Calculate the time (in seconds) needed for a car to accelerate from 0 m/s to 25 m/s at 5 m/s^2?

Answer:

[tex]\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.[/tex]

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges  [tex]\rm q_1[/tex] and [tex]\rm q_1[/tex], separated by a distance [tex]\rm r[/tex], is given by

[tex]\rm F = \dfrac{kq_1q_2}{r^2}.[/tex]

where k is the Coulomb's constant.

Initially,

[tex]\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.[/tex]

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

[tex]\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).[/tex]

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, [tex]\rm F_f = +1.30\ N.[/tex]

The new charges on the two objects are

[tex]\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.[/tex]

The new force is given by

[tex]\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\[/tex]

Using (1),

[tex]\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0[/tex]

[tex]\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0[/tex]

[tex]\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.[/tex]

Using (1),

When [tex]\rm q_1 = \pm 8.00\times 10^{-7}\ C[/tex],

[tex]\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.[/tex]

When [tex]\rm q_1=\pm 4.6\times 10^{-6}\ C[/tex],

[tex]\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.[/tex]

Since, [tex]\rm |q_2|>|q_1|[/tex]

Therefore, [tex]\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.[/tex]

Answer:

(a) [tex]L_{D} = 1.052\times 10^{- 4} m[/tex]

N = [tex]4.87\times 10^{4}[/tex]

(b) [tex]L'_{D} = 5.531\times 10^{- 6} m[/tex]

N' = [tex]7.087\times 10^{- 4}[/tex]

(c) [tex]L''_{D} = 4.43\times 10^{- 13} m[/tex]

N'' = [tex]3.63\times 10^{- 14}[/tex]

Solution:

As per the question, we have to calculate the Debye, [tex]L_{D}[/tex] length and N for the given cases.

Also, we utilize the two relations:

1. [tex]L_{D} = \sqrt{\frac{KT\epsilon_{o}}{ne^{2}}}[/tex]

2. N = [tex]\frac{4}{3}n\pi(L_{D})^{3}[/tex]

Now,

(a) n = [tex]10^{10} cm^{- 3}\times (10^{- 2})^{- 3} = 10^{16} m^{- 3}[/tex]

KT = 2 eV

Then

[tex]L_{D} = \sqrt{\frac{2\times 1.6\times 10^{- 19}\times 8.85\times 10^{- 12}}{10^{16}(1.6\times 10^{- 19})^{2}}}[/tex]

(Since,

e = [tex]1.6\times 10^{- 19} C[/tex]

[tex]\epsilon_{o} = 8.85\times 10^{- 12} F/m[/tex])

Thus

[tex]L_{D} = 1.052\times 10^{- 4} m[/tex]

Now,

N = [tex]\frac{4}{3}\times 10^{16}\pi(1.052\times 10^{- 4})^{3} = 4.87\times 10^{4}[/tex]

(b) n = [tex]10^{6} cm^{- 3}\times (10^{- 2})^{- 3} = 10^{12} m^{- 3}[/tex]

KT = 0.1 eV

Then

[tex]L'_{D} = \sqrt{\frac{0.1\times 1.6\times 10^{- 19}\times 8.85\times 10^{- 12}}{10^{12}(1.6\times 10^{- 19})^{2}}}[/tex]

[tex]L'_{D} = 5.531\times 10^{- 6} m[/tex]

N' = [tex]\frac{4}{3}\times 10^{12}\pi(5.531\times 10^{- 6})^{3} = 7.087\times 10^{- 4}[/tex]

(c) n = [tex]10^{17} cm^{- 3}\times (10^{- 2})^{- 3} = 10^{23} m^{- 3}[/tex]

KT = 800 eV

[tex]L''_{D} = \sqrt{\frac{800\times 1.6\times 10^{- 19}\times 8.85\times 10^{- 12}}{10^{23}(1.6\times 10^{- 19})^{2}}}[/tex]

[tex]L''_{D} = 4.43\times 10^{- 13} m[/tex]

N'' = [tex]\frac{4}{3}\times 10^{23}\pi(4.43\times 10^{- 13})^{3} = 3.63\times 10^{- 14}[/tex]

Answer:

b. It is dropped

Explanation:

If the initial velocity is zero, the object move from rest. That happens if the object is dropped

Answer:

ΔTmin = 3.72 °C

Explanation:

With a 16-bit ADC, you get a resolution of [tex]2^{16}=65536[/tex] steps. This means that the ADC will divide the maximum 10V input into 65536 steps:

ΔVmin = 10V / 65536 = 152.59μV

Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:

[tex]\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C[/tex]

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.

