calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.

Answers

Answer 1
So you would multiply 4 x 16 to an equation of 12

Related Questions

a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time​

Answers

HERE IS YOUR ANSWER!

please help!!!
When a switch is turned from the off to the on position, it is changing the circuit in which of the following ways? O An open circuit is being changed into a closed circuit. A closed circuit is being changed into an open circuit. O A parallel circuit is being changed into a series circuit. A series circuit is being changed into a parallel circuit.​

Answers

Answer:

i Believe the correct answer is "An open circuit being changed into a closed circuit"

Explanation:

The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2

Answers

Answer:

a)    F = 21.16 N,  b)     a = 3.17 10²⁸ m / s

Explanation:

a) The outside between the alpha particles is the electric force, given by Coulomb's law

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

in that case the two charges are of equal magnitude

          q₁ = q₂ = 2q

let's calculate

         F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]

         F = 21.16 N

this force is repulsive because the charges are of the same sign

b) what is the initial acceleration

         F = ma

         a = F / m

         a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27

         a = 3.17 10²⁸ m / s

this acceleration is in the direction of moving away the alpha particles

Balance the following equation:
H3B03 →_B203 +_H20
a. 1, 3,2
b. 2,4,6
C. 4, 2, 6
d. 6, 4,2

Answers

It's c I think ( 4 , 2 , 6 , )

what is the acceleration of a satellite moving in a circular orbit around the earth of radius 2r​

Answers

Explanation:

You do the radius times the circumference of the earth

A long, straight wire has a uniform constant charge with linear charge density, - 3.60 nC/m. The wire is surrounded by a long nonconducting, thin-walled cylindrical shell that is charged on its outside surface, such that the electric field outside the shell is zero. The shell has a radius of 1.50 cm.

Required:
What uniform area charge density rho is needed on the shell for the electric field to be zero outside the shell?

Answers

Answer:

Uniform area charge density rho is needed is 3.82*10^-8 C.m^-2

Explanation:

See the attached files.

To find the rho, I used Gauss law for cylindrical shell which is equation 1 and Gauss law for the rod which is equation 4.

Note that in equation 4, Lamda is the charge per length while L is the length if the rod. Also R is the radius of the shell.

The final answer is 3.82*10^-8 C.m^-2 which is the uniform area charge density rho is needed.

Explain what is happening in this picture

Answers

Answer:

in this video waves are coming up for the BOTTOM to the top of the sandbar

What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)

Answers

Answer:

30.4 s

Explanation:

A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g

a = 4 x 9.8 = 39.2 m /s²

initial velocity u = 0

Final velocity = 3.60 times speed of sound

= 3.6 x 331 = 1191.6 m /s

v = u + at

Putting the values

1191.6 = 0 + 39.2 t

t = 30.4 s .

if the forces on an object are balanced the resultant force is equal to zero true false

Answers

Answer:

If the forces are balanced, the resultant force is zero. If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force. a moving object changes speed and/or direction in the direction of the resultant force.

Explanation:

As the distance between the sun and earth decreases, the speed of the planet

a
increases
b
decreases
c
stays the same

Answers

Answer:

Explanation:

Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.

For this assignment, you should mathematically solve and record a video testing your solution for the following prompt: Two rolls of toilet paper, of equal mass and radius, are dropped from different heights so that they hit the ground at the same time. One roll of toilet paper is dropped normally while the other is dropped while a person holds onto a sheet of toilet paper such that the roll unravels as it descends. Determine the ratio of heights h1/h2, where h1 represents the height of the toilet paper dropped normally and h2 represents the height of the toilet paper that unravels, so that both rolls hit the ground at the same time.

Answers

Answer:

h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

Explanation:

Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )

Determine the ratio of heights  h1/h2

mass of tissues = same

radius of tissues = same

h1 = height of tissue 1

h2 = height of tissue 2

For the first tissue ( Tissue that dropped manually )

potential energy = kinetic energy

mgh = 1/2 mv^2  

therefore the final velocity ( v^2 ) = 2gH  ----- ( 1 )

second tissue ( Tissue that dropped while rotating )

gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )

To determine the ratio of heights we will equate equations 1 and 2

hence :

gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )

∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.

Answers

Answer:

first value+2nd +3rd

Explanation:

thug life and there

Which statement BEST explains why a bouncing basketball will not remain in motion forever?

Group of answer choices

The energy is transferred to sound and heat energy.

The energy is used up and destroyed.

The energy is transferred to light and potential energy.

The energy is transferred to chemical and heat energy.

Answers

Answer:

The energy is transferred to chemical and heat energy.

Explanation:

If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. This is the proof of the answer i Hope this helps :)

The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0. What is the angle?

