A -2.37 µC point charge and a 4.74 µC point charge are a distance L apart.
Where should a third point charge be placed so that the electric force on that third charge is zero? (Hint: Solve this problem by first placing the -2.37 µC point charge at the origin and place the 4.74 µC point charge at x = −L.)

Answer:hello

Explanation:

Answer:

Fx = (783.451 , 373.063 , 240.124 , 240.124) N

Fy = (470.745 , 597.026 , 318.66)

Fres =768.216

Theta = 77.2 degrees from + x

Explanation:

F1 = 914N; F2 = 704N ; F3 = 399N

Part A

Fx1 = F1*cos (A) = 914*cos (31) = +783.451 N

Fx2 = F2*sin (B) = 704*sin (32) = -373.063 N

Fx3 = F3*cos (C) = 399*cos (53) = -240.124 N

                                   Sum of Fx = +170.26 N

Part B

Fy1 = F1*sin (A) = 914*sin (31) = +470.745 N

Fy2 = F2*cos (B) = 704*cos (32) = +597.026 N

Fy3 = F3*sin (C) = 399*sin (53) = -318.66 N

                                     Sum of Fy = + 749.111 N

Part C

Fres = sqrt (Fx^2 + Fy^2 )

Fres = sqrt (170.26^2 + 749.111^2)

Fres = 768.216 N

Part D

Angle = arctan (Fy/Fx)

Angle = arctan (749.111/170.26)

Angle = 77.2 degrees

To find the x and y components of each pull, we use trigonometry. The magnitude of the resultant pull is found by summing up the squares of the x and y components and taking the square root. The direction of the resultant pull is determined using the inverse tangent function.

To find the x components of each pull, we can use trigonometry. The x component of a force is given by the magnitude of the force multiplied by the cosine of the angle it makes with the x-axis. So, the x components of the three pulls are: F1x = F1 × cos(theta1), F2x = F2 × cos(theta2), F3x = F3 × cos(theta3).

To find the y components of each pull, we use the same approach but with the sine function. So, the y components of the three pulls are: F1y = F1 × sin(theta1), F2y = F2 × sin(theta2), F3y = F3 × sin(theta3).

Once we have the x and y components of each pull, we can find the magnitude of the resultant pull by summing up the squares of the x and y components and taking the square root: F = √((F1x)² + (F2x)² + (F3x)² + (F1y)² + (F2y)² + (F3y)²).

To find the direction of the resultant pull, we can use the inverse tangent function. The angle counted from the +x axis in the counterclockwise direction is given by: theta = atan(F1y + F2y + F3y / F1x + F2x + F3x).

So, the answers to the question are: F1x, F2x, F3x = 914N × cos(theta1), 704N × cos(theta2), 399N × cos(theta3); F1y, F2y, F3y = 914N × sin(theta1), 704N × sin(theta2), 399N × sin(theta3); F = √((F1x)² + (F2x)² + (F3x)² + (F1y)² + (F2y)² + (F3y)²); theta = atan(F1y + F2y + F3y / F1x + F2x + F3x).

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Answer:

A. 10^12 meters which is bigger than the distance between Sun and Earth

Explanation:

In a case where we take all the atoms in a single drop of water and put them on a single line as closely packed as they can be, the total length of the line would be a function of the diameter of an atom.

Total length L = Diameter of an atom d × number of atom in a droplet of water N

L = dN

N = 10^22

d ~= 0.1nm = 10^-10m

L = 10^22 × 10^-10 m

L = 10^12 m

So, the length would be approximately 10^12 m

distance between the sun and earth is 147.34million km

D= 1.47 × 10^11 m

L > D

Therefore, the length would be approximately 10^12 m

Which is greater than the distance between the earth and the sun.

Answer:

N = 2500

Explanation:

By definition, the sampling frequency (fs) is:

[tex] f_{s} = N / T [/tex]

where N: is the number of samples and T: is the time

Hence, the number of samples at the sampling frequency, 500 Hz, to get 5 seconds of data is:

[tex] N = f_{s} \cdot T = 500 s^{-1} \cdot 5 s = 2500 [/tex]  

I hope it helps you!                        

To capture a 50Hz signal for 5 seconds with a sampling frequency of 500 Hz, we need 2500 samples.

