You and your friends are pushing a car with a total of 1080 N of force. You are able to push the car for a distance of 218 m. Calculate the work done on the car.

Answers

Answer 1

Answer:

How much power is required to pull a sled if you use 60J of work in 5 seconds? ... Misfortune occurs and Renatta and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push their car 218 m to the ... Calculate the amount of work done when moving a 567N crate a distance of 20 meters.

Explanation:

Misfortune occurs and Renatta and her friends find themselves getting a ... They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel ... Write down what they give you. ... Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m ... Work = Force x Distance = Joules (force = weight).


Related Questions

The density of 1 kilogram of gold is

Answers

Answer:

0.02 kg/cm³

Explanation:

An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k = 450 N/m. The object is determined to have a velocity of 0.3 m/s when passing through the equilibrium.

1. Find the amplitude of the motion

2. Find the total energy of the object at any point of its motion​

Answers

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ([tex]U_{e}[/tex]), in joules, is equal to maximum translational kinetic energy ([tex]K[/tex]), in joules:

[tex]U_{e} = K[/tex]

[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)

Where:

[tex]k[/tex] - Spring constant, in newtons per meter.

[tex]A[/tex] - Amplitude, in meters.

[tex]m[/tex] - Object mass, in kilograms.

[tex]v[/tex] - Speed of the object at equilibrium, in meters per second.

If we know that [tex]k = 450\,\frac{N}{m}[/tex], [tex]m = 0.25\,kg[/tex] and [tex]v = 0.3\,\frac{m}{s}[/tex], then the amplitude of the motion is:

[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]k\cdot A^{2} = m\cdot v^{2}[/tex]

[tex]A = v\cdot \sqrt{\frac{m}{k} }[/tex]

[tex]A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }[/tex]

[tex]A \approx 0.274\,m[/tex]

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ([tex]E[/tex]), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ([tex]k = 450\,\frac{N}{m}[/tex], [tex]A \approx 0.274\,m[/tex])

[tex]E = U_{e}[/tex]

[tex]E = \frac{1}{2}\cdot k\cdot A^{2}[/tex]

[tex]E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}[/tex]

[tex]E = 16.892\,J[/tex]

The total energy of the object at any point of its motion is 16.892 joules.

2(A + B)
15. The resultant of A and B is perpendicular to A
What is the angle between A and B?
(a) cos
(b) cos
La
(c) sin
(d) sin​

Answers

Answer:

θ = cos^(-1) (-A/B)

Explanation:

The image of the reauktant forces A & B are missing, so i have attached it.

Now, from the attached image, we will see that;

Angle between A and B is θ

Also;

A = Bcos(180° − θ)

Now, in trigonometry, we know that;

cos(180° − θ) = -cosθ

Thus;

A = -Bcosθ

cosθ = -A/B

Thus;

θ = cos^(-1) (-A/B)

state newton first law of motion​

Answers

Newton’s first law of motion states that there must be a cause—which is a net external force—for there to be any change in velocity, either a change in magnitude or direction. An object sliding across a table or floor slows down due to the net force of friction acting on the object.

got it off g lol..

Answer:

it state that everybody in the universe is state that" universe continues its state of rest or uniform motion in a straight path unless it is acted upon by external force."

In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 32.0 nm long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring.

Required:
a. Find the ratio of the frequency with the virus attached ( fS+V) to the frequency without the virus (fS) in terms of mV and mS, where mV is the mass of the virus and mS is the mass of the silicon sliver.
b. In some data, the silicon sliver has a mass of 2.13×10^-16 g and a frequency of 2.04×10^15 Hz without the virus and 2.85×1014 with the virus. What is the mass of the virus in grams?