Answers

Answer:

θ = 90º

Explanation:

The velocity is given by

          v = [tex]\frac{dr}{dt}[/tex]

calculate

          v = 3 i ^ + √2 j ^ + 2t k ^

acceleration is defined by

         a = dv / dt

         a = 2 k ^

one way to find the angle is with the dot product

         v. a = | v | | a | cos θ

         cos θ= v.a / | v | | a |

Let's look for the value of each term

       v. a = 4 t

        | v | = [tex]\sqrt{3^2 + 2 + (2t)^2 }[/tex] = [tex]\sqrt{ 11 + 4t^2}[/tex]

        | a | = 2

they ask us for the angle for time t = 0

         v. a = 0

        | v | = √11 = 3.317

we substitute

        cos θ = 0 /√11

        cos θ = 0

therefore the angles must be θ = 90º

John attaches a ball to a spring. The diagram below shows what happens. Which option shows the direction of the force of the ball on the spring?

Answers

The answer is downwards because the ball is pulling down the spring due to its weight

Option C shows the direction of the force of the ball on the spring. The direction of the force of the ball on the spring will be downwards.

What is force?

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

The spring is extended downward because the weight is always act downwards. The direction of the force of the ball on the spring will be downwards.

Hence, option C shows the direction of the force of the ball on the spring

To learn more about the force, refer to the link;

https://brainly.com/question/26115859

#SPJ2

77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m

Answers

Answer:

a

Explanation:

i just took the test

A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.

Answers

Answer:

Explanation:

The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .

The weight component acting on box parallel to incline plane

= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N

This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .

Force exerted by person = 98 N

distance travelled in 5 s

= velocity x time

= 2 x 5 = 10 m

Work done by person

= 98 x 10

= 980 J .

What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.

Answers

Answer:

the sign and magnitude of the charge is - 2 x 10⁻⁵ C.

Explanation:

Given;

mass of the particle, m = 1.43 g = 0.00143 kg

electric field experienced by the particle, E = 700 N/C

The force experienced by the particle is calculated as;

F = mg = EQ

Where;

Q is the magnitude of the charge

[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]

The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.

Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.

The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.

The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.

What is electric charge?

The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.

Given information-

The mass of the particle is 1.43 g or 0.00143 kg.

The magnitude of the downward-directed electric field is 700 N/C.

The magnitude of the free-fall acceleration is 9.80 meter per second squared.

The electric field is defined as the electric force per unit charge. It can be given as,

[tex]E=\dfrac{F}{q}[/tex]

Rewrite the equation for the charge,

[tex]q=\dfrac{F}{E}[/tex]

Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,

[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]

Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.

The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.

Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.

Learn more about the electric charge here;

https://brainly.com/question/14372859

A roller coaster moving along its track rolls into a circular loop of radius r. In the loop, it is only affected by its initial velocity, gravity, and the shape of the track. Let v denote the instantaneous speed and a denote the magnitude of the instantaneous acceleration of the roller coaster in the loop. Which of the following is true in the loop?
a. The roller coaster is not in uniform circular motion, but we still have a=v^2/r everywhere on the loop
b. The roller coaster is not in uniform circular motion, but the tangential acceleration is so small that we can approximate a by v^2/r everywhere on the loop
c. The roller coaster is in uniform circular motion
d. The roller coaster is not in uniform circular motion, and a=v^2/r is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes

Answers

Answer:

c. The roller coaster is in uniform circular motion

Explanation:

Since the loop is circular with radius r, and its instantaneous speed, v is always constant, and also, its centripetal acceleration, a' = v²/r.

Since the angular speed, ω = v/r does not change, the magnitude of its  tangential acceleration is zero although there is a change in its direction because the direction of its initial velocity changes. That is a" = rα and α = Δω/Δt since Δω = 0, α = 0 and a" = r(0) = 0

So, there is no tangential acceleration. Since there is no tangential acceleration, our instantaneous acceleration which is the vector sum of our centripetal acceleration and tangential acceleration is a = √(a'² + a"²) =  √(a'² + 0²) = √a'² = a' = v²/r

So, a is always v²/r.

Since the instantaneous acceleration is always (a = v²/r) constant, the motion is uniform. So, it is uniform circular motion.

The roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.

The given parameters;

radius of the circular path, = r instantaneous speed = v instantaneous acceleration = a

The motion tension on the loop at the lowest point in the circular motion is given as;

[tex]T = mg + \frac{mv^2}{r}[/tex]

The motion tension on the loop at the highest point in the circular motion is given as;

[tex]T = \frac{mv^2}{r} - mg[/tex]

This shows that circular motion is affected by;

acceleration due to gravity, gradius of the circular path, rspeed of the motion, vmass of the object, m

Thus, we can that the roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.

