Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?A) 16F
B) 8F
C) 4F
D) F
E) F/4

Answer:

Protein kinase A

Explanation:

Protein kinase A is also called Cyclic AMP- dependent protein kinase or A kinase. It is an enzyme that enhance protein covalently using the phosphate group. The function of this enzyme is that it helps to end the effect of different hormones working via the Cyclic AMP signalling pathway. Protein kinase A can be found in the cytoplasm which phosphorylate proteins.

Protein kinase A helps the cell in regulating sugar, glycogen and lipids metabolism level.

Answer:

The answer is Protein Kinase A.

Refer below for the explanation.

Explanation:

In cell science, protein kinase A is a group of compounds whose movement is reliant on cell levels of cyclic AMP. PKA is otherwise called cAMP-subordinate protein kinase. Protein kinase A has a few capacities in the cell, including guideline of glycogen, sugar, and lipid digestion.

Answer:

[tex]2.33*10^{15}Hz[/tex]

Explanation:

The relationship between velocity v, frequency f and wavelength for electromagnetic waves is given by;

[tex]v=\lambda f..............(1)[/tex]

Given;

[tex]v=3*10^8m/s\\\lambda=129nm=129*10^{-9}m\\f=?[/tex]

We make substitutions into equation (1)  as follows;

[tex]f=\frac{v}{\lambda}\\f=\frac{3*10^8}{129*10^{-9}}[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

width of slit(a)≅ 0.1mm

Explanation:

Wave length of laser pointer =λ = 685 nm

Distance between screen and slit = L = 5.5 m

Width of bright band = W=8.0cm=0.08m

width of slit=a

recall the formula;

W=(2λL)/a

a=2λL/W

a=(2 *685*10⁻⁹*5.5m)/0.08m

a=7535*10⁻⁹/0.08

a=94187.5 *10⁻⁹

a=0.0000941875m

a=0.0941875mm

a≅0.1mm

Answer:

The wide of silt is a=93.5×10⁻⁶m

Explanation:

Given data

Wavelength λ=680 nm

Length L=5.5 m

Width w=8.0 cm

To find

Width of slit a

Solution

A single slit of width a has a bright central maximum of width

ω=2λL/a

a=2λL/ω

Substitute the given values

[tex]a=\frac{2*(680*10^{-9}m)5.5m}{0.08m} \\a=93.5*10^{-6}m[/tex]

The wide of silt is a=93.5×10⁻⁶m

Final answer:

The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².

Explanation:

To calculate the acceleration of the object, you can use the formula:

acceleration = (final velocity - initial velocity) / time

Using the given information, the final velocity is 8 m/s, the initial velocity is 0 m/s, and the time is 7 seconds - 3 seconds = 4 seconds.

Substituting the values into the formula:

acceleration = (8 m/s - 0 m/s) / 4 s = 2 m/s²

Therefore, the acceleration of the object from 3 seconds to 7 seconds is 2 m/s².

The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².

To calculate the acceleration of the object from 3 seconds to 7 seconds, we use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time Interval

Given:

So, the calculation is:

Acceleration = (8 m/s - 0 m/s) / 4 s = 2 m/s²

The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².

Final answer:

The force exerted by the toy locomotive on the caboose is 3.08 N, which is calculated by adding the force required to overcome friction and the force necessary for acceleration according to Newton's second law.

Explanation:

To determine the magnitude of the force exerted by the toy locomotive on the caboose, we can use Newton's second law, which states that the force is equal to the mass times the acceleration (F = ma).

However, we also need to account for the frictional force acting on the caboose. The total force required to accelerate the caboose includes the force to overcome friction and the force needed for acceleration:

Step 1: Calculate the force needed to overcome friction, which is already given as 0.48 N.

Step 2: Determine the force needed to accelerate the caboose using the formula F = ma, where m is the mass of the caboose (1.0 kg) and a is the acceleration (2.6 m/s2).

Step 3: Add the force to overcome friction to the force required for acceleration to get the total force exerted by the locomotive on the caboose.

Step 2 in detail: F = ma = 1.0 kg × 2.6 m/s2 = 2.6 N.

Step 3 in detail: The total force exerted by the locomotive on the caboose is 2.6 N (to accelerate) + 0.48 N (to overcome friction) = 3.08 N.

Therefore, the force exerted by the toy locomotive on the caboose is 3.08 N.

