \frac{s-V_o}{s-V_s} )[/tex]  

[tex]f1=440 ( \frac{343-17}{343-38} )[/tex]  

f1 = 470.29 Hz

so frequency wren hear when approaches is 470.29 Hz

and

after passing from equation 1 we take both Vo and Vs as positive

[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex]

[tex]f1=440 ( \frac{343+17}{343+38} )[/tex]

f1 = 415.75 Hz

so frequency wren hear after hawk pass is 415.75 Hz

Answer:

a) 31557600 seconds b) 3.168x10⁻⁸ years

Explanation:

1 hour is equivalent to 3600 seconds, that multiplied by 24 hours (1 day) gives 86.400 seconds

a) how many seconds there are in 1.00 year?

[tex]\frac{86400 s}{1 day} x 365,25 days[/tex]  ⇒  31557600 s

So in 1 year there are a total of 31557600 seconds.

b) how many years are there in 1.00 second?

Using the quantity from part a) it is get:

[tex]1 s x \frac{1 year}{31557600 s}[/tex] ⇒ 3.168x10⁻⁸ years

So in 1 second there are a total of 3.168x10⁻⁸ years.

Answer:

0.42 m

Explanation:

For thermal expansion, the formula used is as follows

L = L₀ ( 1 + α t )

L is length after rise of temperature by t , L₀ is length before rise of temperature , α is coefficient of thermal expansion and t is rise in temperature.

L - L₀ = L₀α t

Difference of length = L₀α t

L₀ = 1000 m , α = 10.5 x 10⁻⁶ , t = 40

Difference of length = 1000 x 10.5 x 10⁻⁶x 40

= 0.42 m .

Final answer:

The force of attraction between the divalent cation and monovalent anion can be calculated using Coulomb's law. Plugging the values into the formula gives a force of -1.742 N.

Explanation:

The force of attraction between two ions can be calculated using Coulomb's law. The formula is given by F = k * (Q1 * Q2) / r^2, where F is the force, k is the Coulomb constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges of the ions, and r is the distance between their centers. In this case, since the anion is monovalent and the cation is divalent, the charges would be -1 and +2 respectively.

To calculate the force of attraction, we need to find the equilibrium interionic separation. Given that the atomic radii of the divalent cation and monovalent anion are 0.19 nm and 0.126 nm respectively, their total separation would be the sum of their radii, which is 0.316 nm.

Plugging these values into the formula, we get:

F = (9 x 10^9 Nm^2/C^2) * (-1 C) * (+2 C) / (0.316 x 10^-9)^2 = -1.742 N

Explanation:

Length of house, l = 54 ft = 16.45 m

Breadth of house, b = 48 ft = 14.63 m

Height of house, h = 8 ft = 2.43 m

We need to find the volume of the interior of the house. The house is in the form of cuboid. The volume of cuboid is given by :

[tex]V=l\times b\times h[/tex]

[tex]V=16.45\ m\times 14.63\ m\times 2.43\ m[/tex]

[tex]V=584.81\ m^3[/tex]

Since, [tex]1\ m^3=1000000\ cm^3[/tex]

or the volume of the interior of the house, [tex]V=5.84\times 10^8\ cm^3[/tex]

Hence, this is the required solution.

Answer:

(a) Area = [tex]83.022m^2[/tex] (b)  Area = [tex]9.07m^2[/tex]

Explanation:

We have given radius of the circle r = 5.142 m and r = 1.7 m

We have to find the area of the circle

We know that area of the circle is given  by

[tex]A=\pi r^2[/tex], here r is the radius of the circle

(a) Radius = 5.142 m

So area [tex]A=\pi r^2=3.14\times 5.142^2=83.022m^2[/tex]

(b) Radius = 1.7 m

So area [tex]A=\pi r^2=3.14\times 1.7^2=9.07m^2[/tex]

Explanation:

Given

Refractive Index [tex](n_1)=1.65[/tex]

[tex]n_2=1.35[/tex]

angle of incident[tex]=35 ^{\circ}[/tex]

