A car starts from rest at a stop sign. It accelerates at 3.8 m/s 2 m/s2 for 6.0 s, coasts for 1.6 ss , and then slows down at a rate of 3.3 m/s 2 m/s2 for the next stop sign.
How far apart are the stop signs?

Answer:

It takes 6.37 revolutions to stop.

Explanation:

The constant angular acceleration is negative if we choose the wheel direction of rotation as positive direction, so radial acceleration is α=5.0[tex] \frac{rad}{s^{2}}[/tex].Because the wheel is changing its velocity and we already know radial acceleration we should use the Galileo's kinematic rotational equation:

[tex]\omega^{2}=\omega_{i}^{2}+2\alpha\varDelta\theta [/tex]

with [tex] \omega[/tex] the final angular velocity (is zero because the wheel comes to rest), [tex] \omega_{i}[/tex] the initial angular velocity and Δθ the angular displacement. Solving (1) for Δθ :

[tex]\varDelta\theta=\frac{\omega^{2}-\omega_{i}^{2}}{2\alpha} [/tex]

[tex]\varDelta\theta=\frac{0-20^{2}}{(2)(-5.0)}=40rad [/tex]

The angular displacement can be converted to revolution knowing that 1 revolution is 2π rad:

[tex]40rad=\frac{40rad}{2\pi\frac{rad}{rev}} [/tex]

[tex] 40rad=6.37 rev[/tex]

Number of revolution take place is 6.4 revolution

Velocity of rotating wheel = 20 rad/s

Acceleration of magnitude = 5.0 rad/s²

Number of revolution take place

Using Third equation of motion;

20² - 0² = 2(5)(s)

400 = (10)(s)

Total distance = 40 rad

Number of revolution take place = Total distance / 2π

Number of revolution take place = 40 / 2(3.14)

Number of revolution take place = 40 / 6.28

Number of revolution take place = 6.369

Number of revolution take place = 6.4 revolution

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To solve this problem we will apply the definition of electrostatic force. From the variables present there and explained later we will find the value of the distance reorganizing said expression, that is

[tex]F = \frac{k q_1 q_2}{d^2}[/tex]

Here

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge of each object

d = Distance

Replacing our values we have that

[tex]3 = \frac{(9*10^9)(1)(1)}{d^2}[/tex]

Rearranging and solving for the distance we have

[tex]d = 54772.25m[/tex]

Therefore the distance between the two objects is 54772.25m

The maximum Volume Flow Rate of water the pump can provide, given an efficiency of 82% and an elevation of 15 m, is approximately 0.033 L/s.

First, we must convert the pump's horsepower to a more usable unit in this context - like watts. In physics, 1 horsepower equals roughly 746 watts. Therefore, the pump has power of 8*746 = 5968 watts.

Given the mechanical efficiency (ME) and the height (h), the maximum work the pump can do is given by M.E. * Power. So, the pump does work of 0.82*5968 = 4895.76 watts.

The work done on the water by the pump is equal to the change in potential energy of the water, PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (15 m). With rearranging, you could express m = Power/(g * h). But we're looking for the volume flow rate, not the mass flow rate, so we need to convert mass to volume. Since the density of water (ρ) is 1 kg/L, the volume flow rate = m/ρ = Power/(g * h * ρ).

Substituting all known values, we get: Volume flow rate = 4895.76 W /(9.8 m/s^2 * 15 m * 1 kg/L) = 0.033 L/s.

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Explanation:

Since the comb has a net charge, it attracts the paper, which has a net charge equal to zero. When the paper touches the comb, an electrical interaction is established between the charge of the comb and the neutral paper, because of this, the paper now has a net charge with the same sign of the comb and they repel.

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

Answer:

A) [tex]k=2.63*10^{5} N/m[/tex].

B)[tex]v=2.10m/s[/tex]

C)[tex]a=90.0m/s^{2}[/tex]

Explanation:

This problem is a simple harmonic motion problem. The equation of motion for the SHM is:

[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],

where x is the displacement of the mass about its point of equilibrium, t is time, and [tex]\omega[/tex] is the angular frequency.

A)

First, we need to remember that

[tex]\omega^{2}=\frac{k}{m}[/tex],

where k is the spring constant, and m is the mass.