Answers

Answer:

a)   m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1) ,  b)  m_v = 1.07 10⁻¹⁴ g

Explanation:

a) The angular velocity of a simple harmonic motion is

           w² = k / m

where k is the spring constant and m is the mass of the oscillator

let's apply this expression to our case,

silicon only

         w₉² = [tex]\frac{K}{m_s}[/tex]

         k = w₀² m_s

silicon with virus

         w² = [tex]\frac{k}{m_s + m_v}[/tex]

          k = w² (m_v + m_s)

in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal

           w₀²  m_s = w² (m_v + m_s)

           m_v = ([tex]\frac{w_o}{w}[/tex])²  m_s - m_s

           m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1)

b) let's calculate

          m_v = 2.13 10⁻¹⁶ [([tex]\frac{20.4}{2.85}[/tex])² - 1)]

          m_v = 1.07 10⁻¹⁴ g

How much work will a 500 watt motor do in 10 seconds?

Answers

Answer:

50j

Explanation:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Answer:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Explanation:

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of 0.295 cm above its initial position. What is the velocity of the bullet on leaving the gun's barrel

Answers

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

What does it mean if the reflected beam is above the incident beam? What does it mean if reflected beam is below the incident beam?

Answers

Answer:

aim at prisma and will have all colors

Explanation:

Answer:

If the vision position is above the actual image location then the light travel from the object in such a way that the angle of incidence is less than the angle of reflected ray which means that the reflected beam is above the incident beam.

Explanation:

Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle

Answers

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

Hint : ΔV = Ed

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F[tex]_{net[/tex] = 0

so, F[tex]_E[/tex] = F[tex]_B[/tex]

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

A cart of mass m is moving with negligible friction along a track with known speed v1 to the right. It collides with and sticks to a cart of mass 4m moving with known speed v2 to the right. Which of the two principles, conservation of momentum and conservation of mechanical energy, must be applied to determine the final speed of the carts, and why

Answers

Answer:

conservation of linear momentum

We were told that two objects became stuck together hence we have to use the principle of conservation of  momentum to obtain the final velocities of the carts.

What is conservation of momentum ?

The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.

We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.

Learn more about conservation of momentum: https://brainly.com/question/11256472

Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________

a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.

Answers

Answer:

d. the star is a member and also a part of an eclipsing binary star system.

Explanation:

If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).

The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.

Thus, option D is correct.

How much energy is supplied to a 9 V bulb if it is switched on for 3 minutes and takes a current of
0.2 A ?

Answers

Answer:

0.01j

Explanation:

the energy equals the work done by the bulb.

Workdone=

[tex]workdone = \frac{power}{time} [/tex]

power=voltage×current

=9×0.2

=1.8 W.

THEREFORE,

time=3×60

= 180s

workdone=1.8/180

=0.01 j

You need friction created by your tires and the road ____
to control your speed and direction.

Answers

Answer:

surface

Explanation:

You need friction created by your tires and the road surface

to control your speed and direction.

The nucleus of a certain type of uranium atom contains
92 protons and 143 neutrons. What is the total charge of
the nucleus?

Answers

Answer:

charge = electrons + protons

=92+92

=184

17. 53 A small grinding wheel is attached to the shaft of an electric motor that has a rated speed of 3600 rpm. When the power is turned off, the unit coasts to rest in 70 s. The grinding wheel and rotor have a combined weight of 6 lb and a combined radius of gyration of 2 in. Determine the average magnitude of the couple due to kinetic friction in the bearings of the motor.

Answers

Answer:

0.337 lb-in

Explanation:

From the law of conservation of angular momentum,

L' = L" where L = initial angular momentum of system and L" = final angular momentum of system

Now L = Iω + Mt where Iω = angular momentum of shaft + wheel and Mt = impulse on system due to couple M.

L' = Iω' + (-Mt) (since the moment about the shaft is negative-anticlockwise)

L' = Iω' - Mt where Iω' = angular momentum of shaft at t' = 0 + wheel and Mt = impulse on system due to couple M in time interval t = 70 s.

L" = Iω"  where Iω" = angular momentum of shaft at t" = 70 s.

Now I = moment of inertia of system = mk² where m = mass of system = W/g where W = weight of system = 6 lb and g = acceleration due to gravity = 32 ft/s². So, m = W/g = 6lb/32 ft/s² = 0.1875 lb-s²/ft and k = radius of gyration = 2 in = 2/12 ft = 1/6 ft.