Learn more here:https://brainly.com/question/14672628

A star's emission line of 400 nm appears shifted to 404 nm in the spectrum. What can you conclude from this shift?
A. The star is approaching you with the speed of 3000 km/s.
B. The star is approaching you with the speed of 30300 km/s.
C. The star is receding from you with the speed of 3000 km/s.
D. The star is receding from you with the speed of 30300 km/s.

Answers

Answer:

C. The star is receding from you with the speed of 3000 km/s

Explanation:

To get this answer we use the doppler effect equation . The formula for a receding emissor is given in the attachment.

We solve for V

V = 3x10⁶m/s

V = 3000km/s

We have the wavelength to be shifting towards red. Therefore we conclude that it is receding. We say the star is receding with speed of 3000km/s towards you.

Thank you!

The Earth’s orbit _____.

is an ellipse

goes around the moon

is a circle

causes day and night

Answers

The earths orbit revolves around the sun

Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?

Answers

Answer:

The answer is below

Explanation:

Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.

After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:

(a - v) / 5.68 = 3.87

a - v = 21.9816

v = a - 21.9816

For motorcycle B:

(b - v) / 5.68 = 18.2

b - v = 103.376

v = b - 103.376

Therefore:

a - 21.9816 = b - 103.376

b - a = -21.9816 + 103.376

b - a = 81.3944

a) The difference between their speeds at the beginning was 81.3944 m/s

b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.

Therefore motorcycle B was moving faster

A spring has a spring constant of 25 Newtons per meter. The minimum force required to
stretch the spring 0.20 meter from its equilibrium position is approximately

Answers

Answer:

6.3N

Explanation:

Guessed it right on castle learning

Answer:

6.3 N

Explanation:

F=kx

F=(25N/m)(0.25m)

6.3 N

I need help will mark brainliest

Answers

Answer: ITS 1 TRUST ME MAN BYE K

Explanation: OK BYE TRUST YEAH

PHYSICS QUESTION PLS HELP

Answers

The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is

mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J

The total energy is the same, 970,200 J.

Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.

At point B, the coaster has dropped to a height of 10 m, so it has PE

mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J

which means it must have KE

970,200 J = KE + 294,000 J   →   KE = 676,200 J

which gives the coast a speed v at point B of

1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J   →   v21.2 m/s

At point C, the coaster has a speed of 16.0 m/s, so it has KE

1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J

and hence PE

970,200 J = 384,000 J + PE   →   PE = 586,200 J

This lets us determine the height h at C:

mgh = (3000 kg) (9.80 m/s²) h = 586,200 J   →   h ≈ 19.939 m

which means the loop has diameter h - 10 m ≈ 9.94 m.

At point D, the coaster is 15 m above the ground so its PE at D is

mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J

and so its KE is

970,200 J = KE + 441,000 J   →   KE = 529,200 J

and hence has speed v at D

1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J   →   v18.9 m/s

What are the two rules that light follows.​

Answers

ok so i dont know srry5

Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.

Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?

Answers

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

a device that spreads light into different wavelengths is a what?

Answers

maybe a spectrograph ?

Starting from the front door of your ranch house, you walk 55.0 m due east to your windmill, turn around, and then slowly walk 35.0 m west to a bench, where you sit and watch the sunrise. It takes you 30.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench.

Required:
a. For the entire trip from your front door to the bench, what is your average velocity?
b. For the entire trip from your front door to the bench, what is your average speed?

Answers

Answer:

Explanation:

Average velocity = Total displacement / total time

Average speed = total distance covered / total time

a )

For the entire trip from your front door to the bench

Total displacement = 55 - 35 = 20 m  [ first displacement is positive and second displacement is negative , because second displacement is in opposite direction ]

Total displacement = 20 m

Total time = 30 + 36 = 66 s

Average velocity = 20 / 66

= .303 m / s

b )

For the entire trip from your front door to the bench

Total distance covered  = 55 + 35 = 90  m

Total time = 30 + 36 = 66 s

Average speed  = 90 / 66

= 1.36  m / s

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 520 lines/mm , and the light is observed on a screen 1.4 m behind the grating.

Required:
What is the distance between the first-order red and blue fringes?

Answers

Answer:

0.143 m

Explanation:

Since

d = 1/N = 1/520 = 1.92 * 10^-3 mm

For red light;

θ = sin^-1  (1 * λred/d) =  sin^-1  (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°

L = 1.4 * (tan 19.98) = 0.509 m

For blue light;

θ = sin^-1  (1 * λblue/d) =  sin^-1  (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°

L = 1.4 * (tan 14.66°) = 0.366 m

Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m

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