From here we can simply solve for k, so

[tex]k=\omega^{2}m[/tex].

Now,  we need to make use of an equation that relates the frequency and angular frequency. The equetion is

[tex]\omega=2\pi \nu[/tex],

where [tex]\nu[/tex] is the frequency. This leads us to

[tex]k=(2\pi \nu)^{2}m[/tex],

[tex]k=142(2*6.85*\pi)^{2}[/tex],

[tex]k=2.63*10^{5} N/m[/tex],

B) In simple harmonic motion, the velocity behaves as follow:

[tex]v=\omega Acos(\omega t)[/tex] (this is obtained by solving the equation of motion of the mass for the displacement x and take the derivative),

where A is the amplitude of the motion. Since we want the maximum value for the speed, we make [tex]cos(\omega t)=1[/tex] (this because cosine function goes from -1 to 1). With this, the maximum speed is simply

[tex]v = \omega A\\v=(2\pi \nu)A\\v=(2*6.85*\pi)*0.0486\\v=2.10m/s[/tex]

C) Here we are going to use the equation of motion of SHM

[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],

we know that

[tex]a=\frac{d^{2}x}{dt^{2}}[/tex] , where a is the acceleration,

[tex]a+\omega^{2}x=0\\a=-\omega^{2}x[/tex]

in this case, x goes from -A to A, so for a to be maximum we need that [tex]x=-A[/tex] ,and we get

[tex]a=-\omega^{2}(-A)\\a=\omega^{2}A\\a=(2\pi \nu)^{2}A\\a=(2*6.85*\pi)^{2}(0.0486)\\a=90.0m/s^{2}[/tex]

Answer:

The two tapes will either attract or repel each other, depending on the nature of interaction. This is explained below:

Explanation:

When you rip the two pieces of tape off the table, there is a tug-of-war for electric charges between tape and table. The tape either steals negative charges (electrons) from the table or leaves some of its own negative charges behind, depending on what the table is made of (a positive charge doesn’t move in this situation). In any case, both pieces of tape end up with the same kind of charge, either positive or negative. Since like charges repel, the pieces of tape repel each other.

When the tape sandwich is pulled apart, one piece rips negative charges from the other. One piece of tape therefore has extra negative charges. The other piece, which has lost some negative charge, now has an overall positive charge. Because opposite charges attract, the two pieces of tape attract each other.

Answer:

c.The stopping impulse is the same for either the hard objects or the airbag. Unlike the windshield or dashboard, the air bag gives some increasing the time for the slowing process and thus decreasing the average force on the passenger

Explanation:

The stopping impulse is the same for either the hard objects or the airbag.

Unlike the windshield or dashboard, the airbag gives some increasing the

time for the slowing process and thus decreasing the average force on the

passenger.

Air bags are put in automobiles ion order to prevent life threatening injuries

such as brain damage to individuals.  The air bag contains air helps

to increase the time for the slowing process.

This thereby prevents the average force on the passenger and prevents

injuries which should have occurred.

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Answer:

34.35464 m/s

Explanation:

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

E = Electric field = 0.08 N/C

s = Displacement = [tex]4.2\times 10^{-8}\ m[/tex]

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

Electrical force is given by

[tex]F=qE[/tex]

Work done is given by

[tex]W=Fs\\\Rightarrow W=qEs[/tex]

Work done is also given by the kinetic energy

[tex]\dfrac{1}{2}mv^2=qEs\\\Rightarrow v=\sqrt{\dfrac{2qEs}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 0.08\times 4.2\times 10^{-8}}{9.11\times 10^{-31}}}\\\Rightarrow v=34.35464\ m/s[/tex]

The velocity of the electron is 34.35464 m/s

According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions

Answer:

40%

Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.

Answer:

a) 25F

Explanation:

Assuming that the two small spheres can be modeled as point charges, according to Coulomb's law, the magnitude of the electrostatic force is given by:

[tex]F=\frac{kq_1q_2}{R^2}[/tex]

In this case, we have [tex]R'=\frac{R}{5}[/tex]:

[tex]F'=\frac{kq_1q_2}{R'^2}\\F'=\frac{kq_1q_2}{(\frac{R}{5})^2}\\F'=25\frac{kq_1q_2}{R^2}\\F'=25F[/tex]

Answer: Energy dissipated E = 96J

Explanation:

Given:

Current I = 4A

Resistance R = 2 Ohms

Time t = 3 seconds

The energy dissipation in an electric circuit can be derived from the equation below:

E = IVt ....1

Where;

I = current, V = Voltage (potential difference), t= time and E = energy dissipated

But we know that;

V = I×R .....2

Substituting equation 2 to 1, we have

E = IVt = I(I×R)t = I^2(Rt)

Substituting the values of I,R and t

E = 4^2 × 2 ×3 = 96J

Energy dissipated E = 96J

Final answer:

The energy dissipated in the circuit is 96J.

Explanation:

To calculate the energy dissipated in a circuit, you need to use the formula P = IV, where P is power, I is current, and V is voltage. In this case, the current is 4A and the total resistance is 2Ω. Using Ohm's law (V = IR), we can find the voltage as V = I * R = 4A * 2Ω = 8V. Now, we can calculate the power dissipated as P = IV = 4A * 8V = 32W. Lastly, to find the energy dissipated, we multiply the power by the time, so 32W * 3s = 96J.

Explanation:

Water in the air (humidity) helps to dissipate static charge that builds up. If the air is very dry, the charge can't dissipate, so it builds up until there is enough to spark.

Tropical countries are typically more humid than the United States, but I guess that depends on where you are in the US.

A student's first experience with static electricity in an American winter is due to the dry air, which allows for greater accumulation and discharge of electrical charges, unlike in humid tropical climates.

This is because static electricity is more prevalent in cold, dry environments.

In summary, the primary reason a student may experience static electricity for the first time in an American winter is due to the dry air conditions that favour the accumulation and sudden discharge of electric charges.

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

[tex]m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s[/tex]

So Option d is correct one

The velocity of skater is [tex]0.07m/s[/tex]

Option d is correct.

For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied.

                     [tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

                 [tex]65*v_{1}=0.15*32\\\\v_{1}=\frac{0.15*32}{65} =0.07m/s[/tex]

The velocity of skater is [tex]0.07m/s[/tex]

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Answer:

= 15 m/s

Explanation:

Considering right side(west) as positive x-axis and south as negative y-axis.

velocity of boat in still water [tex]v_b=12\hat{i}[/tex]

velocity of stream [tex]v_s=-9\hat{j}[/tex]

now relative velocity of boat w.r.t. stream [tex]v_{b/s}=12\hat{i}+9\hat{j}[/tex]

this velocity with which ac distance will be covered.

therefore magnitude of [tex]v_{b/s} =\sqrt{12^2+9^2}[/tex]

= 15 m/s

Answer: p = - {1/2} , q = {1/2}

Explanation: The frequency of oscillation of a pendulum is given as

F = 1/2π *√{l/g}

Where √ is square root

l is lenght

g is acceleration due to gravity

But

F = 1/T

Where T is the period of Oscillation

Making T subject of formula we have

T= 1/F

T = 2π√{g/l}

Here the power on l is -[1/2]= p

Also,

Power on g is 1/2 =q

All because of the square root.

Answer:

a much larger slit, the phenomenon of Sound diffraction that slits for light.

this is a series of equally spaced lines giving a diffraction envelope

Explanation:

The diffraction phenomenon is described by the expression

    d sin θ = m λ

Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.

For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.

The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.

When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope

When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns

Answer:

The nearest plant (A) receives 4 times more radiation from the farthest plant

Explanation:

The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of ​​the sphere

                I = P / A

               P = I A

The area of ​​the sphere is

               A = 4π r²

Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets

               P = I₁ A₁ = I₂ A₂

               I₁ / I₂ = A₂ / A₁

 Suppose index 1 corresponds to the nearest planet,

            r2 = 2 r₁

            I₁ / I₂ = r₁² / r₂²

            I₁ / I₂ = r₁² / (2r₁)²

            I₁ / I₂ = ¼

           4 I₁ = I₂

The nearest plant (A) receives 4 times more radiation from the farthest plant

Answer:

[tex]F_H_n=230.04 N[/tex]

The Required  horizontal force is 230.04N

Explanation:

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

[tex]F_H_n=F_f\\F_h=mg*u[/tex]

where:

F_{Hn} is the required Force

u is the friction coefficient

m is the mass

g is gravitational acceleration=9.8m/s^2

[tex]200=mg*u[/tex]                         Eq (1)

Now, weight increases by 42% and friction coefficient decreases by 19%

New weight=(1.42*m*g) and new friction coefficient=0.81u

[tex]F_H=(1.42m*g*.81u)[/tex]          Eq (2)

Divide Eq(2) and Eq (1)

[tex]\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N[/tex]

The Required  horizontal force is 230.04N

To solve this problem we will apply the concepts related to the balance of Forces, in this case the centripetal force of the body must be equal to the gravitational force exerted by the moon on it.

