Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diameter and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.

Answers

Answer 1

Answer:

work done in pumping the entire fuel is 1399761 J

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

we assume that the fuel completely fills the tank.

therefore, the volume of a cylinder =  

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x  x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = 1399761 J


Related Questions

Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra

Answers

Answer:

La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.

Explanation:

Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:

[tex]v = v_{o} + g\cdot t[/tex]

[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]

Donde:

[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.

[tex]t[/tex] - Tiempo, medido en segundos.

[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.

[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.

Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:

[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]

[tex]v = -196.14\,\frac{m}{s}[/tex]

[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]

[tex]y-y_{o} = -1961.4\,m[/tex]

La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.

g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.

Answers

Answer:

frequency =4125Hz

Explanation:

L = 4cm = 0.04m

f =v/2L

f = 330/2 x 0.04

f = 4125Hz

A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?

Answers

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

 [tex]a_{max} = 0.00246 \ m/s^2[/tex]

b

   [tex]k =722.2 \ N/m[/tex]

Explanation:

From the question we are told that

     The  amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]

     The period is [tex]T = 17 \ s[/tex]

    The test weight is  [tex]W = 13 \ N[/tex]

Generally the radial acceleration is mathematically represented as

        [tex]a = w^2 r[/tex]

at maximum angular acceleration

       [tex]r = A[/tex]

So  

       [tex]a_{max} = w^2 A[/tex]

Now [tex]w[/tex] is the angular velocity which is mathematically represented as

      [tex]w = \frac{2 * \pi }{T}[/tex]

Therefore

       [tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]

substituting values

       [tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]

       [tex]a_{max} = 0.00246 \ m/s^2[/tex]

Generally this test weight is mathematically represented as

     [tex]W = k * A[/tex]

Where k is the spring constant

Therefore

        [tex]k = \frac{W}{A}[/tex]

substituting values        

      [tex]k = \frac{13}{0.018}[/tex]

     [tex]k =722.2 \ N/m[/tex]

The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material

Answers

Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]

Explanation:

Given: Speed of red light = 700 nm

= [tex]700\times10^{-9}[/tex] m

[tex]= 7\times10^{-7}[/tex] m

Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]

Speed of light = [tex]3\times10^8[/tex] m

Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]

[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]

Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]

The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].

A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration

Answers

Answer:

The average velocity for this trip is 0 km/hr

Explanation:

We know that average velocity = total displacement/total time.

Now, its displacement is d = final position - initial position.

Since the  car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.

So, its displacement is d = 0 km - 0 km = 0 km.

Since the total time for the round trip is 2 hours, the average velocity is

total displacement/ total time = 0 km/2 hr = 0 km/hr.

So the average velocity for this trip is 0 km/hr  

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?

Answers

Answer:

The glider’s speed after the skydiver lets go is 26 m/s

Explanation:

To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum

Mathematically;

mv = mv + mv

so 680 * 26 = (680-60)v + 60 * 26

17680 = 620v + 1560

17680-1560 = 620v

16120 = 620v

v = 16120/620

v = 26 m/s

A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?

Answers

Answer:

72J

Explanation:

distance moved is equal to 3m.then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

The answer is 72J.

Distance moved is equal to 3m.

Then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

Is there any definition of force?

A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.

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A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?

Answers

Answer:

39.94m/s.

Explanation:

Kinetic energy is expressed as KE = 1/2 mv² where;

m is the mass of the body

v is the velocity of the body.

For the heavier shell;

m = 5kg

KE gained = 100J

Substituting this values into the formula above to get the velocity v;

100 = 1/2 * 5 * v²

5v² = 200

v² = 200/5

v² = 40

v = √40

v = 6.32 m/s

Note that after the explosion, both body fragments will possess the same velocity.

For the lighter shell;

mass = 2.0kg and v = 6.32m/s

KE of the lighter shell = 1/2 * 2 * 6.32²

KE of the lighter shell = 6.32²

KE of the lighter shell= 39.94m/s

Hence, the lighter one gains a kinetic energy of 39.94m/s.

The gain in the kinetic energy of the smaller fragment is 249.64 J.