Answer: 3.889N/m

Explanation:

f=(1/2)*√k/m

f=n/t

f= 21/3.3=6.4Hz

6.4*2=√k/m

12.73=√k/m

12.73^2=k/m

162.0529=k/m

Since m=24g

Convert g to kg

m=24/1000

m=0.024kg

K=162.0529*0.024

K=3.889N/m

Final answer:

The spring constant of the spring is approximately 24 N/m.

Explanation:

When a 24-gram mass is attached to a spring, causing it to make 21 complete vibrations in 3.3 seconds, we are looking at a problem that involves simple harmonic motion and can use the properties of a mass-spring system to determine the spring constant (k).

The formula to calculate the frequency (f) is:

f = number of vibrations / time
f = 21 / 3.3 s
f = 6.36363636 Hz

The frequency is related to the spring constant (k) and mass (m) as follows:

f = (1 / (2π)) * (√(k / m))

Given the mass (m) and frequency (f), we can solve for k:

k = (2πf)² * m
k = (2 * π * 6.36363636 Hz)² *0.024 kg
k ≈ 24 N/m

Therefore, the spring constant of the spring is approximately 24 N/m.

Answer:

Technician A

Explanation:

The stator is a STATIONARY component of a rotary system. The stator is a stationary component in the alternator, therefore, technician B saying that it is a rotating component is WRONG.

Also, slip rings allows the flow of electrical signals and electrical power from a stationary structure to a rrotating structure. Therefore, technician A is correct in his take that the slip rings allow current to flow from the stationary end frame to the moving rotor.

Final answer:

Technician A is correct about slip rings allowing current flow to the rotor, while Technician B is incorrect as the stator is the stationary part of an alternator.

Explanation:

Technician A is correct in saying that slip rings allow current to flow from the stationary part to the moving rotor. Slip rings are made of a ring and brushes; the ring is attached to the rotating part of a machine (like an alternator's rotor), and the brushes are attached to the stationary part (such as the end frame), facilitating electrical contact as the rotor spins.

Technician B is incorrect; the stator is the stationary part of an alternator or electric motor that contains conductors where the electric current is induced by the motion of the rotor. In an alternator, the rotor is the rotating component, not the stator.

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

   mass per unit length ([tex]\mu[/tex]) = [tex]\frac{M}{l}[/tex] = 1 kg/m

      Tension = 4 N

    Speed (v) = [tex]\sqrt{\frac{F}{\mu}}[/tex]

So, when F is doubled then change in value of F will be as follows.

               F = 4 + 4 = 8 N

Therefore, speed will be calculated as follows.

           v = [tex]\sqrt{\frac{8 N}{1 kg/m}}[/tex]

              = 2.8 m/s

Thus, we can conclude that the wave speed is 2.8 m/s.

Answer:

Technician B

Explanation:

Answer:

Check attachment for solution

Explanation:

Given that 12V battery

Answer:

Terminal voltage = 8.36 V

Current = 1.82 A

Explanation:

E.M.F of battery = 12V

Internal resistance of battery (r) = 2Ω

Resistance of resistor (R) = 4.6Ω

Now the formula for terminal voltage across the battery is;

V = ε - Ir

Where ε is EMF and I is electric current

Using ohms law, we know that V = IR and I = V/R.

Thus, let's put V/R for current in the potential difference equation;

V = ε - r(V/R)

Thus, lets make V the subject of the formula ;

V + (rV/R) = ε

V(1 + r/R) = ε

So, V = ε/(1 + r/R)

V = 12/(1 + (2/4.6))

V = 12/(1 + 0.4348)

V = 12/1.4348 = 8.36 V

Thus from V=IR, we can find current. So 8.36 = I(4.6)

I = 8.36/4.6 = 1.82 A

To find the Current in a circuit passing through each specified point in the circuit, we can use Ohm's law and the known resistances and voltages.

To find the current passing through each specified point in the circuit, we can use Ohm's law and the known resistances and voltages.

For example, to find the current passing through resistor R₂, we first need to find the voltage applied to it. This can be done by subtracting the voltage drop across resistor R₁ from the battery voltage. Once we have the voltage, we can use Ohm's law to find the current through R₂.

We can repeat this process for the other specified points in the circuit using the appropriate resistances and voltages.

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The maximum safe speed on a curved road can be calculated using the formula v = √(2.3r), where v represents speed in miles per hour and r is the radius of curvature in feet. By substituting the given radius of curvature into the formula, we can find the maximum safe speed. In this case, the maximum safe speed is approximately 29.18 miles per hour.