The gravitational force is given by the function

[tex]F_g = \frac{GmM}{r^2}[/tex]

Here

G = Gravitational Universal constant

M = Mass of the planet

m = Mass of the satellite

r = Radius(orbit)

Now the centripetal force is given as

[tex]F_c =\frac{mv^2}{r}[/tex]

Here

m = mass of satellite

v = Velocity of satellite

r = Radius (orbit)

Since there must be balance for the satellite to remain in the orbit

[tex]F_c = F_g[/tex]

[tex]\frac{mv^2}{r} = \frac{GmM}{r^2}[/tex]

[tex]v^2= \frac{GM}{r}[/tex]

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

The velocity depends on the mass of the planet and the orbit, and not on the mass, so if the orbit is maintained, the velocity will be the same: 946m/s

Answer:

(a). The final velocity is 1.78 m/s.

(b). The lost kinetic energy is 266.13 J.

Explanation:

Given that,

Mass of first skater = 60 kg

Mass of second skater = 75.0 kg

Initial velocity = 4.00 m/s

(a). We need to calculate the final velocity

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]

Put the value into the formula

[tex]60\times4.00+75.0\times0=(60+75.0)v[/tex]

[tex]v=\dfrac{60\times4.00}{60+75}[/tex]

[tex]v=1.78\ m/s[/tex]

The final velocity is 1.78 m/s.

(b). We need to calculate the lost kinetic energy

Using formula of kinetic energy

[tex]\Delta E=E_{2}-E_{1}[/tex]

[tex]\Delta E=\dfrac{1}{2}(m_{1}u_{1}^2+m_{2}u_{2}^2)-\dfrac{1}{2}(m_{1}+m_{2})v^{2}[/tex]

[tex]\Delta E=\dfrac{1}{2}(60\times4^2)-\dfrac{1}{2}\times(60+75)\times1.78^2[/tex]

[tex]\Delta E=266.13\ J[/tex]

Hence, (a). The final velocity is 1.78 m/s.

(b). The lost kinetic energy is 266.13 J.

a. The final velocity should be considered as the 1.78 m/s.

b. The lost kinetic energy  should be considered as the 266.13 J.

Since

Mass of first skater = 60 kg

Mass of second skater = 75.0 kg

Initial velocity = 4.00 m/s

a.

We know that

[tex]mm1v1 + m2v2 = (m1 + v1)v\\\\60*4.00 + 75*0 = (60 + 75)v[/tex]

v = 60 + 4/60 + 75

= 1.78 m/s

b.

[tex]E = 1/2(m1v1^2 + m2v2^2) - 1/2(m1 + vm2)v^2\\\\= 1/2(60*4^2) - 1/2 * (60 + 75)*1.78^2[/tex]

= 266.13 J

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The ratio between rotational energy and translational kinetic energy is [tex]1.47\cdot 10^{-3}[/tex]

Explanation:

The translational kinetic energy of the ball is given by:

[tex]KE_t = \frac{1}{2}mv^2[/tex]

where

m is the mass of the ball

v is the speed of the ball

The rotational kinetic energy of the ball is given by

[tex]KE_r = \frac{1}{2}I\omega^2[/tex]

where

I is the moment of inertia

[tex]\omega[/tex] is the angular speed

The moment of inertia of a solid sphere through its axis is given by

[tex]I=\frac{2}{5}mR^2[/tex]

where

m is the mass of the ball

R is its radius

Substituting into the previous equation,

[tex]KE_r = \frac{1}{2}(\frac{2}{5}mR^2)\omega^2 = \frac{1}{5}mR^2 \omega^2[/tex]

The ratio between the two energies is

[tex]\frac{KE_r}{KE_t}=\frac{\frac{1}{5}mR^2 \omega^2}{\frac{1}{2}mv^2}=\frac{2R^2 \omega^2}{5v^2}[/tex]

And substituting:

R = 3.91 cm = 0.0391 m

v = 33.6 m/s

[tex]\omega=52.1 rad/s[/tex]

we find:

[tex]ratio = \frac{2(0.0391)^2(52.1)^2}{5(33.6)^2}=1.47\cdot 10^{-3}[/tex]

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The ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere should be [tex]1.47.10^-3[/tex].