Light originates in the medium of higher refractive index

Let [tex]\theta _1[/tex] and [tex]\theta _2 [/tex]be the incident and refraction angles respectively  thus according to snell's law

[tex]n_1sin\theta _1=n_2sin\theta _2[/tex]

[tex]1.65\times sin35=1.35\times sin\theta _2[/tex]

[tex]sin\theta _2=0.701[/tex]

[tex]\theta _2=44.507^{\circ}[/tex]

(b)Ray originates from lighter medium

[tex]1.35\times sin35=1.65\times sin\theta _2'[/tex]

[tex]sin\theta _2'=0.469[/tex]

[tex]\theta _2'=27.96\approx 28^{\circ}[/tex]

Answer:

The atmospheric pressure is [tex]9.845 \times 10^4\ Pa[/tex]

Explanation:

There are two ways of solving this exercise:

1)

In physics you can find that mmHg is a unit of pressure.

Pressure = Force/area.

If you consider the weight of mercury as your force (mass* acceleration of gravity = density*volume* acceleration of gravity ), then

[tex]P =\frac{\rho Vg}{A} = \frac{\rho Ahg}{A} =\rho g h[/tex]

where h is the height of the mercury column and rho its density.

[tex]\Delta P= P_{atm} - P_{hur} = 21.4\ mm Hg[/tex].

if normal atmospheric pressure is [tex]1.013 \times 10^5\ Pa = 759.81\ mmHg[/tex]

then the pressure in the presence of the hurricane is

[tex]P_{hur} = 759.81 - 21.4 = 738.41\ mmHg = 9.845 \times 10^4\ Pa[/tex]

2)

Considering the definition of pressure

[tex]\Delta P = \rho g h[/tex]

where [tex]\rho = 13.6\ g/cm^3[/tex], [tex]g = 9.8\ m/s^2 =980\ cm/s^2[/tex] and [tex]h = 21.4\ mm = 2.14\ cm[/tex].

[tex]\Delta P = P_{atm} - P_{hur} = 28521,92\ g/cms^2 = 2852,192\ kg/ms^2[/tex], where [tex]kg/ms^2 = Pa[/tex].

if

[tex]P_{atm} = 1.013 \times 10^5\ Pa[/tex],

then

[tex]P_{hur} = 1.013 \times 10^5 - 2852,192 =9.845 \times 10^4\ Pa[/tex].

Answer:

[tex]V_{2}=\frac{V_{1}}{2}[/tex]    

Explanation:

The potential of a conducting sphere is the same as a punctual charge,

First sphere:

[tex]V_{1}=k*q_{1}/r_{1}[/tex]    (1)

Second sphere:

[tex]V_{2}=k*q_{2}/r_{2}[/tex]        (2)

But, second sphere's radius is twice first sphere radius, and their charges the same:

[tex]q_{1}=q_{2}[/tex]

[tex]r_{2}=2*r_{1}[/tex]

If we divide the equations (1) and (2), to solve V2:

[tex]V_{2}=V_{1}*r_{1}/r_{2}=V_{1}/2[/tex]    

Answer:

mass of bigger object M= 3 kg and mass of smaller object m= 2 kg

Explanation:

force between two objects F= 1.00×10^-8 N

distance between them R= 20 cm = 0.2 m

total mass M+m= 5 kg

We know that

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

putting the values we get

[tex]10^{-8}= 6.67\times10^{-11}\frac{M\times m}{0.2^2}[/tex]

M×m= 6

and M+m= 5

therefore we can calculate that M= 3 kg and m= 2 kg

mass of bigger object M= 3 kg and mass of smaller object m= 2 kg

Answer:

The answer is [tex]V = 1.785 \times 10^{10}\ Km^3[/tex].

Explanation:

You are asking only for the total volume, so the important data here is the surface area of the earth's crust,

[tex]A = 5.10 \times 10^8\ Km^2[/tex]

and the average thickness of the earth's crust,

[tex]h = 35\ Km[/tex].

Imagine that you can stretch the surface area as if it were a blanket, now if you want to calculate its volume, you just need to multiply its area by its thickness,

[tex]V = A*h = 1.785 \times 10^{10}\ Km^3[/tex].

Answer:

[tex]1.232\times 10^{-5}\ T.[/tex]

Explanation:

Given:

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to [tex]\mu_o[/tex] times the net current threading the loop.

[tex]\oint \vec B \cdot d\vec l=\mu_o I.[/tex]

In case of a circular loop, the directions of magnetic field and the line element [tex]d\vec l[/tex], both are along the tangent of the loop at that point, therefore, [tex]\vec B\cdot d\vec l = B\ dl[/tex].

[tex]\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\[/tex]

[tex]\oint dl[/tex] is the circumference of the Amperian loop = [tex] 2\pi r[/tex]

Therefore,

[tex]B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.[/tex]

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, [tex]r_1 = 3.510\ m.[/tex]

The magnetic field at the given point due to this wire is given by:

[tex]B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.[/tex]

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m,  concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, [tex]r_2 = 5.310\ m.[/tex]

The magnetic field at the given point due to this wire is given by:

[tex]B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.[/tex]

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at [tex]r_r=-3.510\ m[/tex] is given by

[tex]B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.[/tex]

Answer:

The equation which describes the simple harmonic motion is [tex]x=0.5\cos(15.81t)[/tex]

Explanation:

Given that,

Mass = 0.4 kg

Spring constant = 100 N/m

Maximum displacement = 0.5 m

We need to calculate the angular frequency

[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]

[tex]\omega=\sqrt{\dfrac{100}{0.4}}[/tex]

We need to find the equation which describes the simple harmonic motion

Using equation of simple harmonic motion

[tex]x = A\cos\omega t[/tex]

Where, A = amplitude

[tex]\omega [/tex] = angular frequency

Put the value of angular frequency

[tex]x=A\cos\sqrt{\dfrac{k}{m}t}[/tex]

Put the value in the equation

[tex]x=0.5\cos\sqrt{\dfrac{100}{0.4}t}[/tex]

[tex]x=0.5\cos(15.81t)[/tex]

Hence, The equation which describes the simple harmonic motion is [tex]x=0.5\cos(15.81t)[/tex]

Answer:

A) 240 W

B) 864000 J

Explanation:

Hi!

We can easily calculate the required power multiplying the voltage of the process times the circulating current:

[tex]P=12V\times 20A=12(\frac{J}{coulomb})\times20(\frac{coulomb}{s})\\P = 240\frac{J}{s}=240W[/tex]

To calculate the total energy provided we need to multiply the power times the total time:

[tex]E=P\times t=240\frac{J}{s} \times 1 hr =240\frac{J}{s} \times 3600s\\ E=864 000J[/tex]

Hope this helps

Have a nice day!

Final answer:

The power required to charge the battery is 240 watts, and the energy provided during the charging process is 864,000 joules.

Explanation:

A student has asked how much power is required to charge a 12 V storage battery by a current of 20 A for 1 hour, and how much energy has been provided during the process.

A) To find the power required to charge the battery, we use the formula:

Power (P) = Voltage (V)  imes Current (I)

P = 12 V  imes 20 A

P = 240 W

Therefore, the power required to charge the battery at this rate is 240 watts.

B) The energy provided during the charging process can be calculated using the formula:

Energy (E) = Power (P)  imes Time (t)

Since power is given in watts and time in hours, we must convert the time to seconds for consistency in units (1 hour = 3600 seconds).

E = 240 W  imes 1 hour  imes 3600 seconds/hour

E = 240 W  imes 3600 s

E = 864,000 J

The energy provided during the charging process is 864,000 joules (or 864 kJ).