The formula v = √(2.3r) models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r, in feet. To find the maximum safe speed at a given radius of curvature, we can substitute the value of r into the formula and solve for v. In this case, the radius of curvature is given as 370 feet, so we substitute r = 370 into the formula:

v = √(2.3 * 370)

Simplifying the expression, we get:

v = √(851)

Taking the square root, we find that the maximum safe speed is approximately v ≈ 29.18 miles per hour.

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Answer:[tex]a_c=46.24\ m/s^2[/tex]

Explanation:

Given

radius of container [tex]r=13.1\ cm[/tex]

cylinder is rotating with [tex]N=2.99 rps[/tex]

Centripetal acceleration at the outer wall is given by

[tex]a_c=\omega ^2\times r[/tex]

where [tex]\omega [/tex]=Angular speed

[tex]\omega =2\pi N[/tex]

[tex]\omega =2\pi \times 2.99[/tex]

[tex]\omega =18.78\ rad/s[/tex]

[tex]a_c=(18.78)^2\times 0.131[/tex]

[tex]a_c=46.24\ m/s^2[/tex]

Answer:

-7.8×[tex]10^{2}[/tex] N

Explanation:

see the attachment plz

Answer:

Explanation:

1. Impulse, I = F.t

  The statement impulse is the product of Force and distance is false.

2. F = m g

   Force necessary to lift the object depends on the mass of the object.

   statement 2 is false.

3. Joule is equal to Newton times meter.

    Statement 3 is false.

4. Work done to lift an object is correct statement.

   Statement 4 is true.

5. Kinetic energy of an object is due to motion.

  Statement 5 is false.

6. Stopping distance is directly proportional to the square of velocity.

     If velocity is doubled, stopping distance is quadrupled.

    Statement 6 is false.

The correct answer is 1.False; 2.False; 3.False; 4.True; 5.False; 6.False.

1. Impulse is the product of force and time, not force and distance. Therefore, the statement is false.

2. The force necessary to lift an object is not simply g, but weight, which is calculated as mass times the gravitational acceleration g. So, this statement is false.

3. A joule is the unit of work or energy, equivalent to one newton meter (N·m), not a newton times a second. Thus, the statement is false.

4. Work is indeed done to lift an object in the classroom since work is the force applied over a distance in the direction of the force. Therefore, this statement is true.

5. Kinetic energy is the energy of motion, not the energy of position. Thus, this statement is false.

6. If the stopping distance is doubled when speed is quadrupled, it overlooks the fact that stopping distance increases with the square of the speed, meaning a quadrupling of speed typically increases the stopping distance by a factor of sixteen, not two. So, this statement is false.

Full Question:

What is the Lewis structure of the covalent compound that contains one nitrogen atom, one hydrogen atom, and one carbon atom?

Explanation:

This covalent compound is the hydrogen cyanide, HCN.

The following steps are used to obtain it;

Step 1. Draw a skeleton structure of the compound based on the elements

Put the least electronegative atom C in the middle with H and Cl on either side.

H-C-N

Step 2. Count the valence (outermost) electrons you can use

H + C + N = 1 + 4 + 5 = 10

Step 3. Add these electrons to give every atom an octet

You have to put a triple bond between C and N.

Note: Each bond is made up of two electrons.

The lewis structure is given in the image below;

Answer: yes.

Explanation: The light that will be incidented on that metal is visible light.

It depends on 3 factors:

1. The temperature

2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer. 

The photoelectric effect causes the metals to get warm when light is shined on them.

Photoelectric effect:

It states that the metals emit electrons when by absorbing electromagnetic radiation.

When light sine on the metal, electrons present on the metals absorb some energy of the light or radiations and leave their atoms. This causes the metal to heat up when light sine on it.

The formula is,

[tex]h \nu= W + E[/tex]

Where,

[tex]h[/tex]- Plank's constant

[tex]\nu[/tex] -  Threshold frequency

[tex]W[/tex] - Energy needed to remove the electron

[tex]E[/tex]-  Kinetic energy of the emitted electron

Therefore, the metal gets warm due to the Photoelectric effect.

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Answer:

Ranked from left to right based on their abundance in the solar nebula:

Hydrogen and Helium-Hydrogen compounds-Rocks-Metals

I hope you find this information useful and interesting! Good luck!