Since

The translational kinetic energy of the ball should be

[tex]KE_t = 1/2mv^2[/tex]

here

m is the mass of the ball

v is the speed of the ball

Now

The rotational kinetic energy of the ball should be

[tex]KE_r = 1/2Iw^2[/tex]

Here

I is the moment of inertia

w is the angular speed

Now

The moment of inertia of a solid sphere through its axis should be

[tex]I = 2/5mR^2[/tex]

Here

m is the mass of the ball

R is its radius

So,

[tex]KE_r = 1/2(2mR^2)w^2\\\\= 1/5mR62w^2[/tex]

Now

the ratio be

[tex]KE_r/KE_t = (1/2mR^2w^2) / (1/2mv^2)\\\\= (2R^2 w^2) / (5v^2)\\\\= 290.0391)^2 (52.1)^2 / 5(33.6)^2\\\\= 1.47*10^-3[/tex]

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Answer

given,

displacement wave function

s(x, t) = 2.00 cos (15.7 x + 858 t)

now, comparing the wave equation with general equation.

s(x, t) = A cos (k x + ω t)

where A is the amplitude of the wave in micrometer.

now,

a) Amplitude of the wave

   A = 2 x 10⁻⁶ μ m

b) [tex]\lambda = \dfrac{2\pi}{k}[/tex]

       k = 15.7 m

   [tex]\lambda = \dfrac{2\pi}{15.7}[/tex]

             λ = 0.4 m

c) wave speed

    [tex]v = \dfrac{\omega}{k}[/tex]

    [tex]v = \dfrac{858}{15.7}[/tex]

           v = 54.65 m/s

d) For instantaneous displacement

Assuming the position and time is given as

x = 0.05 m and t = 3 m s

now,

s(x, t) = 2.00 cos (15.7 x 0.05 + 858 x 3 x 10⁻³ )

s(x, t) = 2.00 cos (3.359)

s(x,t) = -1.95 μ m

The amplitude of the wave is 2.00 micrometers, the wavelength is 15.7, and the speed of the wave is approximately 54.65 m/s.

The displacement wave function is given as s(x, t) = 2.00 cos (15.7x - 858t). In this wave function, the amplitude is the coefficient in front of the cosine function, which is 2.00 micrometers. The wavelength can be determined by finding the coefficient of the x term inside the cosine function, which is 15.7. The speed of the wave can be calculated by dividing the angular frequency (in this case, 858) by the wave number (in this case, 15.7), resulting in a speed of approximately 54.65 m/s.

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Answer:

A) The energy stored in the capacitor increases.

Explanation:

For a capacitor fully charged by battery, when disconnected from battery and the plate separation is increased. The charge on the plate remain constant because there is no where for it to go( it has been disconnected from battery). But the capacitance would decrease, while also the potential difference would increase.

Q = CV ....1

Q is the charge, C is capacitance, V is the potential difference.

The energy stored in a capacitor is given by:

E = 1/2 CV^2

E = 1/2 QV .......2

E is the energy stored in the capacitor,

Therefore since Q remain constant and V increases when the distance between the plates is increased, then according to the equation 2 above the energy stored in the capacitor increases.

Answer:

Can a room be gravitationally shielded? No, it can't.

Explanation:

the room cannot be gravitationally shielded because there is only one gravitational charge, in this case is mass. Mass can always be positive. the room can be electrically shielded because there are two type of charge, positive and negative charge than can cancel each other out.

The specific kinetic energy of a mass (in kilojoules per kilogram) moving at a velocity of 40 m/s is calculated using the kinetic energy formula, resulting in 0.8 kJ/kg.

The specific kinetic energy of a mass moving with a velocity can be determined using the formula for kinetic energy, K = 0.5*m*v². In this case, where the velocity 'v' is given as 40 m/s, and we want to solve the kinetic energy per kilogram, we can consider the mass 'm' as 1 kg. Hence the specific kinetic energy would be K = 0.5*(1 kg)*(40 m/s)² = 800 J = 0.8 kJ/kg, because 1 kilojoule (kJ) = 1000 joules (J).

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Answer:

Ex = -16.51 N/C Ey = 79.14 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.

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*(3.65 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 = 8.99*10⁹*3.65*10⁻⁹ / (0.600)²m² = 91.15 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 (1.10 m, 0.800 m) and (0,0), as follows:

r₂² = 1.10²m² + (0.800)²m² = 1.85 m²

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

E₂ = k*q₂ / r₂² = 8.99*10⁹*(4.2)*10⁹ / 1.85 = 20.41 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 =- (1.10 m / √1.85 m²) =- (1.10 / 1.36) = -0.809

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

sin θ = - (0.800 m / 1.36 m) = -0.588

⇒E₂ₓ = 20.41 N/C * (-0.809) = -16.51 N/C (pointing to the left) (3)

⇒E₂y = 20.41 N/C * (-0.588) = -12.01 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 + (-16.51 N/C) = -16.51 N/C

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

Ey = E₁y + E₂y =  91.15 N/C + (-12.01 N/C) =79.14 N/C

Answer:

Ex = 15.505 N/C

Ey = 79.144 N/C

Explanation:

Particle 1

[tex]E_{1} = \frac{k*q_{1} }{r^2_{1}} \\\\r^2_{1} = 1.1^2+0.8^2\\\\r^2_{1} = 1.85 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.8}{1.1}) = 36.03 degree\\\\ E_{1} = \frac{(8.99*10^9)*(4.2*10^(-9)) }{1.85}\\\\E_{1} = 20.4096 N/C[/tex]

Away from the particle at (0,0) due to + charge

Particle 2

[tex]E_{2} = \frac{k*q_{2} }{r^2_{2}} \\\\r^2_{2} = 0.36 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.6}{0}) = 90 degree\\\\ E_{2} = \frac{(8.99*10^9)*(3.65*10^(-9)) }{0.36}\\\\E_{2} = 91.149 N/C[/tex]

Towards from the particle at (0,0) due to - charge

Resultant Electric field in y direction

[tex]E_{res,y} = E_{2} -E_{1}*cos(Q)\\E_{res,y} = (91.149) - (20.4096)*sin(36.03)\\\\E_{res,y} = 79.144 N/C[/tex]

Resultant Electric field in x direction

[tex]E_{res,x} = E_{1}*cos(Q)\\E_{res,y} =(20.4096)*cos(36.03)\\\\E_{res,x} = 16.505 N/C[/tex]

Yes, it is possible. The ball would fall straight down from the perspective of a person standing at the side of the road. The person who threw the ball would see it initially move backward and then fall vertically downward.

Yes, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road while someone is riding in the back of a pickup truck and throws a softball straight backward. This would occur under the condition that the ball is thrown with the same initial velocity as the truck's speed. When the ball is thrown straight backward with the same initial speed as the truck's, it will continue to move with the same speed in the backward direction relative to the truck. From the perspective of the person who threw the ball, they would see the ball initially move straight backward and then fall vertically downward due to the force of gravity.

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The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.

To the person who threw the ball, it would appear to travel straight backward relative to them.

If someone is riding in the back of a pickup truck and throws a softball straight backward, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road. This will occur if the velocity at which the ball is thrown backward is equal to the velocity of the truck moving forward. In this case, the forward motion of the truck and the backward motion of the ball will cancel each other out.

For the person who threw the ball, the motion would still appear as if the ball was thrown straight backward relative to their frame of reference, assuming the truck is moving with constant velocity. The ball would move backward with the same speed at which it was thrown.

To summarize:

The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.

To the person who threw the ball, it would appear to travel straight backward relative to them.

Answer:

A) At 2.2 s the position of the particle is 9.4 m.

B) At t =2.2 s the velocity is 9.7 m/s.

C) At t = 2.2 s the acceleration of the particle is 8.8 m/s²

Explanation:

Hi there!

A)The velocity of the particle is given by the variation of the position over time. If the time interval is very small, we get the instantaneous velocity that can be expressed as follows:

dx/dt = 2 · t²

Separating variables, we can find the equation of position as a function of time:

dx = 2 · t² · dt

Integrating both sides between x0 = 2.3 m and x and from t0 = 0 and t:

∫ dx = 2 ∫ t² · dt

x - 2.3 m = 2/3 · t³

x = 2.3 m + 2/3 m/s³ · t³

Replacing t = 2.2 s:

x = 2.3 m + 2/3 m/s³ · (2.2 s)³

x = 9.4 m

At 2.2 s the position of the particle is 9.4 m

B) Now, let´s evaluate the velocity function at t = 2.2 s:

v = 2 · t²

v = 2 m/s³ · (2.2 s)²

v = 9.7 m/s

At t =2.2 s the velocity is 9.68 m/s

C) The acceleration is the variation of the velocity over time (the derivative of the velocity):

dv/dt = a

a = 4 · t

At t = 2.2 s:

a = 4 m/s³ · 2.2 s

a = 8.8 m/s²

At t = 2.2 s the acceleration of the particle is 8.8 m/s²

(A) The particle's position at time, t = 2.2 s is 7.1 m.

(B) The velocity of the particle at 2.2 s is 9.68 m/s.

(C) The acceleration of the particle at 2.2 s is 8.8 m/s².

The given parameters:

The particle's position at time, t = 2.2 s is calculated as follows;

[tex]x = \int\limits^{t_1}_{t_0} {v} \, dt\\\\ x = \int\limits^{t_1}_{t_0} {2t^2}\\\\ x = [\frac{2t^3}{3} ]^{2.2}_0\\\\ x = \frac{2(2.2)^3}{3} \\\\ x = 7.1 \ m[/tex]

The velocity of the particle at 2.2 s is calculated as follows;

[tex]v = 2t^2\\\\ v = 2(2.2)^2\\\\ v = 9.68 \ m/s[/tex]

The acceleration of the particle at 2.2 s is calculated as follows;

[tex]a = \frac{dv}{dt} \\\\ a = 4t\\\\ a = 4(2.2)\\\\ a = 8.8 \ m/s^2[/tex]

Learn more about velocity and acceleration here: https://brainly.com/question/139187

Answer:

1 / 2

Explanation:

This problem is a 1 - D steady state heat conduction with only one independent variable (x).

1 - D steady state:

Q = dT / Rc

Q = heat flow

dT = change in temperature between a pair of node

Rc = thermal resistance

Rc = L / k*A

Since in both cases Rod A and Rod B have identical boundary conditions:

dT_a = dT_b

So,

R_a =  L / k*(pi*r^2)

R_b = 2L / k*(pi*(2r)^2) = L / k*(2*pi*r^2)

Compute Q_a and Q_b:

Q_a = k * dT *(pi * r^2 * / L)

Q_a = k * dT*(2*pi * r^2 * / L)

Ratio of Q_a to Q_b

Q_a / Q_b = [k * dT *(pi * r^2 * / L)] / [k * dT*(2*pi * r^2 * / L)] = 1 / 2

Final answer:

The ratio of heat flow through rod A to rod B is 1:2 when rod B has double the length and radius of rod A, both rods being made of the same material and subjected to the same temperature difference. This conclusion is derived from Fourier's law of thermal conduction.

Explanation:

The question asks about the comparison of heat flow through two rods of different dimensions but made from the same material and exposed to the same temperature difference. To find the ratio of heat flow through rod A to that through rod B, we use Fourier's law of thermal conduction, which states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the cross-sectional area of the material, and inversely proportional to the length of the material's path. Mathematically, we write this as Q = (kAΔT) / L, where Q is the heat transfer per unit time, k is the thermal conductivity of the material, A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the material's path.

For rod A, assuming a unit thermal conductivity for simplicity, the rate of heat transfer QA = (kπr2(85.00 - 5.00)) / L. For rod B, with double the radius and length, QB = (kπ(2r)2(85.00 - 5.00)) / 2L. Simplifying these expressions, we find that the ratio of heat flow through rod A to that through rod B is QA/QB = 1/2. Thus, rod A transfers heat at half the rate of rod B under the given conditions.