Answer:

B. Work one on one with children in need

Explanation:

Whole group time allows early childhood teachers to introduce new material. It ensures that each student is presented and reviewed with uniform key concepts. It also sets expectations in planning and developing the lessons because it provides baseline assessments.The only thing that seems impossible in this activity is that teacher can give individual or one-to-one attention to each student.

hence the correction will be option B Work one on one with children in need

Answer:

[tex]r_2 = 976.65 m[/tex]

Direction is 19 degree South of East

Explanation:

Let say initial position is our reference

so we will have campsite position given as

[tex]r = 1.50 km[/tex] at 20 degree E of N

now we will have

[tex]r = 1500 sin20\hat i + 1500 cos20\hat j[/tex]

[tex]r = 513 \hat i + 1409.5\hat j[/tex]

now our displacement to walk around is given as

[tex]d_1 = 600 \hat j[/tex]

then we move 20 degree W of N and move 1200 m

so we will have

[tex]d_2 = 1200 sin20(-\hat i) + 1200cos20\hat j[/tex]

so our final position is given as

[tex]r_1 = d_1 + d_2[/tex]

[tex]r_1 = 600\hat j - 410.4 \hat i + 1127.6\hat j[/tex]

[tex]r_1 = -410.4 \hat i + 1727.6\hat j[/tex]

now we know that

[tex]r_1 + r_2 = r[/tex]

so final leg of the displacement is given as

[tex]r_2 = r - r_1[/tex]

[tex]r_2 = (513 \hat i + 1409.5\hat j) - (-410.4 \hat i + 1727.6\hat j)[/tex]

[tex]r_2 = 923.4\hat i - 318.1 \hat i[/tex]

so magnitude is given as

[tex]r_2 = \sqrt{923.4^2 + 318.1^2}[/tex]

[tex]r_2 = 976.65 m[/tex]

direction is given as

[tex]\theta = tan^{-1}\frac{y}{x}[/tex]

[tex]\theta = tan^{-1}\frac{-318.1}{923.4}[/tex]

[tex]\theta = -19 degree[/tex]

so it is 19 degree South of East

Answer:same

Explanation:

Given

ball A initial velocity=3 m/s(upward)

Ball B initial velocity=3 m/s (downward)

Acceleration on both the balls will be acceleration due to gravity which will be downward in direction

Both acceleration is equal

For ball A

maximum height reached is [tex]h_1=\frac{3^2}{2g}[/tex]

After that it starts to move downwards

thus ball have to travel a distance of h_1+h(building height)

so ball A final velocity when it reaches the ground is

[tex]v_a^2=2g\left ( h_1+h\right )[/tex]

[tex]v_a^2=2g\left ( 0.458+h\right )[/tex]

[tex]v_a=\sqrt{2g\left ( 0.458+h\right )}[/tex]

For ball b

[tex]v_b^2-\left ( 3\right )^2=2g\left ( h\right )[/tex]

[tex]v_b^2=2g\left ( \frac{3^2}{2g}+h\right )[/tex]

[tex]v_b=\sqrt{2g\left ( 0.458+h\right )}[/tex]

thus [tex]v_a=v_b[/tex]

Final answer:

A. The magnitudes of the accelerations of both balls are equal, but they have opposite directions. B. The final speeds of both balls when they reach the ground will be the same.

Explanation:

A. The acceleration of both balls will be the same in magnitude but in opposite directions. Since the acceleration due to gravity acts downward, ball A will have a negative acceleration while ball B will have a positive acceleration. Therefore, the magnitudes of their accelerations will be equal.

B. When both balls reach the ground, their final speeds will also be the same. This is because the vertical motion of the balls is independent of their initial speeds when air resistance is ignored. The time it takes for them to reach the ground will be the same, and hence their final velocities will also be equal.

Answer:

Explanation:

For elestic collision

v₁ = [tex]\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2}[/tex]

[tex]v_2 = [tex]\frac{(m_2-m_1)u_2}{m_1+m_2} +\frac{2m_1u_1}{m_1+m_2}[/tex][/tex]

Here u₁ = 0 , u₂ = 22 m/s , m₁ = 77 kg , m₂ = .15 kg ,  v₁ and v₂ are velocity of goalie and puck after the collision.

v₁ = 0 + ( 2 x .15 x22 )/ 77.15  

= .085 m / s

Velocity of goalie will be .085 m/s in the direction of original velocity of ball before collision.

v₂ = (.15 - 77)x 22 / 77.15 +0

= - 21.91 m /s

=Velocity of puck will be - 21.91 m /s  in the direction opposite  to original velocity of ball before collision.