The materials in the solar nebula can be ranked from highest to lowest abundance as hydrogen and helium gas, icy and rocky planetesimals, dust and vapor, and meteorites, comets, and asteroids.

The materials that made up the solar nebula can be categorized into four general types based on their abundance. These types, ranked from highest to lowest abundance, are:

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Answer:

A. [tex]T_2=M_2g[/tex]

B. [tex]T_2=(M_1+M_2)g[/tex]

Explanation:

Since the only forces acting on the blocks are the tensions and the weights (both in the vertical direction), and the system has acceleration zero, we can write the equilibrium equations for M₁ and M₂ as:

[tex]T_1-T_2-M_1g=0\\ \\T_2-M_2g=0[/tex]

From the second equation, we get:

[tex]T_2=M_2g[/tex],

Which is the answer to the part A.

Next, we substitute this result in the first equation and obtain:

[tex]T_1-M_2g-M_1g=0\\ \\T_1=(M_1+M_2)g[/tex],

Which is the answer to the part B.

Final answer:

To find the tensions in ropes supporting hanging blocks, one must draw free-body diagrams for each mass, and apply Newton's Second Law. The tension in the lower rope equals the weight of the lower mass or this weight adjusted for acceleration. The tension in the upper rope is the sum of the weight of the upper mass and the tension in the lower rope.

Explanation:

The student's question revolves around the concept of tension in ropes in a physics context, where two blocks are hanging one under the other. The physics involved here includes Newton's second law and the interplay of gravitational and tensional forces in a system. To find the tensions T₁ and T₂, we need to draw free-body diagrams for each block and apply Newton's Second Law, F = ma, where F is the net force, m is the mass and a is the acceleration of the block.

For block M₂, which hangs at the bottom, the tension T₂ is the force counteracting the gravitational pull downward. Hence, assuming the system is at equilibrium or moving at constant speed (acceleration = 0), T₂ would equal the weight of M₂, given by the formula T₂ = M₂ × g.

For block M₁, which is above M₂, the tension T₁ is slightly more complicated to calculate as it must support its own weight as well as the tension from the lower rope T₂. Therefore, the tension T₁ in the upper rope would be T₁= (M₁ + M₂) × g again assuming no net acceleration.

If the entire system is accelerating, we must incorporate the net force necessary to accelerate both masses. This changes the calculations respectively:
For the lower block (M₂), T₂ = M₂ × (g + a) if accelerating upward, or T₂ = M₂ × (g - a) if accelerating downward.
For the upper block (M₁), T₁ = (M₁ × g) + T₂ because T₁ has to support both the weight of M₁ and the tension from the lower rope.

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is   29.345 m from the spotlight.

Explanation:

To solve the question we note that Snell's law states that

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ =  14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ =  [tex]\frac{n_1sin\theta_1}{n_2}[/tex]  = [tex]\frac{1.003*0.251}{1.33}[/tex] = 0.1885 and θ₂ = 10.86 °

Since the water depth is 4.0 m we have tanθ₂ = [tex]\frac{4}{x_2}[/tex] or x₂ = [tex]\frac{4}{tan\theta_2 }[/tex] =[tex]\frac{4}{tan(10.86)}[/tex] = 20.845 m

d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.

Answer:

NO

Explanation:

Acceleration is change in velocityΔv in respect to timeΔt

so if the velocity of the car is greater than the truck it does not mean that the car acceleration is greater than the truck.

Sometimes with constant velocity it means no accelaration ,but the truck may have accelaration

so, higher velocity of the car does not mean higher acceleration

Answer:

The answer to the question is;

(a) The amount of work required to withdraw the mica sheet is 4.0×10⁻⁵ J

(b) The potential difference across the capacitor after the mica is withdrawn is 500 V.

Explanation:

To solve the question, we note that

The Capacitance of the capacitor is proportional to the dielectric constant, that is removal of the dielectric material will reduce the capacitance of the capacitor by a factor of the dielectric constant as follows

Energy stored in a capacitor is given by

[tex]E = \frac{1}{2} CV^{2}[/tex]

Where:

C = Capacitance of the capacitor = 2.00 nF

V = Voltage = 100 V

Therefore before the dielectric material is removed, we have

E = [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 1.0×10⁻⁵ J

Asfter the dielectric material is removed we have

E = 5× [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 5.0×10⁻⁵ J

Since energy is neither created nor destroyed, we have

Change in energy of system  = 5.0×10⁻⁵ J - 1.0×10⁻⁵ J = 4.0×10⁻⁵ J

Work done on the system to remove the mica sheet = energy change of the system = 4.0×10⁻⁵ J.