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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:

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:

4500 J and 3000 J

Explanation:

According to conservation of momentum

      [tex]0 = m_1 V_1 + m_2 V_2[/tex]

Given that m_2 = 1.5 m_1 , so

    [tex]V_1 = -1.5 V_2[/tex]

  the kinetic energy of each piece is

    [tex]K_2= \frac{1}{2} m_2v_2^2[/tex]

    [tex]K_1= \frac{1}{2} m_1v_1^2[/tex]

substituting the value of V1 in the above equation

    [tex]K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2[/tex]

  Given that

         K_1 + k_2 = 7500 J

       1.5 K_2 + K_2 = 7500

         K_2 = 7500 / 2.5

               = 3000 J

this is the KE of heavier mass

      K_1 = 7500 - 3000 = 4500 J

this is the KE of lighter mass

The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.

The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.

Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.

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Answer with Explanation:

The position of the particle as a function of time is given by [tex]s(t)=t^2-26t+133[/tex]

Part 1) The position as a function of time is shown in the below attached figure.

Part 2) By the definition of velocity we have

[tex]v=\frac{ds}{dt}\\\\\therefore v(t)=\frac{d}{dt}\cdot (t^2-26t+133)\\\\v(t)=2t-26[/tex]

The velocity as a function of time is shown in the below attached figure.

Part 3) The displacement of the particle in the first 19 seconds is given by [tex]\Delta x=s(19)-s(0)\\\\\Delta x=(19^2-26\times 19+133)-(0-0+133)=-133millimeters[/tex]

Part 4) The distance covered in the first 19 seconds can be found by evaluating the integral

[tex]s=\int _{0}^{19}\sqrt{1+(\frac{ds}{dt})^2}\\\\s=\int _{0}^{19}\sqrt{1+(2t-26)^2}\\\\\therefore s=207.03meters[/tex]

Part 4) As we can see that the position-time graph is parabolic in shape hence we conclude that the motion is uniformly accelerated motion.

Answer:

[tex]E=2.0*10^{11}N/m^{2}[/tex]

Explanation:

Relation between stress and Force:

[tex]\sigma=\frac{F}{A}=\frac{F}{\pi*d^{2}/4}[/tex]

Relation between stress and strain:

Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:

[tex]E=\frac{\sigma}{\epsilon}[/tex]

[tex]\epsilon=\frac{\Delta l}{l}[/tex]

So:

[tex]E=\frac{F*l}{\pi*d^{2}/4*\Delta l}=\frac{50*10^{6}*3}{\pi*(0.4^{2}/4)*5.967*10^{-3}}=2*10^{11}N/m^2[/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

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The distance between the two ships after two hours can be found using vector addition and the law of cosines, and the relative speed of ship A as seen by ship B is determined by subtracting the velocity vector of ship B from that of ship A.

The question asks to determine the distance between two ships two hours after their departure from the same port and the relative speed of ship A as seen from ship B. To solve this, we need to apply the concepts of vector addition and use trigonometry.

First, we find the vectors of the individual ships' movements. Ship A travels at 40 mph in a direction 35° west of north, and Ship B travels at 20 mph 80° east of north.

To find the distance between the two ships after two hours, we'll calculate each ship's displacement (speed multiplied by time) for two hours and then use trigonometry to determine the resultant displacement vector between the two ships. Using the law of cosines, we can find the distance between the ships. To find the relative speed of ship A from B, we will subtract the velocity vector of B from A.

These steps will yield the answers to the question after proper calculations.

To convert 1g of ice at absolute zero to 1g of boiling water, we need to calculate the heat required in three stages: heating the ice to melting point, melting the ice to water, and heating the water to boiling point. Adding the heat required for these three stages, we find the total amount of heat required is approximately 316.38 cal.

To find the total heat needed to turn 1 g of ice at absolute zero to boiling water, we calculate it in three parts. First, we calculate the heat required to raise the temperature of ice from absolute zero (-273.15 degrees Celsius) to 0 degrees Celsius using the formula: Q = mcΔT where m = 1g, c = 0.5 cal/(g*C), and ΔT = 273.15 °C. This gives us 136.58 cal.

Second, we need to factor in the heat needed to convert the ice at 0 degree Celsius to water at 0 degree Celsius. The heat of fusion for ice is known as 79.8 cal/g, so for 1 gram, the heat would be 79.8 cal.

Third, we calculate the heat needed from water at 0 degrees to water boiling point (100 degrees C). Here the formula remains the same but specific heat capacity changes to 1 cal/(g*C), which results in 100 cal.

