What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

Answer:

The two masses are 3.39 Kg and 1.75 Kg

Explanation:

The gravitational force of attraction between two bodies is given by the formula;

F = Gm₁m₂/d²

where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²

m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects

Further calculations are provided in the attachment below

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]

==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

Where K indicates spring constant

m indicates mass

For the new time period

[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]

Now, we will take 2 ratios of the time period

[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]

[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]m' = \frac{0.500}{0.5625}[/tex]

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.

Given :

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.

The formula of the time period is given by:

[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex]   ---- (1)

where m is the mass and K is the spring constant.

The new time period is given by:

[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex]   ---- (2)

where m' is the total mass after the addition and K is the spring constant.

Now, divide equation (1) by equation (2).

[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]

Now, substitute the known terms in the above expression.

[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]

Simplify the above expression in order to determine the value of m'.

[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]

m' = 0.889 Kg

Now, the mass added to the object to change the period to 2.00 s is given by:

m" = 0.889 - 0.500

m" = 0.389 Kg

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Answer:

  K_{f} / K₀ =1.12

Explanation:

This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.

Initial moment. With arms outstretched

         L₀ = I₀ w₀

the wo value is 5.0 rad / s

final moment. After he shrugs his arms

         [tex]L_{f}[/tex] = I_{f}  w_{f}

indicate that the moment of inertia decreases by 11%

        I_{f} = I₀ - 0.11 I₀ = 0.89 I₀

        L_{f} = L₀

        I_{f} w_{f}  = I₀ w₀

        w_{f} = I₀ /I_{f}    w₀

let's calculate

        w_{f} = I₀ / 0.89 I₀   5.0

        w_{f} = 5.62 rad / s

Having these values ​​we can calculate the change in kinetic energy

         [tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)

         K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²

         K_{f} / K₀ =1.12

Answer:

A. Her total angular momentum has decreased

Explanation:

Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.

The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.

Answer:

95.5 m

Explanation:

The displacement is the position of the ending point relative to the starting point.

In this case, the magnitude of the displacement is the diameter of the circular track.

d = 300 m / π

d ≈ 95.5 m

Answer:

 θ  = 180

Explanation:

When an electric dipole is placed in an electric field, there is a torque due to the electric force

           τ = p x E

by rotating the dipole there is a change in potential energy

        ΔU = ∫ τ dθ

        ΔU = p E (cos θ₂ - cos θ₁)

when the dipole starts from an angle to the equilibrium position for θ = 0

          ΔU = pE (cos θ  - cos 0)

           cos θ  = 1 + DU / pE)

let's apply this expression to our case, the change in potential energy is ΔU = -0.2J

let's calculate

          cos θ  = 1 -0.2 / 0.1

          cos θ  = -1

           θ  = 180

Answer:

a) Energy stored in the capacitor, [tex]E = 1.0125 *10^{-3} J[/tex]

b) Q = 45 µC

c) C' = 1.5 μF

d)  [tex]E = 6.75 *10^{-4} J[/tex]

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J[/tex]

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J[/tex]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  [tex]f_e = 0.027 \ m[/tex]

Explanation:

From the question we are told that

    The focal length is  [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]

This negative sign shows the the microscope is diverging light

     The  angular magnification is [tex]m = 300[/tex]

     The  distance between the objective and the eyepieces lenses is  [tex]Z = 19 \ cm = 0.19 \ m[/tex]

Generally the magnification is mathematically represented as

        [tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]

Where [tex]f_e[/tex] is the eyepiece focal length of the microscope

  Now  making [tex]f_e[/tex] the subject  of the formula

         [tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]

substituting values

        [tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]

         [tex]f_e = 0.027 \ m[/tex]

Answer:

-    Force on A = 0.870N

-    charge of the object B = q = 2.1 μC

    charge of the object A = 2q = 4.2 μC

-    a = 0.966 m/s^2

Explanation:

- In order to determine the force on the object A, you take into account the third Newton law, which states that the force experienced by A has the same magnitude of the force experienced by B, but with an opposite direction.

Then, the force on A is 0.870N

- In order to calculate the charge of both objects, you use the following formula:

[tex]F_e=k\frac{q_Aq_B}{r^2}[/tex]         (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

r: distance between the objects = 30.0cm = 0.30m

A has twice the charge of B. If the charge of B is qB=q, then the charge of A is qA=2qB = 2q.

You replace the expression for qA and qB into the equation (1), solve for q, and replace the values of the parameters.

[tex]F_e=k\frac{(2q)(q)}{r^2}=2k\frac{q^2}{r^2}\\\\q=\sqrt{\frac{r^2Fe}{2k}}\\\\q=\sqrt{\frac{(0.30m)^2(0.870N)}{2(8.98*10^9Nm^2/C^2)}}=2.1*10^{-6}C\\\\q=2.1\mu C[/tex]

Then, you have:

charge of the object B = q = 2.1 μC

charge of the object A = 2q = 4.2 μC

- In order to calculate the acceleration of A, you use the second Newton law with the electric force, as follow:

[tex]F_e=ma\\\\a=\frac{F_e}{m}[/tex]

m: mass of the object A = 900g = 0.900kg

[tex]a=\frac{0.870N}{0.900kg}=0.966\frac{m}{s^2}[/tex]

The acceleration of A is 0.966m/s^2

Answer:

The gray spheres is negatively charged while the red is positively charged

Explanation:

This is because theelectrophorus becomes less positive once it pulls some electrons away from the red sphere, but, the electrophorus is replaced on the slab and recharged by grounding it before it proceeds to charge the grey sphere, thereby giving it electrons and making it negatively charged

Answer:

The gray sphere has a positive charge and the red sphere has a positive charge.