(b) From the equation [tex]C = \frac{Q}{V}[/tex]

Where

Q = Charge of the capacitor

V= Voltage

C = Capacitance of the capacitor, we have

Q, the charge of the capacitor is constant

and the final capacitance = [tex]\frac{Initial .capacitance}{5} = \frac{C_1}{5}[/tex]

Therefore we have

V₁ = [tex]\frac{Q}{C_1}[/tex]

V₂ = [tex]\frac{Q}{\frac{C_1}{5} }[/tex] = [tex]{5} *\frac{Q}{C_1}[/tex] = 5·V₁

The velocity increases by a factor of 5 and

V₂ = 5×V₁ =  5 × 100 V = 500 V.

The annual electricity cost of the compressor is $15,386.25.

To calculate the annual electricity cost of the compressor, we first need to find the energy consumption of the motor. We can use the formula:

Energy Consumption = Power x Time

Given that the compressor operates at full load for 2500 hours a year and has a power of 75 hp, we need to convert the power to watts:

1 hp = 746 watt

So, the power of the motor is 75 x 746 = 55,950 watts.

Now, we can calculate the energy consumption:

Energy Consumption = 55,950 watts x 2500 hours

Next, we need to convert the energy consumption to kilowatt-hours (kWh):

1 kWh = 1000 watt-hours

So, the energy consumption in kWh is 55,950 watts x 2500 hours / 1000 = 139,875 kWh.

Finally, we can calculate the annual electricity cost by multiplying the energy consumption by the unit cost of electricity:

Annual Electricity Cost = 139,875 kWh x $0.11/kWh

Therefore, the annual electricity cost of this compressor is $15,386.25.

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Answer:

x=61

y=35

Explanation:

vertical component= 70sin(30)

Horizontal component=70cos(30)

Answer:

A. CHARLES' LAW

B. GAY-LUSSAC'S LAW

C. BOYLE'S LAW

Explanation:

THE QUESTION SEEKS ANSWER BY MATCHING THE PROPERTIES COMPARED BY A GAS LAW TO THE GAS LAW IN QUESTION.

A. CHARLES' LAW COMPARES TEMPERATURE AND VOLUME.

IT ESTABLISHED THE FACT THAT VOLUME AND TEMPERATURE ARE DIRECTLY PROPORTIONAL AT A FIXED PRESSURE

B. GAY-LUSSAC'S LAW COMPARE PRESSURE AND TEMPERATURE.

IT ESTABLISHED THE FACT THAT PRESSURE AND TEMPERATURE ARE DIRECTLY PROPORTIONAL FOR A GIVEN VOLUME OF GAS

C. BOYLE'S LAW COMPARES PRESSURE AND VOLUME

IT ESTABLISHED THAT BOTH ARE INVERSELY PROPORTIONAL AT A GIVEN TEMPERATURE

Answer:

1) D

2) B

3) E

Explanation: Edge 2020

Answer:

(a)

[tex]\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s[/tex]

(b)

1120 N

Explanation:

Change in velocity, [tex]\triangle v[/tex] is given by subtracting the initial velocity from the final velocity and expressed as [tex]\triangle v= v_f -v_i[/tex]

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

[tex]\triangle v=0-8=-8\ m/s[/tex]

To find [tex]m\triangle v[/tex] then we substitute 7 kg for m and -8 m/s for [tex]\triangle v[/tex] therefore [tex]\triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s[/tex]

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

[tex]a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}[/tex]

Substituting t with 0.05 s then [tex]a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}[/tex]

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

Answer:

7.85 m/s

Explanation:

We are given that

Mass of object=m=0.900 kg

[tex]F(x)=\alpha x-\beta x^2[/tex]

[tex]\alpha=60 N/m[/tex]

[tex]\beta=18N/m^2[/tex]

[tex]F(x)=-60x-18x^2[/tex]

U=0 when x=0

Potential energy=[tex]-\int F(x)dx[/tex]

Substitute the values

[tex]U(x)=-\int (-60x-18x^2)dx[/tex]

[tex]U(x)=60(\frac{x^2}{2})+18(\frac{x^3}{3})+C[/tex]

Using the formula

[tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]