So, I = mk² = (0.1875 lb-s²/ft) × (1/6 ft)² = ‭0.00521‬ lb-ft-s², ω' = initial angular speed of system = 3600 rpm = 3600 × 2π/60 = 120π  rad/s = 377 rad/s,  ω" = final angular speed of system = 0 rad/s (since it stops), t' = 0 s, f" = 70 s and M = couple on system

So,

Iω' - Mt" = Iω"

Substituting the values of the variables into the equation, we have

Iω' - Mt" = Iω"

0.00521‬ lb-ft-s² ×  377 rad/s - M × 70 s = 0.00521‬‬ lb-ft-s² × 0 rad/s"

0.00521‬ lb-ft-s² ×  377 rad/s - 70M = ‭0

1.964‬ lb-ft-s = 70M

M = 1.964‬ lb-ft-s/70 s

M = 0.0281‬ lb-ft

M = 0.0281 lb × 12 in

M = 0.337 lb-in

Saved Which of the following is NOT an important function of facial display? Multinio Choic
A. emotion
B. attractiveness
c. Primacy
d. identity

Answers

Answer:

C

Explanation:

Primacy means being first or important so thats not an important facial display as the others.

10 POINTS!! SPACE QUESTION!

Answers

B gas giants in fact Jupiter has more moons than all the inner planets combined thank me later

Answer:

The Gas Giants have more moons.

Explanation:

Mercury-0

Venus-0

Earth-1

Mars-2

Jupiter-66

Saturn-62

Uranus-27

Neptune-13

Which of the following best describes what occurs in a fission reaction?

A.
Two low mass nuclei are joined to form one nucleus.

B.
Electrons are shared between the nuclei.

C.
A single nucleus divides into two or more nuclei and gives off energy.

D.
A chemical reaction occurs between the nuclei.

Answers

Answer:

C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.

Answer:

C.

A single nucleus divides into two or more nuclei and gives off energy.

hope it is helpful to you

When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV.
What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface?
Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.

Answers

Answer:

Explanation:

energy of photon having wavelength of 400 nm = 1237.5/400 eV

= 3.1 eV.

Maximum kinetic energy of photoelectrons = 1.1 eV .

Threshold energy Ф = 3.1 - 1.1 = 2 eV .

energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .

Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy

= 4 - 2 = 2 eV .

Required kinetic energy K₀= 2 eV.

A car is moving with speed 30 m/s and acceleration 4 m/s2 at a given instant. (a) Using a second-degree Taylor polynomial, estimate how far the car moves in the next second.

Answers

Answer:

68 meters moved in the next seconds

Explanation:

Given

[tex]u= 30m/s[/tex]

[tex]a = 4m/s^2[/tex]

Required

Distance covered by the car in the next second

At a point in time t, the current distance is calculated as:

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

Substitute values for a and u in the above equation.

[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]

[tex]s(t) =30t + 2t^2[/tex]

Next, we generate the second degree Taylor polynomial as follows;

Calculate velocity (s'(t))

Differentiate s(t) to get velocity

[tex]s(t) =30t + 2t^2[/tex]

[tex]s'(t) =30 + 4t[/tex]

Calculate acceleration (s"(t))

Differentiate s'(t) to get acceleration

[tex]s'(t) =30 + 4t[/tex]

[tex]s"(t) =4[/tex]

When t = 0

We have:

[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]

[tex]s'(0) =30 + 4*0 = 30[/tex]

[tex]s"(0) = 4[/tex]

So, the second degree tailor series is:

[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]

To see the distance moved in the next second, we set t to 1

So, we have:

[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]

Solving s(1), s'(1) and s"(1)

We have:

[tex]s(1) =30*1 + 2*1^2 = 32[/tex]

[tex]s'(1) =30 + 4*1 = 34[/tex]

[tex]s"(1) =4[/tex]

Hence:

[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]

[tex]T_2(1) = 32 + 34 + 2[/tex]

[tex]T_2(1) = 68[/tex]

g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.