To capture the 50Hz signal for 5 seconds, we need to calculate the number of samples (N) at the assumed sampling frequency (fs) of 500 Hz. According to the Nyquist theorem, the highest frequency we can accurately monitor is half the sampling frequency. In this case, the Nyquist frequency would be 250 Hz. Since the signal frequency is 50 Hz, it is well below the Nyquist frequency. Therefore, we need to sample the signal at a rate greater than twice the signal frequency, which is 100 samples per second. Multiplying this by the desired duration of 5 seconds, we get the total number of samples required:

N = fs * duration = 500 Hz * 5 s = 2500 samples

Answer:

f=qB/2[tex]\pi[/tex]m

Explanation:

The cyclotron frequency is the number of revolutions the particle makes in one second. calculate the cyclotron frequency for this particle

solution

The Lorentz force  [tex]F_{lorentz} =F_{centripetal}[/tex]

is the centripetal force  

and makes the particles path to revolve in a circle:

qvB=mv^2/r

radius=r

m=mas of the particle

B=magnetic flux

q=quantity of charge

v=velocity of the particle

v=qBr/m

where v is the velocity of the particle

recall the velocity v=rω

v=2*Pi*f*r

[tex]2\pi *f*r[/tex]=qBr/m

f=qB/2[tex]\pi[/tex]m

the cyclotron frequency is therefore f=qB/2[tex]\pi[/tex]m

the period(the time it take to make a complete oscillation) will be the inverse of frequency

T=2[tex]\pi[/tex]m/qB

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

[tex]v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s[/tex]

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

[tex]\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m[/tex]

The locations are +5.9 cm or -5.9 cm

Final answer:

The maximum speed of the object is approximately 1.06 m/s. When the object's velocity is one-third of the maximum speed, the object is located approximately 0.060 m from the equilibrium position.

Explanation:

To calculate the maximum speed of the object, the equation for the period of oscillation can be used. The period, T, can be calculated using the formula T = 2π√(m/k), where m is the mass of the object and k is the force constant of the spring. Substituting the given values, we have T = 2π√(0.225/74.5) = 0.942 s. The maximum speed, vmax, can be determined using the formula vmax = 2πA/T, where A is the amplitude of the oscillation. In this case, the amplitude is given as 6.25 cm, which goes to 0.0625 m. Substituting the values, we have vmax = 2π(0.0625)/0.942 = 1.06 m/s.

To find the locations of the object when its velocity is one-third of the maximum speed, we can use the equation for the displacement of an object undergoing simple harmonic motion. The equation is given as x = Acos(ωt + φ), where x is the displacement, A is the amplitude, t is the time, ω is the angular frequency, and φ is the phase constant. Since the object is released from rest, the phase constant is 0. The angular frequency can be calculated using the formula ω = 2π/T, where T is the period. Substituting the given values, we have ω = 2π/0.942 = 6.69 rad/s. The time when the velocity is one-third of the maximum speed can be found by rearranging the formula for velocity, v = ωAsin(ωt + φ), to T/6.69 = Asin(ωt + φ). Solving for t, we find that t ≈ 0.303 s. Substituting this value into the equation for displacement, we have x = Acos(ωt + φ) = 0.0625cos(6.69(0.303)) ≈ 0.060 m.

Answer:

ms^(-3)

Explanation:

V = kt^2

m/s = k (s)^2

k = m/ s^(3)

k = ms^(-3)

Answer:

Ex= -23.8 N/C  Ey = 74.3 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  88.6 N/C + (-14.3 N/C) =74.3 N/C

In this exercise we have to use the knowledge of electric field to calculate the components, in this way we have that:

[tex]E_x= -23.8 N/C \\ E_y = 74.3 N/C[/tex]

As the energetic force exist undeviating, and the energetic field, essentially, is just this energetic force for one charge, we can use the superposition standard to take the energetic field caused by two together charges at any time, as the other charge exist absent.

Due to q₁ exist negative, and situated ahead of the y point around which something revolves, the field on account of this charge will happen indicating upward, in this manner:

[tex]E_1 = k*(4.55 nC) / r_1^2\\ E_{1x} = 0 \\ E_{1y} = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C \\ r_2^2 = 1^2m^2 + (0.6)^2m^2 = 1.36 m^2\\ E_2 = k*q_2 / r^2_2 = 9*10^9*(4.2)*10^9 / 1.36 = 27.8 N/C\\ E_{2x} = E_2* cos\theta\\ E_{2y} = E_2*sin \theta\\ cos \theta = -0.855\\ sin \theta = -0.513\\ E_{2x} = 27.8 N/C * (-0.855) = -23.8 N/C \\ E_{2y} = 27.8 N/C * (-0.513) = -14.3 N/C [/tex]