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.

A certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.

Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.

[tex]\sigma = \dfrac {Q}{A}[/tex]

[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]

[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]

Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].

Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.

[tex]Q' = \sigma A'[/tex]

Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.

For the closed Gaussian Surface, Flux is given below.

[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]

Hence the charge can be written as,

[tex]Q' = \phi\epsilon_\circ[/tex]

So the charge can be given as below.

[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]

Then the surface area of the Gaussian surface is given below.

[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]

Substituting the values in the above equation,

[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]

[tex]A' =0.006076\;\rm m^2[/tex]

[tex]A' = 60.76 \;\rm cm^2[/tex]

Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.

To know more about the charge and charge density, follow the link given below.

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Answer:

Explanation:

Check the diagram attached below for the diagram.

Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark

Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.

Moment = Force * perpendicular distance

The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction

Clockwise moment = 1kg * 25 = 25kg/cm

Anticlockwise moment = W * 25cm = 25W kg/cm

Equating both moments of forces

25W = 25

W = 25/23

W = 1 kg

The mass of the rock is also 1 kg

Answer:

the answers are provided in the attachments below

Explanation:

Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface

Applying Gauss law to the long solid cylinder

A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) E = 2K λ / r

C) Answers from parts a and b are the same

D) attached below

Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.

A ) The expression for the electric field inside the volume at a distance r

Gauss law :  E. A = [tex]\frac{q}{e_{0} }[/tex]  ----- ( 1 )

where : A = surface area = 2πrL ,  q = p(πr²L)

back to equation ( 1 )

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) Electric field at point Outside the volume in terms of charge per unit length  λ

Given that:  linear charge density = area * volume charge density

                                            λ    =  πR²P

from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]

∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex]    ----- ( 2 )

where : πR²P = λ

Back to equation ( 2 )

E = λ  / 2e₀π*r              where : k = 1 / 4πe₀

∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ

E = 2K λ / r

C) Comparing answers A and B

Answers to part A and B are similar

Hence we can conclude that Applying Gauss law to the long solid cylinder

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.

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Answer:

It's more likely that the trailer is heavily loaded

Explanation:

Due to the fact that the frequency is proportional to the square root of the force constant and inversely proportional to the square root of the mass, it is very likely that the truck would be heavily loaded because the force constant would be the same whether the truck is empty or heavily loaded.

Answer:

[tex]F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex]

Explanation:

You have the following potential energy function:

[tex]U(x,y)=A[\frac{1}{x^2}+\frac{1}{y^2}}][/tex]           (1)

A > 0 constant

In order to find the force in terms of the unit vectors, you use the gradient of the potential function:

[tex]\vec{F}=\bigtriangledown U(x,y)=\frac{\partial}{\partial x}U\hat{i}+\frac{\partial}{\partial y}U\hat{j}[/tex]         (2)

Then, you replace the expression (1) into the expression (2) and calculate the partial derivatives:

[tex]\vec{F}=A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]} \hat{i}+A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]\hat{j}\\\\\vec{F}=A(-2x^{-3})\hat{i}+A(-2y^{-3})\hat{j}\\\\F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex](3)

The result obtained in (3) is the force expressed in terms of the unit vectors, for the potential energy function U(x,y).

Answer:

The initial velocity of spaceship P was u₁ = 5.06 m/s

Explanation:

In an elastic collision between two bodies the expression for the final velocity of the second body is given as follows:

[tex]V_{2} = \frac{(m_{2}-m_{1}) }{(m_{1}+m_{2})}u_{2} + \frac{2m_{1} }{(m_{1}+m_{2})}u_{1}[/tex]

Here, subscript 1 is used for spaceship P and subscript 2 is used for spaceship T. In this equation:

V₂ = Final Speed of Spaceship T = 8.1 m/s

m₁ = mass of spaceship P = 4 M

m₂ = mass of spaceship T = M

u₁ = Initial Speed of Spaceship P = ?

u₂ = Initial Speed of Spaceship T = 0 m/s

Using these values in the given equation, we get:

[tex]8.1 m/s = \frac{M-4M }{4M+M}(0 m/s) + \frac{2(4M) }{4M+M}u_{1}[/tex]

8.1 m/s = (8 M/5 M)u₁

u₁ = (5/8)(8.1 m/s)

u₁ = 5.06 m/s

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

[tex]t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}[/tex]

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

Answer:

Explanation:

Given data

length of cube= 0.2015 m

density = 19300 kg/m^3.