The given parameters;

Mass of the shell, m = 7.0 kgMass of one fragment, m₁ = 2.0 kgMass of the second fragment, m₂ = 5.0 kgKinetic energy of heavier fragment, K.E₁ = 100 J

The velocity of the heavier fragment is calculated as follows;

[tex]K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\v^2 = \frac{2K.E}{m} \\\\v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 100}{5} }\\\\v = 6.32 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;

[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\-6.32(5) \ + 2u_2 = 0(7)\\\\-31.6 + 2u_2 = 0\\\\2u_2 = 31.6\\\\u_2 = \frac{31.6}{2} \\\\u_2 = 15.8 \ m/s[/tex]

The gain in the kinetic energy of the smaller fragment is calculated as follows;

[tex]K.E_2 = \frac{1}{2} mu_2^2\\\\K.E_2 = \frac{1}{2} \times 2 \times (15.8)^2\\\\K.E_2 = 249.64 \ J[/tex]

Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.

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At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s, how fast is the current changing

Answers

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

The rate at which the current is changing is;

di/dt = 0.2 A/s

We are given;

Inductance; L = 5 H

Current; I = 3 A

Rate of Increase of energy; dE/dt = 3 J/s

Now, the formula for energy stored in inductor is given as;

E = ¹/₂LI²

Since we are looking for rate at which current is changing, then we differentiate both sides of the energy equation to get;

dE/dt = LI (di/dt)

Plugging in the relevant values gives;

3 = (5 × 3)(di/dt)

di/dt = 3/(5 × 3)

di/dt = 0.2 A/s

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A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?

a. Wave A has an amplitude of E0 and a phase constant of zero.
b. Wave B has an amplitude of E0 and a phase constant of π.
c. Wave C has an amplitude of 2E0 and a phase constant of zero.
d. Wave D has an amplitude of 2E0 and a phase constant of π.
e. Wave E has an amplitude of 3E0 and a phase constant of π.

Answers

Answer:

the greatest intensity is obtained from   c

Explanation:

An electromagnetic wave stagnant by the expression

           E = E₀ sin (kx -wt)

when two waves meet their electric fields add up

           E_total = E₁ + E₂

the intensity is

           I = E_total . E_total

           I = E₁² + E₂² + 2E₁ E₂ cos θ

where θ  is the phase angle between the two rays

       

Let's examine the two waves

in this case E₁ = E₂ = E₀

          I = Eo2 + Eo2 + 2 E₀ E₀ coasts

         I = E₀² (2 + 2 cos θ )

         I = 2 I₀ (1 + cos θ )

     let's apply this expression to different cases

a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀

b) cos π = -1     this implies that     I_total = 0

c) the cosine is  1,

         I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ

         I = E₀² (5 +4 cos θ)

         I = E₀² 9

         I = 9 Io

d) in this case the cos pi = -1

          I = E₀² (5 -4)

          I = I₀

e) we rewrite the equation

         I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ

         I = Eo2 (10 + 6 cos θ)

         cos π = -1

         I = E₀² (10-6)

         I = 4 I₀

the greatest intensity is obtained from   c

The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.

What is an amplitude?

An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.

In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.

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You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.

Answers

Answer:

a. The primary turns is 60 turns

b. The secondary voltage will be 360 volts.

Explanation:

Given data

secondary turns N2= 40 turns

primary turns N1= ?

primary voltage V1= 120 volts

secondary voltage V2= 8 volts

Applying the transformer formula which is

[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]

we can solve for N1 by substituting into the equation above

[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]

the primary turns is 60 turns

If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)

[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]

the secondary voltage will be 360 volts.

(a) In the primary coil, you have "60 turns".

(b) The emf across the secondary coil would be "360 volts".

Transformer and Voltage

According to the question,

Primary voltage, V₁ = 120 volts

Secondary voltage, V₂ = 8 volts

Secondary turns, N₂ = 40 turns

(a) By applying transformer formula,

→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]

or,

   N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]

By substituting the values,

        = [tex]\frac{40\times 120}{8}[/tex]

        = [tex]\frac{4800}{8}[/tex]

        = 60

(2) Again by using the above formula,

→ V₂ = [tex]\frac{60\times 240}{40}[/tex]

       = [tex]\frac{14400}{40}[/tex]

       = 360 volts.