The total x and y part on account of two together charges exist just the total of the part of Ex and Ey:

[tex]E_x = E_{1x} + E_{2x} = 0 + (-23.8 N/C) = -23.8 N/C\\ E_y = E_{1y} + E_{2y} = 88.6 N/C + (-14.3 N/C) =74.3 N/C [/tex]

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Answer:

[tex]v_2[/tex]≅-4.1 m/s (-ve for left)

[tex]v_2\\[/tex] ≅4.1 m/s (To the left)

Explanation:

According to the conservation of momentum:

Momentum before collision=Momentum After Collision

[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]

Where:

[tex]m_1[/tex] is the 15g marble

[tex]m_2[/tex] is the 22 g marble

[tex]v_1[/tex] is the velocity of 15g marble before collision

[tex]v_2[/tex] is the velocity of 22g marble before collision

[tex]v'_1[/tex] is the velocity of 15g marble after collision

[tex]v'_2[/tex] is the velocity of 22g marble after collision

Note: -ve sign for left, +ve sign for right

We have to calculate v_2:

Above equation on rearranging will become:

[tex]v_2=\frac{m_1v'_1+m_2v'_2-m_1v_1}{m_2} \\v_2=\frac{(0.015kg)(-5.4m/s)+(0.022kg)(2m/s)-(0.015kg)(3.5)}{0.022kg} \\v_2=-4.06 m/s\\[/tex]

[tex]v_2[/tex]≅-4.1 m/s (-ve for left)

[tex]v_2\\[/tex] ≅4.1 m/s (To the left)

Final answer:

The initial velocity of the 22 g marble before the elastic collision can be calculated using the conservation of momentum. After setting up the momentum equation with the given masses and velocities and solving for the unknown initial velocity, it is found to be 4.1 m/s to the left. So the correct option is d.

Explanation:

To calculate the initial velocity of the 22 g marble in an elastic collision, we use the conservation of momentum and conservation of kinetic energy principles. The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

The initial momentum of the system can be calculated by the sum of the momentum of each marble before they collide. Since we know the mass and final velocity of each marble, and the initial velocity of the 15 g marble, we can set up the equation:

(Mass of 15 g marble) × (Initial velocity of 15 g marble) + (Mass of 22 g marble) × (Initial velocity of 22 g marble) = (Mass of 15 g marble) × (Final velocity of 15 g marble) + (Mass of 22 g marble) × (Final velocity of 22 g marble)

Plugging in the known values and solving for the initial velocity of the 22 g marble, we get:

(0.015 kg) × (3.5 m/s) + (0.022 kg) × (Initial velocity of 22 g marble) = (0.015 kg) × (-5.4 m/s) + (0.022 kg) × (2.0 m/s)

0.0525 kg·m/s + (0.022 kg) × (Initial velocity of 22 g marble) = -0.081 kg·m/s + 0.044 kg·m/s

0.0525 kg·m/s + (0.022 kg) × (Initial velocity of 22 g marble) = -0.037 kg·m/s

(0.022 kg) × (Initial velocity of 22 g marble) = -0.037 kg·m/s - 0.0525 kg·m/s

(0.022 kg) × (Initial velocity of 22 g marble) = -0.0895 kg·m/s

Initial velocity of 22 g marble = -0.0895 kg·m/s / 0.022 kg

Initial velocity of 22 g marble = -4.068 m/s (to the left, since it's negative)

The initial velocity of the 22 g marble is 4.1 m/s to the left, which corresponds to option (d).

To find the electric field strength that would store 12.5 J of energy in every 6.00 mm3 of space, one must use the formula for energy density related to electric field strength, considering the permittivity of free space.

To solve for the electric field strength that would store 12.5 J of energy in every 6.00 mm3 of space, we need to use the relationship between the electric field (E), the energy density (u), and the permittivity of free space (ε0). The energy density in an electric field is given by u = ½ε0E2. Rearranging for E, and substituting in the given values allows us to solve for the electric field strength.

Assuming ε0 = 8.85 x 10-12 C2/N·m2, and converting the volume from mm3 to m3 (6.00 mm3 = 6.00 x 10-9 m3), we can substitute these values into the rearranged formula E = √(2u/ε0) to find the electric field strength needed to store 12.5 J of energy.

To solve this problem we will use Henry's law. This law states that at a constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure exerted by that gas on the liquid. Mathematically it is formulated as follows:

[tex]C = K_H*P[/tex]

Where,

[tex]K_H[/tex] = Henry's constant for C02 at 25°C is equal to [tex]3.6*10^{-2}M/atm[/tex]

C = Gas concentration is 0.19M

Replacing we have,

[tex]0.19 M = (3.6*10^-2 M/atm)*P[/tex]

[tex]P = 5.277 atm[/tex]

Therefore the pressure of carbon dioxide is 5.277 atm

To calculate the pressure of CO2 needed to maintain a 0.19 M concentration in a can of soda, you apply Henry's law, which results in the pressure equalling 0.00684 atm.

The subject of this question is Henry's law, which is used in Chemistry to relate the solubility of a gas in a liquid to the pressure of that gas above the liquid. In this case, we're asked to calculate the pressure of the CO2 gas needed to maintain a certain concentration in a can of soda. Henry's law is defined as: P = kH × C, where P is the gas pressure, kH is Henry's law constant, and C is the concentration.

To solve this problem, we rearrange Henry's law to find the pressure: P = kH × C. Given that the Henry's law constant for CO2 is 3.6 × 10−2 M/atm and the CO2 concentration is 0.19 M, we can apply these values and get P = (3.6 × 10−2 M/atm) × (0.19 M) which equals 0.00684 atm.

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Explanation of finding the net electric force on a charge Q using Coulomb's Law in an arrangement with charges on the square's corners.

To find the net electric force on a charge Q located at the center of a square with charges on its corners, we must consider the forces exerted by each charge using Coulomb's Law.

The net electric force on charge Q can be calculated by adding the individual force vectors from the charges Q₁ and Q₂.

By determining the force vectors and summing them up, we can express the net electric force on charge Q in terms of q, Q, a, and appropriate constants.

The magnitude of the net electric force exerted on the charge Q at the center of the square is [tex]\(\frac{4kqQ}{a^2}\).[/tex]

To find the magnitude of the net electric force exerted on the charge Q at the center of the square, we can calculate the electric force exerted by each individual charge and then sum them up vectorially.

Let q be the magnitude of each corner charge and a be the side length of the square.

The electric force between two charges q separated by a distance r is given by Coulomb's law:

[tex]\[F = \frac{k \cdot q \cdot Q}{r^2}\][/tex]

Considering symmetry, the horizontal components of the electric forces from the charges on the left and right sides cancel out, as do the vertical components from the top and bottom charges.

Thus, the net electric force acting on Q is the sum of the forces from the top and bottom charges, which are equal in magnitude:

[tex]\[F_{\text{net}} = 2 \cdot F_{\text{top}}\][/tex]

[tex]\[F_{\text{net}} = 2 \cdot \frac{k \cdot q \cdot Q}{(\frac{a}{\sqrt{2}})^2}\][/tex]

[tex]\[F_{\text{net}} = 2 \cdot \frac{k \cdot q \cdot Q}{\frac{a^2}{2}}\][/tex]

[tex]\[F_{\text{net}} = \frac{4kqQ}{a^2}\][/tex]

Therefore, the magnitude of the net electric force exerted on the charge Q at the center of the square is [tex]\(\frac{4kqQ}{a^2}\).[/tex]

Answer:

T = approximately 24 hs.

Explanation:

In order to keep the satellite over a fixed point on the equator, as the earth rotates, the satellite must have the same angular velocity that Earth has, which means that it must have a period equal to the time used by Earth to complete an entire rotation on itself, which is almost exactly 24 Hs.

Mathematically, this can be obtained taking into account that the force that keeps the satellite in orbit is the centripetal force, which is actually the gravitational force exerted by Earth, so we can write the following equality:

Fg = Fc ⇒ G*ms*me / (re +rsat)² = ms*ω²*(re +rsat)

By definition, ω =ΔФ / Δt

For a complete revolution, ΔФ = 2*π, and Δt = T (period of the rotation),

so we can replace ω by (2*π/T), solving then for T:

T= 86,313 sec. (24 hs are exactly 86,400 sec, so the value is actually very close to the theorical one).

The period of a satellite in a geosynchronous orbit is 24 hours, matching the Earth's rotation. The satellite follows a circular orbit keeping it in the same place relative to Earth's surface. Satellites in this orbit provide various services due to their continuous presence above a particular location.

The period of a satellite in a geosynchronous orbit is 24 hours. It's called geosynchronous because it matches the Earth's rotation period which is also 24 hours, this is how they are able to stay over the same point on the Earth's surface. The satellites move in elliptical orbits, but those in geosynchronous orbit follow a circular path so they maintain a constant altitude and remain in constant position relative to the Earth's surface.

As there are many satellites launched each year, being in a geosynchronous orbit helps these satellites to provide continuous services like weather tracking, communications, global positioning systems, etc. These satellites also avoid being captured by Earth's gravity and pulled into the atmosphere because they are at a distance of approximately 36000km from the earth's surface, where gravity is much weaker.

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Answer:

(a) 0.017m/s^2

(b) 17/100,000

(c) 0.17m, 0.558ft

Explanation:

(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s

Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2

(b) g = 9.8m/s^2, a = 0.0017m/s^2

a/g = 0.0017/9.8 = 0.00017 = 17/100,000

(c) Distance = speed × time = 0.017m/s × 10s = 0.17m

Distance in foot = 0.17 × 3.2808ft = 0.558ft

Answer:

a) acceleration a = 2.68 m/s^2

b) fraction of g = 0.273 or 27.3%

c) distance travelled d = 134m (in SI unit)

d = 439.5ft (in feet)

Explanation:

a) converting 60mph (miles per hour) to m/s

∆v = 60 miles/hour × 1609.344meters/mile × 1/3600 seconds/hour

∆v = 26.82m/s

t = 10s

acceleration = change in velocity/time = ∆v/t

Acceleration a = 26.82/10 = 2.68m/s^2

b) acceleration due to gravity is g = 9.8m/s^2

Fraction of g = a/g = 2.68/9.8 = 0.273

Fraction = 0.273 or 27.3%

c) distance travelled can be calculated using the equation below.

d = ut + 0.5at^2 ....1

Initial speed u = 0

time t = 10s

Acceleration a = 2.68m/s^2

d = distance covered.

From equation 1, putting u = 0

d = 0.5at^2

d = 0.5(2.68 × 10^2)

d = 134m

Converting meters to feet

d = 134m × 3.28ft/m

d = 439.5ft

The force on an ion interacting with an induced dipole can be found using Coulomb's law. The force can be represented as Fionondipole = qEinduced, where q is the charge of the ion and Einduced is the induced electric field. The magnitude of the force can be expressed as (1/(4πε0)) * (2αEext)/(r3) * q, where α is the polarizability of the molecule, Eext is the external electric field, and r is the distance between the ion and the molecule.

The force experienced by an ion interacting with an induced dipole can be found using Coulomb's law. The force can be represented as Fionondipole = qEinduced, where q is the charge of the ion and Einduced is the induced electric field. The induced electric field can be expressed as Einduced = (1/(4πε0)) * (2αEext)/(r3), where α is the polarizability of the molecule, Eext is the external electric field, and r is the distance between the ion and the molecule. Combining these expressions, the magnitude of the force Fionondipole can be written as:

Fionondipole = (1/(4πε0)) * (2αEext)/(r3) * q

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Answer

70 m/s,  140 m/s, 28 m/s²

Explanation:

Average velocity = total distance travel / time = 350 / 5 = 70 m/s

average velocity = (initial velocity + final velocity) / 2

70 × 2 = 140 m/s

change in velocity = final velocity - initial velocity = 140 m/s - 0 = 140 m/s

acceleration = (final velocity - initial velocity) / t = 140 m/s / 5 = 28 m/s²

Answer:

1. Final velocity is 140m/s

2. Change in velocity is 140m/s

3. Acceleration is28m/s^2

Explanation:

Initial velocity u=0,

Distance travelled S=350m

Time taken t=5 seconds

The car move constant acceleration, then we can use any of equations of motion

v=u+at

v^2=u^2+2as

s=ut+(at^2)/2

Using equation 3

S=ut+(at^2)/2

350= 0×5+ (a ×5^2)/2

350= 0+(a×25)/2

350=25a/2

350×2=25a

700=25a

a=700/25

a=28m/s^2. Answer

1. Now, using equation 1

v=u+at

v=0+28×5

v=0+140

v=140m/s. Answer

2. The changed in the velocity of the car is final velocity minus the initial velocity

Change in velocity = v-u

Change in velocity = 140-0

Therefore,

Change in velocity is 140m/s. Answer

3. The acceleration has been answered before solving and it is

a= 28m/s^2

Answer:

a)The acceleration of the airplane is 2.5 m/s².

b)The distance AB is 1125 m.

Explanation:

Hi there!

a)The equation of velocity of an object moving in a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time "t".

v0 = initial velocity.

a = acceleration.

t = time

We have the following information:

The airplane starts with zero velocity (v0 = 0) and its velocity after 30 s is 270 km/h (converted into m/s: 270 km/h · 1000 m/1 km · 1 h /3600 s = 75 m/s). Then, we can solve the equation to obtain the acceleration:

75 m/s = a · 30 s

75 m/s / 30 s = a

a = 2.5 m/s²

The acceleration of the airplane is 2.5 m/s².

b)The distance AB can be calculated using the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time "t".

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at A, then, x0 = 0. Since the airplane is initially at rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

Let´s find the position of the airplane after 30 s:

x = 1/2 · 2.5 m/s² · (30 s)²

x = 1125 m

The position of B is 1125 m away from A (the origin), then, the distance AB is 1125 m.

(a) the acceleration of the plane is 2.5 m/s²

(b) The distance AB is 1125m

(a) Given that the initial velocity of the plane at point A is u = 0

and the final velocity of the plane at point B of take-off is v = 270 km/h

[tex]v=\frac{270\times1000}{3600}m/s=75m/s[/tex]

the time taken to teach point B is t = 30s

So from the first equation of motion, we get:

[tex]v=u+at[/tex]

where a is the acceleration.

[tex]75=a\times30\\\\a=2.5\;m/s^2[/tex]

(b) From the second equation of motion we get,

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

where s is the distance

[tex]s = \frac{1}{2}\times2.5\times30^2\\\\s=1125m[/tex]

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Answer:

Coefficients are 1/2, -z and +1

Explanation:

The step by step explanation is in the attachment given below.

Answer:

0.556m

Explanation:

unit conversion

58.2 cm = 0.582 m -> the outer radius is 0.582/2 = 0.291 m

[tex]7.87 g/cm^33 = 7.87 g/cm^33 * 1/1000 (kg/g) * 100^3 cm^3/m^3 = 7870 kg/m^3[/tex]

Let r be the inner diameter we can find the volume of the hollow part of the sphere using the following formula

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

The volume of the iron shell is the volume of the whole sphere subtracted by volume of the hollow part

[tex]V_s = V - V_h = \frac{4}{3}\pi 0.291^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.0246 - r^3)[/tex]

Let water density [tex]\rho_w = 1000 kg/m^3[/tex], then the buoyant force is the weight of water displaced by the shell

[tex]F_b = g\rho_w V = g1000\frac{4}{3}\pi(0.291^3) = g1000\frac{4}{3}\pi(0.0246)[/tex]

For the shell to almost completely submerged in water, its buoyant force equal to the weight of the shell

[tex]F_b = W_s[/tex]

[tex]g1000\frac{4}{3}\pi(0.0246) = g\rho_sV_s[/tex]

[tex]g1000\frac{4}{3}\pi(0.0246) = g7870\frac{4}{3}\pi(0.0246 - r^3)[/tex]

[tex]1000*0.0246 = 7870(0.0246 - r^3)[/tex]

[tex]24.6 = 193.602 - 7870r^3[/tex]

[tex]r^3 = (193.602 - 24.6)/7870 = 0.02147[/tex]

[tex]r = \sqrt[3]{0.02147} = 0.278 m[/tex]

So the inner diameter is 0.278*2 = 0.556 m

To find the inner diameter of the hollow sphere, one must take steps of calculating the total volume of the sphere, then the weight, the displaced water volume and finally the total internal volume. With this information, the inner diameter of the sphere can be worked out.

To solve this physics problem, we need to find the inner diameter of a hollow spherical iron shell that is floating almost completely submerged in water. The student has provided the outer diameter of the shell (58.2 cm) and the density of iron (7.87 g/cm3). Using these values, we'll use some concepts from physics, specifically principles of buoyancy and formulas for volume and density.

The first step will be to calculate the total volume of the sphere using the outer diameter. This can be done with the formula for the volume of a sphere: V=(4/3)πR^3, where R is the radius, which is half the diameter.

Since the sphere is floating almost completely submerged, it's displacing a volume of water equivalent to its own weight. Now that we know the sphere's volume, we can calculate the sphere's weight using the density of iron.