But the volume of cube is given as [tex]l*l*l= l^3[/tex]

[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]

The density is expressed as = mass/volume

[tex]mass=19300*0.00818= 157.87m^3[/tex]

Answer:

I believe it is they will weigh the same

Explanation:

Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics

The head side is heavier than the handle side. - this will be discovered.

Theoretically, the body's center of gravity is where all of the weight is believed to be concentrated. Knowing the centre of gravity is crucial because it may be used to forecast how a moving object will behave when subjected to the effects of gravity. In designing immobile constructions like buildings and bridges, it is also helpful.

We know that center of gravity is  close to some particular point refers the mass of the point is greater then others. It is given that: The center of gravity of an ax is on the centerline of the handle, close to the head.

So, we can conclude that the head side of the ax is heavier than the handle side of it.

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Answer:

Ф = 28.9°

Explanation:

given:

radius (r) = 117m

velocity (v) = 25.1 m/s

required: angle Ф

Ф = inv tan (v² / (r * g))      we know that g = 9.8

Ф = inv tan (25.1² / (117 * 9.8))

Ф = 28.9°

Answer:

335.79cm/s

Explanation:

When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;

v = fλ        ------------------(ii)

From the question;

f = 6.3Hz

λ = 53.3cm

Substitute these values into equation (i) as follows;

v = 6.3 x 53.3

v = 335.79cm/s

Therefore, the speed of the wave pulses along the wire is 335.79cm/s

Answer:

c.they have opposite charges.

Explanation:

because the protons have a positive charge and the neutrons have no charge.

Answer:

The apparent weight of the pilot is 3311 N

Explanation:

Given;

radius of the vertical circle, r = 600 m

speed of the plane, v = 150 m/s

mass of the pilot, m = 70 kg

Weight of the pilot due to his circular motion;

[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]

Real weight of the pilot;

[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]

Apparent weight - Real weight of pilot = weight due to centripetal force

[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]

Therefore, the apparent weight of the pilot is 3311 N

Answer:

16 years.

Explanation:

Using Kepler's third Law.

P2=D^3

P=√d^3

Where P is the orbital period and d is the distance from the sun.

From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.

P=?

P= √4^3

P= √256

P= 16 years.

Orbital period is 16 years.

Answer:

Yes.

Explanation:

Law of relativity in relation to light states that the speed of light in a vacuum does not depend on all the motion of the observers and that all motion must be defined relative to a frame of reference and that space and time are relative, rather than absolute concepts. This was formulated by Albert Einstein in 1905.

Light travels more slowly in gas than in air because it interacts with atoms of glass that made it way through it and the refractive index of glass is more than air. This does contradict the theory of relativity as the speed of lights travel slower in glass because it's motion is slow and it is not relative.

Answer:

a)  v_correcaminos = 22.95 m / s ,  b)  x = 512.4 m ,

c) v = (45.83 i ^ -109.56 j ^) m / s

Explanation:

We can solve this exercise using the kinematics equations

a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff

         v² = v₀² + 2 a x

they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²

         v = √ (2 15 70)

         v = 45.83 m / s

with this value calculate the time it takes to arrive

        v = v₀ + a t

        t = v / a

        t = 45.83 / 15

        t = 3.05 s

having the distance to the cliff and the time, we can find the constant speed of the roadrunner

         v_ roadrunner = x / t

         v_correcaminos = 70 / 3,05

         v_correcaminos = 22.95 m / s

b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.

Let's start by looking for the time to reach the cliff floor

            y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

in this case y = 0 and the height of the cliff is y₀ = 100 m

          0 = 100 + 45.83 t - ½ 9.8 t²

          t² - 9,353 t - 20,408 = 0

we solve the quadratic equation

         t = [9,353 ±√ (9,353² + 4 20,408)] / 2

         t = [9,353 ± 13] / 2

         t₁ = 11.18 s

        t₂ = -1.8 s

Since time must be a positive quantity, the answer is t = 11.18 s

we calculate the horizontal distance traveled

        x = v₀ₓ t

        x = 45.83 11.18

        x = 512.4 m

c) speed when it hits the ground

         vₓ = v₀ₓ = 45.83 m / s

we look for vertical speed

         v_{y} = [tex]v_{oy}[/tex] - gt

         v_{y} = 0 - 9.8 11.18

         v_{y} = - 109.56 m / s

         v = (45.83 i ^ -109.56 j ^) m / s

Answer:

the correct answer is d Beats

Explanation:

when two sound waves interfere time has different frequencies, the result is the sum of the waves is

       y = 2A cos 2π (f₁-f₂)/2    cos 2π (f₁ + f₂)/2

where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.

When examining the correct answer is d Beats

Answer:

Coulomb's law is:

[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]

First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:

N = (1/{e0})*C^2/m^2

then we have:

{e0} = C^2/(m^2*N)

And we know that N = kg*m/s^2

then the dimensions of e0 are:

{e0} = C^2*s^2/(m^3)

(current square per time square over cubed distance)

And knowing that a Faraday is:

F = C^2*S^2/m^2

The units of e0 are:

{e0} = F/m.

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

Answer:

superscript

Explanation: