What initial speed v is required if the blocks m1 =2.5 kg and m2=1.5 kg are to travel a distance d =7.0cm before coming to rest? Assume the coefficient of kinetic friction between m1 and the tabletop is ųk=0.21

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

Explanation:

We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:

Vt = ωr with ω being the angular velocity.

We now calculate ω:

ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s

Now that we have ω we can calculate the centripetal aceleration:

a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2

Answer:

a = Δv/Δt = (0 - 20) / (5 - 2) = -6⅔ m/s²

The ratio of the primary turns to the secondary turns is 1/3

The correct answer to the question is Option A. 1/3

From the question given above, the following data were obtained:

Ratio = Nₚ/Nᵣ

Nₚ/Nᵣ = 225 / 675

Nₚ/Nᵣ = 1/3

Therefore, the ratio of the primary turns to the secondary turns is 1/3

Complete question:

See attached photo

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Answer:

Explanation:

Relative to an origin at the bottom of the hill,

PE = mgh = 10(9.8)(15) = 1470 J

Answer:

One force will be gravity & inertia.

Explanation:

Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth

Answer:

Without the full content of your question, I will have to GUESS at the context and assume

E = Energy

released when glucose is broken down.

Answer:

The total amount of energy, kinetic plus potential, remains the same.

Explanation:

Answer:

Explanation:

30 = e^5k

ln30 = lne^5k

ln30 = 5k

k = ln30/5

k = 0.68023947...

round to your heart's content.

The law of reflection allows to find the results for the questions about ray reflection in a plane mirror are:

    a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.

    b)  The extension of the reflected rays is what forms the image.

    c)  The image's distance  is 20 cm behind the flat mirror.

    d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C  

    e) the Rays at points A, B, C cannot perceive in the point E.

    f) attachment we see the rays that come out from the pencil eraser.

    g) The image is behind the mirror at 15 cm.

The geometric interaction describes the interaction of light rays with surfaces, looking for where the rays are directed, it is described by two phenomenological laws:

             [tex]\theta_i = \theta_{r}[/tex]  

From these two general laws, geometric optics establishes a relationship for the formation of the image, called the constructor's equation.

            [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

Where f is the focal length, p and q are the distance to the object and the image, respectively.

In this exercise, the medium is a mirror, which is why it must comply with the law of reflection.

a) In the attachment we see a diagram of the incident and reflected rays for the three points.

According to the law of reflection, the incident and reflected angles are equal.

b) From the diagram we can see that the extension of the reflected rays is what forms the image, which is called virtual and is located behind the mirror.

c) In the diagram we see two rays to form the image, we see that the distance to the object is equal to the distance to the image.

From the constructor's equation a plane mirror has an infinite radius.

      p = -q

Therefore the image's distance  is 20 cm behind the flat mirror.  Therefore the distance to the object and the image are the same, the negative sign indicates that the image is behind the mirror.

d) A person located at point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C since their angle of reflection is not equal to the incident angle.

To perceive a ray it must have an angle of incidence of 25º.

e) Point E is located very far from the pencil, so the incident angle increases as does the reflected angle.

the Rays at points A, B, C cannot perceive.

f) In the attachment we see the rays that come out from the pencil eraser, they indicate that the distance to the plane mirror is 15.0 cm,

g) The image is behind the mirror at 15 cm.

In conclusion using the law of reflection we can find the results for the questions are:

     a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.

     b)  The extension of the reflected rays is what forms the image.

     c)  The image's distance  is 20 cm behind the flat mirror.

     d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C  

    e) the Rays at points A, B, C cannot perceive in the point E.

    f) attachment we see the rays that come out from the pencil eraser.

   g) The image is behind the mirror at 15 cm.

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Answer:

60w

Explanation:

The power required is 30 Watt.

Let us recall that power is defined as the rate of doing work. Hence, we can write as follows;

Power = Work done/ time taken

Now;

work done =  Force × distance

Force = 30.0 N

Distance = 10.0 m

work done = 30.0 N × 10.0 m = 300 J

The power expended = 300 J/10.00 s = 30 Watt

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[tex]\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12 gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}[/tex]

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

The Universe is often studied by Scientist. The primary form of evidence that has led astronomers to conclude that the expansion of the universe is accelerating is the Observations of white dwarf supernovae.

The key evidence for that made scientist to talk about the expansion of the universe is accelerating comes from viewing and studying of white dwarf supernovae. It has been found that most distant stars all orbit at approximate speed as stars found about 30,000 light-years from their center.

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Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

   Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

  vo = 15.5 m/s

The speed of car driving is 15.5 m/s as the car is parked and drive away.

Speed is defined as a measurement of the length of time it takes for an object to travel a certain distance. You can determine an object's speed if you know how far it moves in a given amount of time. Time does not move, hence there is no concept of a speed of time. Time refers to how we move through the temporal realm. Speed is a unit of measurement for how quickly something is moving. A change in velocity results in a change in speed.

To calculate the speed we use the formula

f₀ = f (v + vo) / (v + vs)

Δf = f approach - f depart

69.5 = (769(343 + vo) / (343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

vo = 15.5 m/s

Thus, the speed of car driving is 15.5 m/s as the car is parked and drive away.

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#SPJ2

a.

The direction of the total momentum is 45°

The momentum of the first car is given by p = mv where m = mass of car = 1200 kg and v = velocity of car = 15 m/sj (since it moves due north).

So, p = mv

= 1200 kg × (15 m/s)j

= (18000 kgm/s)j

Also, the momentum of the identical car, p' = mv' where m = mass of car = 1200 kg and v' = velocity of car = (15 m/s)i (since it moves due east).

So, p' = mv'

= 1200 kg × (15 m/s)i

= (18000 kgm/s)i

So, the total momentum of the system P = p + p'

=  (18000 kgm/s)j +  (18000 kgm/s)i

=  (18000 kgm/s)i +  (18000 kgm/s)j

The direction of the total momentum of the system P is gotten from

tanФ = p'/p

= 18000 kgm/s ÷ 18000 kgm/s

= 1

Ф = tan⁻¹(1)

= 45°

The direction of the total momentum is 45°

b.

The magnitude of the total momentum of the system is 25455.84 kgm/s

The magnitude of the total momentum of the system P = √(p'² + p²)

= √[(18000 kgm/s)² + (18000 kgm/s)²]

= (18000 kgm/s)√(1 + 1)

= (18000 kgm/s)√2

= 25455.84 kgm/s

The magnitude of the total momentum of the system is 25455.84 kgm/s

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This question involves the concepts of the thin lens formula, focal length, and image distance.

The image of the house is "16 ft" away from the lens.

According to the thin lens formula:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]

where,

f = focal length = 8 ft

p = object distance = 16 ft

q = image distance = ?

Therefore,

[tex]\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\[/tex]

q = 16 ft

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The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

The mass of the baseball, m = 0.15 kg

The speed at which it moves, v = 30 m/s

Time at which the baseball was in contact with the bat, t = 0.75 ms

t = 0.75/1000 s

t  = 0.00075 s

The impulsive force is given by the formula:

[tex]F=\frac{mv}{t}[/tex]

Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:

[tex]F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N[/tex]

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

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Explanation:

decomposition reaction.....

Answer:

caco₃--cao+co₂

calcium oxide + carbon die oxide gives us calcium carbonate

this is the reaction of Acidic oxide(cao) and Basic oxide (co₂) to form salt.

Answer:120.3676

Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!

The locations where the riders feel less than their normal weight are Location A, Location C and Location D.

The given parameters;

The normal weight of the riders is calculated by applying Newton's second law of motion as follows;

W = mg

W = 9.8m

The apparent weight of the riders for the upward acceleration is calculated  as follows;

[tex]R = m(g + a)[/tex]

The apparent weight of the riders for the downward acceleration is calculated  as follows;

[tex]R = m(g - a)[/tex]

The apparent weight of the riders at location A is calculated as follows;

[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]

The apparent weight of the riders at location B is calculated as follows;

[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]

The apparent weight of the riders at location C is calculated as follows;

[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]

The apparent weight of the riders at location D is calculated as follows;

[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]

The apparent weight of the riders at location E is calculated as follows;

[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]

Thus, the locations where the riders feel less than their normal weight are;

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Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.

All metals are solid at (our) living temperatures except for mercury which is a liquid. Metals are shiny for a while at least when properly finished. Metals are good conductors of both heat and electricity. This is due to the excess electrons in their valence clouds that facilitate energy transfer.What are 5 characteristics of metal?

Metals are lustrous, malleable, ductile, good conductors of heat and electricity

Silver conducts electricity better than any other metal.

Brass is an alloy made from zinc and copper.

Pure gold is too soft for many things so most gold is combined with other metals to make it stronger. ...

World time periods are often listed by the metal used.

Answer:

a. was zero while stationary and increased when the person stood.

Explanation:

momentum is mass times velocity.

initial velocity was zero

final velocity was NOT zero.

Answer:

1. An object is in motion when its distance from another object is changing.

2.Speed is given direction is called velocity.

Speed can be calculated.......

Answer:

Explanation:

Not sure what your options are but anything that says something like

"at the block surface in contact with the ramp along the line from V to Z"  is probably a good shot.

Answer:  A) It has the same types of atoms on both sides of the reaction

arrow.

Explanation:  A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.

Answer:

Explanation:

v² = u² + 2as

0² = u² + 2(-4.7)(235)

u² = 2209

u = 47 m/s

Hi there!

Recall Newton's Law of Universal Gravitation:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]