The weight then can be used to calculate the volume of water displaced, which leads us to the total internal volume of the sphere (total volume - displaced volume). Then the inner diameter can be found by rearranging the formula of the volume of the sphere.

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Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²[tex]\sqrt{((17.4)^{2} +(16.9)^{2}}[/tex] = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

Final answer:

Scientific notation is essential in handling large numbers in physics. The gravitational constant is crucial for calculating gravitational forces. Understanding Earth's mass involves utilizing known values such as its radius and the gravitational constant.

Explanation:

Scientific Notation: Scientific notation is commonly used in astronomy and other sciences to represent very large or small numbers efficiently.

Gravitational Constant: The gravitational constant, denoted by G, is a fundamental constant used to calculate gravitational forces.

Earth's Mass Calculation: Earth's mass, represented as 5.97 x 10²⁴ kg, can be calculated using known values like its radius and the gravitational constant.

Answer:

16916.4 V

Explanation:

Electric potential: This is the work done in bringing a unit positive charge from infinity to that point in against the action of the field. The S.I unit  of Electric potential is V.

mathematically, Electric potential can be expressed as

P = E×d ....................................... Equation 1.

Where P = Electric potential, E = Electric Field, d = distance/height of the level at the top of the Washington  Monument.

Given: E = 100 V/m, d = 555 ft = 555×0.3048 m = 169.164 m.

Substitute into equation 1

P = 100×169.164

P = 16916.4 V.

Thus the potential difference = 16916.4 V.

The magnitude of the potential difference between the ground and the top of the Washington Monument, which is 555 feet or approximately 169.164 meters tall, caused by Earth's electric field of about 100 V/m, is roughly 16,916.4 volts.

The question refers to calculating the electric potential difference caused by Earth's electric field. The provided value of Earth's electric field is approximately 100 V/m near its surface. To find the potential difference (ΔV) between a point on the ground and a point at the height of the top of the Washington Monument, we can use the equation:

ΔV = E * d

where E is the electric field strength and d is the distance in meters. Since the height of the Washington Monument is given in feet, it needs to be converted to meters (1 foot = 0.3048 meters). The height in meters is:

555 ft * 0.3048 m/ft = 169.164 m

Now, we can calculate the potential difference:

ΔV = 100 V/m * 169.164 m = 16916.4 V

Thus, the magnitude of the potential difference between a point on the ground and a point at the level of the top of the Washington Monument is approximately 16,916.4 volts.

To solve this problem we will apply the concept designed to generate the conversion from one unit to another. Basically, what is sought is to eliminate the units of the denominator and the numerator of the conversion factor and the unit to be converted respectively, leaving the units of the new unit. In mathematical terms this is,

[tex]1 in = 2.54cm[/tex]

If we want to convert 28.4in then the conversion factor versus the unit would be

[tex]x = 28.4 in (\frac{2.54cm}{1in})[/tex]

[tex]x = 72.136 cm[/tex]

Therefore the factor of conversion will be 2.54 and the final units for the value given is 72.136cm

Answer:

26.6 cm

Explanation:

We are given that

Spring constant=23.6 N/m

Force exert=6.28 N

We have to find the displacement.

We know that Hooke's law

[tex]F=kx[/tex]

Where k= Spring constant

x=Displacement

Using the formula

Then, we get

[tex]6.28=23.6 x[/tex]

[tex]x=\frac{6.28}{23.6}=0.266m[/tex]

We know that 1 m=100 cm

[tex]x=0.266\times 100=26.6 cm[/tex]

Hence, the displacement =26.6 cm

Answer:The horizontal area enclosed by the waterline of the ship is 12,797.88m2

Explanation: Mass of the ship=29000kg

Total mass of 50 ships= 50×29000

Total Mass=1450000kg

Volume of water with mass= mass/ density

Assuming density of ocean water =1030kgm3

Volume=1450000/1030

Volume=1407.77m3

Area enclosed by the ship =A=V/h

h=11cm=0.11m

Area= 1407.77/0.11

Area=12,797.88m2

Answer:Kobe

Explanation:

The small difference at the time of reporting the length should be the random uncertainty.

The following information related to the random certainty is:

There should be a difference in the length as compared to the real one for various members.

Therefore we can conclude that The small difference at the time of reporting the length should be the random uncertainty.

