a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

Answer:

The reason for the increase or decrease of the moon is due to the angular perception of the moon.

Explanation:

Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.

The zenith is the highest part of the sky and the horizon the lowest.

When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.

But when looking up at the sky as there is no reference point there will be a failure in the perception of size.

Answer:

Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.

Instead the physical changes implies that if you can return to the same substance through a reverse process.

Explanation:

A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.

A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.

A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.  

• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.  

• In a chemical change always a chemical reaction takes place.  

• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.  

• In a physical change no new chemical species form.  

• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.  

• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.  

Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.  

To know more about:

https://brainly.com/question/7279398

Answer:

-50N

Explanation:

F=ma=m(Vf-Vi)/t

F=(10)(-10)/(2)=-50N

So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]

where [tex]M_{0}[/tex] is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]

where

[tex]\beta = V/c[/tex]

c  is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

the equation reduces to

1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]

square both sides of the equation, we have

3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]

3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1

2.0625 = 3.0625[tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] = 0.67

β = 0.819

substitute for  [tex]\beta = V/c[/tex]

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

Answer:

 Good Diet: ! gallon of water a day, Fruits, Vegetables, White meats(Chicken), Don't eat past 3 PM.

Bad Diet: Pizza, Red meat, Baked goods, Eating at late hours.

Explanation: I know the difference because, When you drink water first thing in the morning it gets your metabolism running. Than means you can digest foods better, you want to feed your body good foods but you should not eat until you feel stuffed. You should eat until you are no longer starving. Than you should drink a cup of water in between meals. I know you should not eat past 3 pm because your body needs time to digest foods because you should never go to sleep with a full stomach. I know the difference between good food and bad food because when you eat healthy food and a balanced diet, your body will have more energy and you wont feel tired afterwards. Eating bad foods and food with artificial sugars will clump up in your kidneys, and your body will have small bursts of energy but you will feel lazy afterwards...Your body is supposed to stay energized from a healthy meal in order to give you the energy your body needs to exercise. If you feel droopy all the time and you don't want to do anything, than you are unhealthy.

Answer:

A vegetarian diet is an example of a good fad diet if you do it correctly. It can help you get lots of veggies and good nutrients from them while still following the non-meat diet you want. This can be effective and good for weight loss becasue you are still eating and getting all the good nutrients and calories from less fatty foods. 

Vegan diet (some can be successful but many people fail and do not do good that is why I choose this) The problem with this fad diet is that it can cause nutritional deficiencies and lead to a host of additional health problems, including negatively impacting hormonal health and metabolism. Many people also struggle to find healthy vegan food and end up eating bad and fatty foods instead. 

Explanation:

Got a 100

Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

Answer

The effect is that it Decreases the field current IF and increases slope K1

Answer:

A. magnetic field

Explanation:

The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.

As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.

The magnetic field is produced as a result when an electrical current is passed through the conducting wire .

Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .

Learn more about the magnetic fields here, refer to the link given below;

brainly.com/question/23096032

#SPJ6

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

Answer:

your question answer is 22°

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              [tex]D = \frac{1.22 \lambda }{d }[/tex]

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

Answer:

0.6 m/s

Explanation:

2Hz = 2^-1 = 2 /s

30cm = .3m

Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.

(.3m)*(2 /s) = 0.6 m/s

Answer:

The intensity of light from the first polarizer  is [tex]I_1 = 26 W/m^2[/tex]

Explanation:

  The intensity of the unpolarized light is  [tex]I_o = 52.0 \ W/m^2[/tex]

Generally the intensity of light that emerges from the first polarized light is

            [tex]I_1 = \frac{I_o}{2 }[/tex]

 substituting values

             [tex]I_1 = \frac{52. 0}{2 }[/tex]

             [tex]I_1 = 26 W/m^2[/tex]

Answer:

145060km

Explanation: Given that

speed = dx/dt = 2t^2 +1

integrate

x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)

given that at t=0, x = 1000

so 1000 = 2/3 X (0)^3 + 0 + c

or c = 1000

So x = 2/3t^3 + t + 1000

for t = 1 min = 60s

x = 2/3 X 60^3 + 60 + 1000

x = 2/3×216000+ 1060

x = 144000+1060

= 145060km

At one minute, it will be 145060km far from the earth

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

Answer:

a

   [tex]\lambda = 1.667 nm[/tex]

b

     [tex]\theta = 0.8681^o[/tex]

Explanation:

From the question we are told that

   The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]

    The  is distance of the screen from the slit is  [tex]D = 2.60 \ m[/tex]

    The distance between the central bright fringe and either of the adjacent bright   [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]

Generally  the condition for constructive interference is  

      [tex]d sin \tha(\theta ) = n \lambda[/tex]

From the question we are told that small-angle approximation is valid here.

So    [tex]sin (\theta ) = \theta[/tex]

=>        [tex]d \theta = n \lambda[/tex]

=>        [tex]\theta = \frac{n * \lambda }{d }[/tex]

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         [tex]y = D * sin (\theta )[/tex]

From the question we are told that small-angle approximation is valid here.

So

       [tex]y = D * \theta[/tex]

=>   [tex]\theta = \frac{ y}{D}[/tex]

So

     [tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]

     [tex]\lambda =\frac{d * y }{n * D}[/tex]

substituting values

       [tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]

        [tex]\lambda = 1.667 *10^{-6}[/tex]

        [tex]\lambda = 1.667 nm[/tex]

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     [tex]dsin (\theta ) = n \lambda[/tex]

    [tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]

    [tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]

    [tex]\theta = 0.8681^o[/tex]

Complete question is;

The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

13 m

Explanation:

We are given;

Distance between two nearly parallel mirrors; d = 6.5 m

Distance between the face and the nearer mirror; x = 3 m

Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m

Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.