Thus the above approach is correct.  

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To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface

Answers

Answer:

Explanation:

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Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?

Answers

Answer:

0.46N

Explanation:

See attached file

Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is

Answers

Answer:

If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE

Explanation

This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.

A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?

Answers

Answer:

1.3515x10^5pa

Explanation:

Plss see attached file

Why would physics be used to study light emitted by a star?
O A. Stars form interesting shapes in the sky.
B. Light is very pretty.
O C. The positions of stars control our lives.
O D. Light is a form of energy.

Answers

Answer:

O D.

Explanation:

Physics has an aspect that deals with the study of energy

Answer:

D. Light is a form of energy

Explanation:

The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine

Answers

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?

Answers

Answer:

Explanation:

Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least  and kinetic energy is maximum .

In this way , there is conservation of mechanical energy .

A ​46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this​ causeway?

Answers

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

The required force parallel to the incline to hold the monolith on this​ causeway will be "2.9 tons".

Angle and Force

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), [tex]F_t[/tex] = F Sin a

By substituting the values,

                                       = 46 × 0.064

                                       = 2.9 tons

Thus the response above is appropriate answer.

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An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?

Answers

Answer:

The highest rms voltage will be 8.485 V

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]

where [tex]V_{rms}[/tex] is the rms voltage

[tex]V_{p}[/tex] is the peak or maximum voltage

substituting values into the equation, we'll have

[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V

Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.

Answers

Answer:

Explanation:a 1

what is transmission of heat?​

Answers

Answer:

Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are

Conduction ConvectionRadiation.

Example of Conduction:

Heating a metal

Example of Convection:

Sea Breeze

Example of Radiation:

Sun

Hope this helps ;) ❤❤❤

Answer:

Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.

There are three(3) different ways heat can transfer and they are:

a) Conduction (through direct contact).

b) Convection (through fluid movement).

c) Radiation (through electromagnetic waves).

Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).

2. Baking a pie in an oven(radiation).

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You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.

Answers

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = [tex]\frac{W}{F}[/tex]

d = [tex]\frac{3}{9}[/tex]

d = 0.33 m

The distance d the block ice moved is 33 cm.

The intensity of sunlight at the Earth's distance from the Sun is 1370 W/m2. (a) Assume the Earth absorbs all the sunlight incident upon it. Find the total force the Sun exerts on the Earth due to radiation pressure. N (b) Explain how this force compares with the Sun's gravitational attraction.

Answers

Answer:

F= 3.56e22N

Explanation:

Using the force of radiation acting on the earth which is

force = radiation pressure x area = (intensity/c)xpi R^2

force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s

force = 5.82x10^8 N

But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:

F=GMm/r^2

G=6.67x10^(-11)=6.67e-11

M=mass sun = 2x10^30kg=2e30

m=mass earth = 6x10^24kg

r=earth sun distance = 1.5x10^11m

F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =

F= 3.56e22N

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.

Required:
a. How much work is done by the tension?
b. How much work is done by gravity?

Answers

B)is pills everything to the surface of the earth not sure about A

The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

The given parameters;

mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 N

The work done by the tension is calculated as follows;

W = Fd

W = 15 x 0.15

W = 2.25 J

The work done by gravity is calculated as;

W = (25 x 9.8) x 0.15

W = 36.75 J

Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

Learn more here: https://brainly.com/question/19498865

What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC

Answers

Answer:

E = 17.64 x 10⁶ N/C = 17.64 MN/C

Explanation:

The electric field is given by the following formula:

E = F/q

E= W/q

E = mg/q

where,

E = magnitude of electric field = ?

m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

= charge = 3 nC = 3 x 10⁻⁹ C

Therefore,

E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)

E = 17.64 x 10⁶ N/C = 17.64 MN/C

what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates

Answers

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

             

the electric force is

             F_{e} = q E

   

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

         

               m = 2e E / g

     

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

 

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed

Answers

Answer:

Explanation:

d

if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c​

Answers

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/s
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