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Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

Answer:

Power will be 356.90 watt  

Explanation:

We have given total number of steps in the stair = 1600

And height id the stair h = 320 m

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

Mass is given m = 75 kg

So work done in climbing the stairs [tex]W=mgh=75\times 9.8\times 320=235200J[/tex]

Time is given t = 10 min 59 sec

So time [tex]t=10\times 60+59=659sec[/tex]

We know that power is rate of doing work

So power [tex]P=\frac{Work}{time}=\frac{235200}{659}=356.90watt[/tex]

So power will be 356.90 watt

The man generated approximately 357.26 watts or 0.479 horsepower while climbing the stairs.

To calculate the power generated by the man while climbing the stairs, we can use the work-energy principle. The work done by the man in climbing to a height  against the force of gravity is equal to the change in his gravitational potential energy. The power is then the work done divided by the time taken to do it.

 The gravitational potential energy (PE) is given by:

[tex]\[ PE = mgh \][/tex]

where:

-[tex]\( m \)[/tex] is the mass of the man (75 kg),

-[tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \) on Earth),[/tex]

- [tex]\( h \)[/tex] is the height climbed (320 m).

 First, we calculate the work done (PE):

[tex]\[ PE = 75 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 320 \, \text{m} \][/tex]

[tex]\[ PE = 235,440 \, \text{J} \][/tex]

 Now, we convert the time from minutes to seconds because power is typically calculated in joules per second (watts). The time taken is 10 minutes and 59 seconds, which is:

[tex]\[ 10 \times 60 \, \text{s} + 59 \, \text{s} = 659 \, \text{s} \][/tex]

 Power (P) is the work done divided by the time taken:

[tex]\[ P = \frac{PE}{t} \][/tex]

[tex]\[ P = \frac{235,440 \, \text{J}}{659 \, \text{s}} \][/tex]

[tex]\[ P \approx 357.26 \, \text{W} \][/tex]

 To convert watts to horsepower, we use the conversion factor [tex]\( 1 \, \text{hp} = 746 \, \text{W} \):[/tex]

[tex]\[ P_{\text{hp}} = \frac{P}{746} \][/tex]

[tex]\[ P_{\text{hp}} \approx \frac{357.26 \, \text{W}}{746} \][/tex]

[tex]\[ P_{\text{hp}} \approx 0.479 \, \text{hp} \][/tex]

 Therefore, the man generated approximately 357.26 watts or 0.479 horsepower while climbing the stairs.

 The final answer in the correct format is:

[tex]\[ \boxed{357.26 \, \text{W}} \][/tex]

[tex]\[ \boxed{0.479 \, \text{hp}} \][/tex]

 The answer is:[tex]0.479 \, \text{hp}.[/tex]

Answer:

DC = 3 div. AC= 0

Explanation:

When the input is directed to the input circuitry, and the "A input" switch is set to DC, the pure 15 V DC signal will be showed on the screen, so, if the vertical setting is at 5 V /div, the trace will deflect exactly 3 div.

If the switch is on AC, as this setting inserts a capacitor in series (which is located here to block any unwanted DC component superimposed to an AC signal) the DC signal will be blocked, so no trace will be deflected on the screen (after completed the transient period).

In this case, there are 3 divisions if the switch is on direct current (DC), whereas there are 0 divisions if the switch is on alternating current (AC).

In conclusion, there are 3 divisions if the switch is on direct current, whereas there are 0 divisions if the switch is on alternating current.

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Answer:

A. (200i + 85j - 30k)m

B.  345m

C. 219.37 m

Explanation:

UP: +VE Z

NORTH: +VE Y

EAST: +VE X

A.

DISPLACEMENT : (0.2*1000 i + (-15+0.1*1000)j + (-3*10)k) m

                            = (200i + 85j - 30k)m

B. total distance = 10*3+15+200+100 = 345m

C.

Magnitude = [tex]\sqrt{{200^{2} } +85^{2} +(-30)^{2}[/tex] = 219.37 m

unit vector = [tex]\frac{200i+85j-30k}{219.4}[/tex]

The displacement from the apartment to the restaurant is given by [tex]\Delta x =200\hat{i}+85\hat{j}-30\hat{k}[/tex].

The distance from the apartment to the restaurant is 345 m

The magnitude of the displacement from the apartment to the restaurant is 219.37 m.

Given that;

Part A

To find the final position, let us find the position in each direction;

Part B

Here, we are asked to find the total distance covered.

For that, we add all the distances irrespective of the direction.

Part C

To find the magnitude of the displacement,

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