Answer:b-both objects hit the ground together

Explanation:

Given

Two objects of different mass are released simultaneously from top of a tower and fall to the ground provided air resistance is negligible

then both the object hit the ground together

Because in equation of motion terms there is no unit of mass

But if the air resistance is present then the answer would have been different.

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, [tex]E=\frac{F}{q}[/tex]

[tex]E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}[/tex]

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

Final answer:

The electric field at the position of the electron is approximately 2.43 x 10^4 N/C, and the potential difference between the plates is about 2916 V.

Explanation:

The electric field (E) within two charged parallel plates is given by the expression E = F/q, where F is the force experienced by a charge q. Since an electron has a charge of approximately 1.602 x 10^-19 C, and it is experiencing a force 3.9 x 10^-15 N, we can calculate the electric field as E = (3.9 x 10^-15 N) / (1.602 x 10^-19 C) ≈ 2.43 x 10^4 N/C. The direction of the electric field is from the positively charged plate to the negatively charged plate.

The potential difference (V) between the plates can be found using the relationship V = E × d, where d is the distance between the plates. Given that the plates are 12 cm apart, this distance in meters is 0.12 m, so the potential difference is V = (2.43 x 10^4 N/C) × 0.12 m ≈ 2916 V.

Answer:

the bearing of the line AC will be 154° 51' 48"

Explanation:

given,

bearing of the line AB = 234° 51' 48"

an anticlockwise measure of an angle to the point C measured 80°

to calculate the bearing of AC.

As the bearing of line AB is calculated clockwise from North direction.

the angle is moved anticlockwise now the bearing of AC will be calculated by

                  =  bearing of line AB - 80°

                  =  234° 51' 48"  - 80°

                 =  154° 51' 48".

                  = 154.5148

so, the bearing of the line AC will be 154° 51' 48"

Answer:(a)0,(b)1.061 kg-m/s

Explanation:

Given

mass of ball is 0.10 kg

Initial speed is 15 m/s

Maximum height reached by ball is h

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

final velocity =0

[tex]-\left ( 15\right )^2=-2\times 9.81\times s[/tex]

[tex]s=\frac{15^2}{2\times 9.81}=11.467 m[/tex]

thus momentum of ball at maximum height is 0 as velocity is zero

For halfway to maximum height

[tex]v^2-u^2=2as_0[/tex]

where [tex]s_0=\frac{11.467}{2}[/tex]

[tex]s_0=5.73 m[/tex]

[tex]v^2=15^2-2\times 9.81\times 5.73[/tex]

v=10.61 m/s

Thus its momentum is

[tex]mv=0.10\times 10.61=1.061 kg-m/s[/tex]

Answer:

F = 2 N

Explanation:

Let the rod is made up of large number of point charges

so here force due to one small part which is at distance "x" from the end of the rod on the given charge is

[tex]dF = \frac{kdq q}{(d + x)^2}[/tex]

here we know that

[tex]dq = \frac{Q}{L} dx[/tex]

so we have

[tex]F = \int \frac{k Q q dx}{L(d + x)^2}[/tex]

[tex]F = \int_0^L \frac{kQq dx}{L(d + x)^2}[/tex]

[tex]F = \frac{kQq}{L} ( \frac{1}{d} - \frac{1}{d + L})[/tex]

now plug in all data

[tex]F = \frac{(9\times 10^9)(2C)(10^{-9})}{2.5} (\frac{1}{2} - \frac{1}{(2 + 2.5)})[/tex]

[tex]F = 2N[/tex]

Answer:

ρ=0.0102lbm/ft^3

Explanation:

To solve this problem we must take into account the equation of continuity, this indicates that the sum of the mass flows that enter a system is equal to the sum of all those that leave.