Adding the calculated heat amounts for each phase gives the total heat required. Hence, the total amount of heat needed is 136.58 cal + 79.8 cal + 100 cal = 316.38 cal.

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To calculate the magnitude of the charge that creates a 4.70 N/C electric field at a point 3.70 m away, we can use Coulomb's law. The magnitude of the charge is approximately 2.58 * 10^(-6) C.

To calculate the magnitude of the charge that creates a 4.70 N/C electric field at a point 3.70 m away, we can use Coulomb's law. Coulomb's law states that the electric field is equal to the force exerted by the charge, divided by the distance squared. The equation is:

E = k * (q / r^2)

where E is the electric field, q is the charge, r is the distance, and k is the electrostatic constant.

In this case, we are given the electric field (4.70 N/C) and the distance (3.70 m). Plugging these values into the equation, we can solve for the magnitude of the charge:

4.70 N/C = k * (q / (3.70 m)^2)

Simplifying the equation, we have:

q = (E * r^2) / k

q = (4.70 N/C * (3.70 m)^2) / k

Using the given values of the electrostatic constant and performing the calculation, we find that the magnitude of the charge is approximately 2.58 * 10^(-6) C.

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = [tex]\frac{\textup{V}}{\textup{R}}[/tex]

or

I₀ = [tex]\frac{\textup{6.53}}{\textup{594}}[/tex]

or

I₀ = 0.0109 A

also,

I = [tex]I_0[1-e^{-\frac{t}{\tau}}][/tex]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = [tex]0.0109[1-e^{-\frac{\tau}{\tau}}][/tex]

or

I = [tex]0.0109[1-2.717^{-1}][/tex]

or

I = 0.00688 A

or

I = 6.88 mA

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:

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.

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Answer:[tex]\theta =27.26^{\circ}[/tex]

Explanation:

Given

Vector A has magnitude of 5 units

Vector B has magnitude of 9 units

Dot product of A and B is 40

i.e.

[tex]A\cdot B=|A||B|cos\theta [/tex]

[tex]40=5\times 9\times cos\theta [/tex]

[tex]cos\theta =\frac{8}{9}=0.889[/tex]

[tex]\theta =27.26^{\circ}[/tex]

Answer:

The weight required on the piston equals 420 Newtons.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}[/tex]

where

'F' is the force that acts on the piston

'A' is the area of the piston.

Since the force in the question is 21000 Newtons thus upon putting the values in the above equation we get

[tex]\frac{21000}{3.0m^{2}}=\frac{W}{0.06m^{2}}[/tex]

Solving for [tex]W[/tex] we get

[tex]W=\frac{21000}{3}\times 0.06\\\\\therefore W=420Newtons[/tex]

To lift an object weighing 21,000 N using a piston with an area of 0.060 m² and a platform of 3.0 m², you need a force of 420 N on the smaller piston.

To solve this problem, we need to use the principle of hydraulic systems which states that the pressure in one piston is equal to the pressure in the other piston.

The formula for pressure is:

P = F / A

where P is pressure, F is force, and A is area.

Given:

First, calculate the pressure exerted by the object on the larger platform:

P = F / A = 21,000 N / 3.0 m² = 7,000 Pa

Since the pressure is the same in the smaller piston, we can now find the force needed on the smaller piston:

F = P × A = 7,000 Pa × 0.060 m² = 420 N

Thus, a force of 420 N is required on the smaller piston to lift the object weighing 21,000 N.

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck  = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

              time =[tex]\dfrac{distance}{speed}[/tex]

              time =[tex]\dfrac{545}{28}[/tex]

              time =19.46 s

velocity of the car when it crosses the truck

[tex]S = ut + \dfrac{1}{2}at^2[/tex]

[tex]545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2[/tex]

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

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: a) is the vector sum of the electric field that would be caused by each charge separately.

Explanation: In order to explain this problem we have to know the electric field can be calculated by using the superposition principle so in this case this the right answer.

All the other are false, b) never can be the field zero along a line that connect the charges because it depend on teh sign of the charge and the distance between them. c) always the electric field can be determined for charged points. d) It is possible to find a zero electric field in any point if the points charges have different sign.  

The electric field surrounding two point charges is the vector sum of the electric field that would be caused by each charge separately. So the correct answer is A.

The correct answer is a) is the vector sum of the electric field that would be caused by each charge separately. When two point charges are present, the electric field at any point is the vector sum of the electric fields caused by each charge individually. The superposition principle states that the total electric field is the sum of all individual electric fields. Therefore, the electric field surrounding two point charges is the vector sum of the electric field caused by each charge separately.