Answers

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

[tex]A_1v_1 = A_2v_2[/tex]

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]

v₂ = 30 m/s

b)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]

m₁ = 12600 kg

c)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]

m₂ = 12600 kg

please help, no links please! I dont understand stand this question and im going to cry

Which part of this system is in the gas phase?

Answers

Answer:

I'm pretty sure its helium

The air Inside the balloon is in the gas phase

An object’s
✔ mass
will remain constant throughout the universe, but its
can change from planet to planet.

If you increase the mass of a planet, what happens to its gravity?

If the gravity on a planet decreases, what happens to the weight of an object on that planet?

Answers

Answer:

mass, weight, strength of gravity increases, weight decreases

Explanation:

got it on edge

Answer:

An object’s

✔ mass

will remain constant throughout the universe, but its

✔ weight

can change from planet to planet.

If you increase the mass of a planet, what happens to its gravity?

✔ strength of gravity increases

If the gravity on a planet decreases, what happens to the weight of an object on that planet?

✔ weight decreases

Explanation:

right on edge 22

In 2-3 complete sentences, analyze how scientists know dark matter and dark energy exist.

Answers

Answer:

It doesn't interact with baryonic matter and it's completely invisible to light and other forms of electromagnetic radiation, making dark matter impossible to detect with current instruments. But scientists are confident it exists because of the gravitational effects it appears to have on galaxies and galaxy clusters.

Explanation:

A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it

Answers

Answer:

 w = 0.319 rad / s

Explanation:

This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.

         

initial instant. Before the squirrel jumps

           L₀ = m v r

final instant. After the trough and the squirrel are together

          L_f = (I_fetter + I_ardilla) w

angular momentum is conserved

          L₀ = L_f

          m v r = (I_fetter + I_ardilla) w

          w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]

the moment inercial ofbody is

         I_thed = 2.00 kg m²

We approach the squirrel to a specific mass

          I_ardilla = m r²

we substitute

            w = m v r / ( I_[feefer  + m r²)

             

           

let's calculate

              w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )

              w = 0.6426 / 2.0119

               w = 0.319 rad / s

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft

Answers

Answer:

[tex]523269.9\ \text{N/m}[/tex]

Explanation:

q = Charge

r = Distance

[tex]q_1=25\ \text{C}[/tex]

[tex]r_1=3000\ \text{m}[/tex]

[tex]q_2=40\ \text{C}[/tex]

[tex]r_2=850\ \text{m}[/tex]

The electric field is given by

[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]

The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]

Which form of energy increases when a spring is compressed?

Answers

Answer:

When the spring compresses, elastic potential energy increases.

Answer:

the answer is b

Explanation:

elastic potential energy

While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.

Answers

Answer:

the smallest separation between two objects is 0.8067 m

Explanation:

Given the data in the question;

Altitude h = 5.75 km = 5750 m

Diameter D = 4.0 mm = 0.004 m

λ = 460 nm = 4.6 × 10⁻⁷ m

Now, Using Rayleigh criterion for Airy disks resolution.

we know that, Minimum angular separation for resolving two points is;

θ = 1.22λ / D

so we substitute

θ = (1.22 × 4.6 × 10⁻⁷)  / 0.004

θ = 5.612 × 10⁻⁷ / 0.004

θ = 1.403 × 10⁻⁴ rad  

so minimum separation [tex]d_{min[/tex] =  θh

so we substitute

[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m

[tex]d_{min[/tex] = 0.8067 m

Therefore, the smallest separation between two objects is 0.8067 m

A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled?

Answers

Answer:

A boat travels for three hours with a... A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in still water?

Explanation:

A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled? *
A)60km
B)60km[W]
C)17km[W]
D) 6.6km[W]

Answers

Answer:

B)60km[W]

Explanation:

The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].

Hope that helps!

Correct answer is (B) 60Km
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