Therefore, to find the mass flow of exhaust gases we must add the mass flows of air and fuel.

m=0.59+60=60.59lbm/s( mass flow of exhaust gases)

The equation that defines the mass flow (amount of mass that passes through a pipe per unit of time) is as follows

m=ρVA

Where

ρ=density

V=velocity

m=mass flow

A=cross-sectional area

solving for density

ρ=m/VA

ρ=60.59/{(1485)(4)}

ρ=0.0102lbm/ft^3

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

Answer:

Option B

Solution:

As per the question:

Heat produced at the rate of 10 W  

The resistor R and 2R are in series.

Also, in series, same current, I' passes through each element in the circuit.

Therefore, current is constant in series.

Also,

Power,[tex] P' = I'^{2}R[/tex]

When current, I' is constant, then

P' ∝ R

Thus

[tex]\frac{P'}{2R} = \frac{10}{R}[/tex]

P' = 20 W

The power dissipated in the resistor 2R is B. 20 W.

When resistors are connected in series, the current through each resistor is the same. Given two resistors, R and 2R, connected in series across a battery, we know that the power dissipated by resistor R is 10 W.

In a series circuit, the voltage across each resistor is different, but the current is the same.

The power dissipated by a resistor in a circuit is given by the formula:

P = I²R

Since the resistors are in series, the same current (I) flows through both resistors. Let's denote the current flowing through the series circuit as I. Using the given power dissipation for resistor R, we get:

10 W = I²R

To find the current:

I = √(10/R)

Next, we calculate the power dissipated by the resistor 2R:

P = I²(2R)

Substituting the value of I:

P = (10/R) × (2R) = 20 W

Thus, the power dissipated in resistor 2R is 20 W.

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

[tex]\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow  \text{Time taken by light}=2.8583\times 10^{-6}[/tex]

Time taken by light = 2.8583×10⁻⁶ seconds

We want to find the distance between the student and the lighting given that we know the time that passes between the moment he sees the lighting and the moment he hears it.

The solutions are:

Now, notice that there is an intrinsic problem with this question.

The student sees the "flash" when the light impacts on his eyes. Thus, the chain of events is:

So there is a little time between when the lighting happens and the student sees it, that will also have an impact on the distance to the lighting.

then we have:

(3*10^8 m/s)*t₁ = D

(343m/s)*t₂ = D

Where t₁ is the time that the light needs to reach the student, t₂ is the time the sound needs to reach the student, and D is the distance between the student and the ligthing.

We also know that:

t₂ - t₁ = 2.5s

Then we can write:

t₂ = t₁ + 2.5s and replace it on the second equation:

(343m/s)*(t₁ + 2.5s) = D

Then we have a system of equations:

(3*10^8 m/s)*t₁ = D

(343m/s)*(t₁ + 2.5s) = D

We can see that both of these left sides are equal to the same thing, then we must have:

(3*10^8 m/s)*t₁ = (343m/s)*(t₁ + 2.5s)

Then we solve it for t₁

t₁*(3*10^8 m/s - 343m/s) =  (343m/s)*2.5s = 857.5m

t₁ = (857.5m)/(3*10^8 m/s - 343m/s) = 2.858*10^(-6) s

Now that we know the value of t₁, we can find the value of D.

(3*10^8 m/s)*t₁ = D

(3*10^8 m/s)* 2.858*10^(-6) s = D = 857.501 m

Then the solutions are:

If you want to learn more, you can read:

https://brainly.com/question/4598472

Answer:

[tex]a_{avg} = 180.85 m/s^{2}[/tex]

Given:

Initial velocity, u = 0 m/s

Final velocity, v = 34 m/s

Time interval, [tex]\Delta t = 0.188 s[/tex]

Solution:

Acceleration is the rate at which the velocity of an object changes.

Thus

Average acceleration, [tex]a_{avg} = \frac{\Delta v}{\Delta t}[/tex]

[tex]a_{avg} = \frac{v - u}{t - 0} = \frac{34 - 0}{0.188} = 180.85 m/s^{2}[/tex]