Thus;

Distance of the first reflection of the back of the head in the rear mirror from the object head is;

y' = 2y

y' = 2 × 3.5

y' = 7

The total distance of this image from the front mirror would be calculated as;

z = y' + x

z = 7 + 3

z = 10

Finally, the second reflection of this image will be 10 meters inside in the front mirror.

Thus, the total distance of the image of the back of the head in the front mirror from the person will be:

T.D = x + z

T.D = 3 + 10

T.D = 13m

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

Complete Question

At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.

Answer

a. Approximately [tex]5*10^{10}[/tex] fissions per second.

b. Approximately [tex]6*10^{12 }[/tex]fissions per second.

c. Approximately [tex]4*10^{11}[/tex] fissions per second.

d. Approximately [tex]3*10^{12}[/tex] fissions per second.

e. Approximately[tex]3*10^{14}[/tex] fissions per second.

Answer:

The correct option is  d

Explanation:

From the question we are told that

       The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]

Thus to generated  [tex]100 \ J/s[/tex] i.e  (100 W  ) the rate of fission is  

              [tex]k = \frac{100}{3.2 *10^{-11} }[/tex]

              [tex]k =3*10^{12} fission\ per \ second[/tex]

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

Explanation:

Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,

[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]

[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]

[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]

[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]

after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,

[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]

[tex]\qquad \Phi_{B, f}=0[/tex]

(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,

[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]

[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]

[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]

(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb

(b) The induced EMF in the coil is 0.583 mV

Given that the coil has N = 250 turns

and an area of A = 14cm² = 1.4×10⁻³m².

It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T

(a) The flux passing through the coil is given by:

Ф = NBAcosθ

where θ is the angle between area vector and the magnetic field

The area vector is perpendicular to the plane of the coil.

So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil

So the initial flux is:

Φ = NABcos0° = NAB

Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb

Ф = 1.75 × 10⁻⁵ Wb

Finally, θ = 90°, and since cos90°, the final flux through the coil is 0

(b) The EMF induced is given by:

E = -ΔФ/Δt

E = -(0 - 1.75 × 10⁻⁵)/0.030

E = 0.583 × 10⁻³ V

E = 0.583 mV

Learn more about magnetic flux:

https://brainly.com/question/15359941?referrer=searchResults

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

[tex]v = \sqrt{\frac{T}{\mu} }[/tex]

where;

μ is mass per unit length

[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]

The final tension T₂ will be calculated as;

[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]

Therefore, the tension on string when the speed was raised is 134.53 N

Answer:

Explanation:

magnetic field due to an infinite current carrying conductor

B = k x 2I / r where k = 10⁻⁷  , I is current in conductor and r is distance from wire

putting the given data

B = 10⁻⁷ x 2 x 100 / 8

= 25 x 10⁻⁷ T .

B )

earth's magnetic field = .5 gauss

= .5 x 10⁻⁴ T

= 5 x 10⁻⁵ T

percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100

= 5 %

C )

ii.  No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.

Answer:

b. The current in the loop always flows in a counterclockwise direction.

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.

Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

Learn more about the magnetic field here:

https://brainly.com/question/14848188

Answer:

a

   [tex]m = 0.169 \ kg[/tex]

b

  [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Explanation:

From the question we are told that

    The  spring constant is  [tex]k = 14 \ N/m[/tex]

     The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]

     The number of oscillation is  [tex]n = 30[/tex]

      The  time taken is  [tex]t = 20 \ s[/tex]

Generally the the angular speed of this oscillations is mathematically represented as

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

where T is the period which is mathematically represented as

     [tex]T = \frac{t}{n}[/tex]

substituting values

     [tex]T = \frac{20}{30 }[/tex]

     [tex]T = 0.667 \ s[/tex]

Thus  

       [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]

       [tex]w = 9.421 \ rad/s[/tex]

this angular speed can also be represented mathematically as

       [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]m =\frac{k }{w^2}[/tex]

substituting values

      [tex]m =\frac{ 15 }{(9.421)^2}[/tex]

      [tex]m = 0.169 \ kg[/tex]

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        [tex]v = - Awsin (wt)[/tex]

The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]

     [tex]v_{max} = - A* w[/tex]

=>   [tex]|v_{max} |= A* w[/tex]

=>    [tex]|v_{max} |= 0.06 * 9.421[/tex]

=>   [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Answer:

A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.

Answer:

μ = 1

F = P√2

Explanation:

The parabola equation is: y = ½ x².

The slope of the tangent is dy/dx = x.

The angle between the tangent and the x-axis is θ = tan⁻¹(x).

At x = 1, θ = 45°.

Draw a free body diagram of the block.  There are three forces:

Weight force P pulling down,

Normal force N pushing perpendicular to the surface,

and friction force Nμ pushing up tangential to the surface.

Sum of forces in the perpendicular direction:

∑F = ma

N − P cos 45° = 0

N = P cos 45°

Sum of forces in the tangential direction:

∑F = ma

Nμ − P sin 45° = 0

Nμ = P sin 45°

μ = P sin 45° / N

μ = tan 45°

μ = 1

Draw a new free body diagram.  This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.

Sum of forces in the tangential direction:

∑F = ma

F − Nμ − P sin 45° = 0

F = Nμ + P sin 45°

F = (P cos 45°) μ + P sin 45°

F = P√2