A ball is thrown horizontally to the right, from the top of a vertical cliff of height h. A wind blows horizontally to the left, and assume (simplistically) that the effect of the wind is to provide a constant force to the left, equal in magnitude to the weight of the ball. How fast should the ball be thrown so that it lands at the foot of the cliff

ok so i dont know srry5

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

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 :)

Answer: 30 Years Old

Explanation: The constitution has around three qualifications for service in the U.S. Senate, Your age must be at least 30 years.

Answer:

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

Explanation:

Answer:

Explanation:

Angle of incidence at the longer face = 45⁰ ( see the figure in the attached file )

For total reflection from longer face

i = C where C is critical angle .

And relation between critical angle and refractive index is as follows .

μ = 1 / sinC

= 1 / sin45

= √2

= 1.414 .

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = [tex]\frac{1}{2}[/tex] mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             [tex]K_e = K_p[/tex]

              K_p = ½ m v_e²

              K_p = [tex]\frac{1}{2}[/tex]  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]

        v/c= 2.33 10⁻⁴

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]

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

Answer:

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

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

Answer:

c

Explanation:

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

Answer:

a

Explanation:

i just took the test

maybe a spectrograph ?

Answer:

       y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Explanation:

Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation

             λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]

Let's use conservation of energy for speed

starting point

             Em₀ = U = e V

final point

             Em_f = K = ½ m v²

             Em₀ = Em_f

             eV = ½ m v²

             v =[tex]\sqrt{\frac{2eV}{m} }[/tex]

we substitute

             λ=  [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]

the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum

            a sin θ = m λ

first zero occurs at m = 1, also these experiments are performed at very small angles

            sin θ = θ

            θ = λ / a

This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain

           θ = 1.22 λ / D

where D is the diameter of the opening

we substitute

          θ = [tex]\frac{1.22}{D}[/tex]   \sqrt{ \frac{h^2 m}{2eV}}  

this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians

            θ = y / L

when substituting

where y is the minimum distance that can be resolved for this acceleration voltage

            y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Answer:

first value+2nd +3rd

Explanation:

thug life and there

Answer:

(A)  the acceleration  experienced by the proton 2.821 x 10¹² m/s²

(B) the speed of the proton is 2.67 x 10⁵ m/s

Explanation:

Given;

electric field experienced by the proton, E = 2.95 x 10⁴ N/C

charge of proton, Q = 1.6 x 10⁻¹⁹ C

mass of proton, m = 1.673 x 10⁻²⁷ kg

distance moved by the proton, d = 1.26 cm = 0.0126 m

(a)

The force experienced by the proton is calculated as;

F = ma = EQ

where;

a is the acceleration  experienced by the proton

[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]

(b) the speed of the proton is calculated;

v² = u² + 2ad

v² = 0 + (2 x 2.821 x 10¹² x 0.0126)

v² = 7.109 x 10¹⁰

v = √7.109 x 10¹⁰

v = 2.67 x 10⁵ m/s

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:

Answer: ITS 1 TRUST ME MAN BYE K

Explanation: OK BYE TRUST YEAH

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.

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;

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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;

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;

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

Answer:

a) 36°

b) 0.109 kg

Explanation:

Heat flows from brass to copper with the brass having its temperature

Length of brass = 0.4

Length of copper = 0.8

Temperature of = 36.15

See attachment for calculation

The temperature at the joint is 36.15°C

The amount of ice melted is 1.086 kg

The rate of transfer of thermal energy,

   H = Q/t = KAΔT/L

  where, K is the thermal conductivity of the substance, A is cross-sectional area, ΔT is temperature difference at the ends and L is the length

As given in the question,

  the length of the brass section   [tex]L_{1}[/tex] = 0.4 m

  it's thermal conductivity      [tex]K_{b}[/tex] = 109 J[tex]s^{-1}m^{-1}K^{-1}[/tex]

  the temperature at the brass end  [tex]T_{1}[/tex] = 373K

  the length of the copper section [tex]L_{2}[/tex] = 0.8 m

  it's thermal conductivity      [tex]K_{c}[/tex] = 385 J[tex]s^{-1}m^{-1}K^{-1}[/tex]

  the temperature at the brass end  [tex]T_{1}[/tex] = 273K

 cross-sectional area of both the substance is same A = 0.007 [tex]m^{2}[/tex]

  Let the temperature at the joint be T

The rate of heat flow must be constant across the whole length of the setup.

Hence at the joint,

   [tex]\frac{K_{b}A(T_{1}-T) }{L_{1} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]

⇒ [tex]\frac{ 109*A*(373-T)}{0.4} =\frac{385*A(T-273)}{0.8}[/tex] ⇒ T=309.15 K

⇒ T = 36.15°C is the temperature at the joint.

Now we have to calculate the equivalent thermal conductivity K of the setup in order to calculate the amount of heat transfer.

considering equivalent thermal conductivity K throughout the setup we can form the following equation to calculate its value

   [tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]

   ⇒   [tex]\frac{ K*A*(100)}{1.2} =\frac{385*A(36.15)}{0.8}[/tex]

   ⇒  K = 208.76 J[tex]s^{-1}m^{-1}K^{-1}[/tex]

the amount of heat transferred at the copper end in ice-water mixture in 5 minutes(300 seconds) :

        Q = [tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} }[/tex] × t =  [tex]\frac{208.76*0.007*100}{1.2}[/tex] × 300 = 36533 J

        latent heat of fusion of ice [tex]L_{f}[/tex] = 33600 J/kg

         [tex]Q=mL_{f}[/tex]

         [tex]m=\frac{Q}{L_{f} }[/tex]

         [tex]m=\frac{36533}{33600}[/tex] ⇒ m = 1.086 kg of ice is melted in 5 minutes

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HERE IS YOUR ANSWER!

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   →   v ≈ 21.2 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   →   v ≈ 18.9 m/s

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 .

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

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