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Answer:

boy average velocity to catch the ball just before it hits the ground is 8.53 m/s

Explanation:

given data

vertical distance = 75.1 m

horizontal distance = 33.4 m

to find out

Find the average velocity to catch the ball just before it hits the ground

solution

we know here ball is coming downward and boy is running to catch it

so first we calculate the time of ball to reach at ground that is express by equation of motion

s = ut + 0.5 × at²    .....................1

here u is initial speed that is zero and a = 9.8 and s is distance

put here all value to get time t  in equation 1

75.1 = 0 + 0.5 × (9.8)t²  

t = 3.914 s

so

boy speed is = [tex]\frac{distance}{time}[/tex]

speed = [tex]\frac{33.4}{3.914}[/tex]

speed = 8.53 m/s

so average velocity to catch the ball just before it hits the ground is 8.53 m/s

Answer:

- 1.65 × 10⁻⁷ C

Explanation:

Given:

Sides of the cube = 30 cm long

Electric flux, φ = - 3.1 kN.m²/C = - 3.1 × 10³ N.m²/C

Now,

For the cube,

The charge using the Gauss law is given as:

q = ε₀ × (nφ)

Here,

q is the charge

ε₀ = 8.85 × 10⁻¹² C²/N.m²

n is the number of sides for the cube = 6

Thus,

q = 8.85 × 10⁻¹² × 6 × (- 3.1 × 10³ )

or

q = - 1.65 × 10⁻⁷ C

Answer:

option(d)

Explanation:

The frequency of a wave is the property of the source of wave.

The velocity of all the electromagnetic waves is same as the speed of light. It only changes as the light passes through one medium to another.

The frequency is defined as the number of waves coming out from the source in 1 second.

As the frequency of wave increases, the number of wave coming per second increases.

So, the number of waves passing by increases but the speed remains same.

Option (d)

Answer:

[tex]3.33\times 10^{-4}[/tex] C

Explanation:

[tex]E[/tex] = Maximum electric field strength = [tex]3\times 10^{6}[/tex] N/C

[tex]r[/tex] = Radius of the sphere  = [tex]1 [/tex] m

[tex]Q[/tex] = maximum charge stored by the sphere  = ?

Considering that the total charge is stored at the center of the sphere, the electric field at the surface of sphere can be given as

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

Inserting the values for the variables in the above equation

[tex]3\times 10^{6}=\frac{(9\times 10^{9})Q}{1^{2}}[/tex]

[tex]3\times 10^{6}=(9\times 10^{9})Q[/tex]

Dividing both side by [tex](9\times 10^{9})[/tex]

[tex]\frac{3\times 10^{6}}{9\times 10^{9}}= \frac{9\times 10^{9}}{9\times 10^{9}}Q[/tex]

[tex]Q = \frac{3\times 10^{6}}{9\times 10^{9}}[/tex]

[tex]Q = 3.33\times 10^{-4}[/tex] C

Answer:

a) The magnitude of the car's total displacement (T) from the starting point is T = 82.67 Km

b) The angle (θ) from east of the car's total displacement measured from the starting direction is θ = 40.88 °

Explanation:

Attached you can see a diagram of the problem.

a) Find the magnitude of the vector T that goes from point A to point D (see the diagram).

The x and y components of this vector are

[tex]T_x=48+29*sin(30)=62.5 Km\\T_y=29+29*cos(30) = 54.11 Km[/tex]

The magnitude of the vector is find using the pythagoras theorem:

[tex]a^2=b^2+c^2[/tex], being a, b and c the 3 sides of the triagle that forms the vector:

[tex]T^2=T_x^2+T_y^2\\T=\sqrt{T_x^2+T_y^2}[/tex]

Replacing the values

[tex]T=\sqrt{(62.5)^2+(54.11)^2} \\T=82.67 Km[/tex]

b) Find the angle θ that forms the vector T and the vector AB (see diagram).

To find this angle you can use the inverse tangent

θ[tex]=tan^{-1}(\frac{T_y}{T_x})[/tex]

θ[tex]=tan^{-1}(\frac{54.11}{62.5})[/tex]

θ=40.88°

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:

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:

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:

W_planet=(W_earth*a)/g

Explanation:

The mass of the object does not change. It is the same at Earth and at the other planet.

W_earth=Mg       weight at Earth

W_planet=Ma       weight at other planet

If we divide the last equations:

W_planet/W_earth=a/g

W_planet=(W_earth*a)/g