Substitute x=0

[tex]U(0)=C\implies C=0[/tex]

[tex]U(x)=30x^2+6x^3[/tex]

[tex]x_1=0.5,x_2=1[/tex]

[tex]v_2=0[/tex]

Using law of conservation energy

[tex]\frac{1}{2}mv^2_1+U(x_1)=\frac{1}{2}mv^2_2+U(x_2)[/tex]

Substitute the values

[tex]\frac{1}{2}(0.9)v^2_1+30(0.5)^2+6(0.5)^3=0+30(1)^2+6(1)^3[/tex]

[tex]\frac{1}{2}(0.9)v^2_1+8.25=36[/tex]

[tex]\frac{1}{2}(0.9)v^2_1=36-8.25=27.75[/tex]

[tex]v^2_1=\frac{27.75\times 2}{0.9}[/tex]

[tex]v_1=\sqrt{\frac{27.75\times 2}{0.9}}[/tex]

[tex]v_1=7.85 m/s[/tex]

Answer:

[tex]v=87.46m/s[/tex]

Explanation:

Objects moving in circular path would be have either centripetal or centrifugal  force.The force is either to center or away from center. When the object is moving along the circular path the centripetal force is

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

Here m is mass, v is velocity and r is radius of circular path

The acceleration is given by:

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

The point of interest is lowest point on circle.The acceleration of plane at  this position point up.The speed of plane from radial acceleration equation is:

[tex]v=\sqrt{a.r}\\[/tex]

Substitute 17 m/s² for a and 450m for r

So

[tex]v=\sqrt{17m/s^{2}*450m }\\ v=87.46m/s[/tex]

Answer:

Magnitude of electric field = E = q/Aε0

Explanation:

Consider plates are placed at a distance of d. As given in the question the charge stored on the plates have magnitude q and given by:

                                          q = CV

And  

                                          V = q/C    ……. (i)

The capacitance is given by the following equation:  

                                         C = Aε0/d ……. (ii)

Put equation (ii) in (i) ,

                                          V = qd/ Aε0 …..(iii)    

The electric field is defined as:  

                                            E = V/d   …… (iv)

Put equation (iii) in (iv),

                                            E = qd/ Aε0d

                                            E = q/Aε0

Hence, the magnitude of electric field will be q/Aε0 .

The electric field between two plates with charges −q and +q is directed from the positive to the negative plate and can be calculated using the charge density and vacuum permittivity. For systems of parallel conducting plates, the electric field can be found using potential differences and distances, while the interaction of a charge with a dipole requires considering forces from each dipole component.

The magnitude of the electric field E between two plates with charges −q and +q can be found using the formula E = σ/ε₀, where σ is the surface charge density (σ = q/A, A is the area of the plate) and ε₀ is the vacuum permittivity. The direction of the electric field is from the positive to the negative plate. Considering a scenario with two parallel conducting plates 15 cm apart and each with a charge of −0.225 C and +0.225 C, you would first calculate the electric field strength using the aforementioned formula and then state that its direction is from the positive to the negative plate.

For the case involving three parallel conducting plates with specified potentials, you calculate the average electric field experienced by an electron by finding the potential difference between the plates and dividing it by the distance between the plates. The electric field between the plates is uniform if we assume the plates are infinitely large and parallel.

The net force on a charge −q placed equidistant from a dipole with charges +2q and −2q can be found using Coulomb's law. Since the forces due to the +2q and −2q charges will be equal in magnitude but opposite in direction, the result will be a null net force in the direction along the line connecting the two dipole charges. However, there will be a resultant force perpendicular to this line, attracting the −q charge towards the dipole's center.

A negative test charge moves toward regions of higher electric potential and lower electric potential energy, which is the opposite direction of the electric field defined by positive charges.

A negative test charge will accelerate toward regions of higher electric potential and lower electric potential energy. This happens because a negative charge moves oppositely to the electric field direction, which is defined from high to low potential. When a negative charge, such as an electron, moves toward a higher potential, it is moving towards a region where it would have a lower potential energy if it were positive; however, since it is negative, its potential energy actually decreases as it moves in this direction.

Understanding electric fields and potentials, consider that a positive test charge is repelled by positive charges and attracted to negative charges. Since the field lines point away from positive charges and toward negative charges, a negative test charge would move in the direction opposite to the field lines, meaning it moves from lower to higher potential but in doing so, it lowers its electric potential energy.