In which state of matter are molecules fastest?

Answer:

Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].

Explanation:

Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].

If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].

Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.

Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)

By Newton's second law of motion, the acceleration of this vehicle would be:

[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].

In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:

[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].

Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.

Answer:

In 1924 Louis de Broglie introduced the idea that particles, such as electrons, could be described not only as particles but also as waves. This was substantiated by the way streams of electrons were reflected against crystals and spread through thin metal foils.

Explanation:

I know I probably didn't answer your question, I just used all of my knowledge that I learned about Louie De Broglie. Hope it helps!

Answer:

They take the form of heat, I think thats it but I cant see if you put down any answers

Explanation:

Answer:a

Explanation:

they form hear

L=1m

=2mm²=(2/1000²)=2(10^-6)m

y=4x10¹¹N/m²

∆l=2mm=2/10.00=0.002m

=(4x10¹¹x2x10^-6x2x10^-3)÷1m

=16x10¹¹-⁶-³

=16x10¹¹-⁹

=16x10²

=1600N

where a=cross sectional area=2x10^-6m

C=2mm= 2x10^-3m

L=1m

hope it helps

Answer:

D1 FG 12 15×AG+5T×G7+3F

Answer:

Explanation:

F = ma

a = F/m

a = mBg / (mB + mA)

a = 3.5(9.8)/(3.5 + 3.4)

a = 4.971014...

a = 5.0 m/s²

If you want to use individual Free Body Diagrams

mass A will have downward weight and upward normal forces equal at mAg

and a horizontal force of string tension T

F = ma

T = mAa

mass B will have a downward force of mBg and an upward force of T

mBg - T = mBa

substitute for T

mBg - mAa = mBa

mBg = a(mB + mA)

a = mBg / (mB + mA)   which is identical to the above answer.

Answer:

d. The truck that left later was travelling faster

Explanation:

Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;

They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;

The only variable that can account for this difference is speed;

The one that left later, therefore, must have been going faster.

Answer:

D = 25g/cm³

Explanation:

1ml = 1cm³

D = m/V

D = 500g/20cm³

D = 25g/cm³

Answer:

it says pdf only i dont knowwhat u want me to do

Answer:

Explanation:

A CD has an OD of 120 mm and an ID of 15 mm and has a mass between 14 and 33 grams. Let's call it m

Lets call the outer and inner radii R and r respectively

Find the moment of inertia about a line perpendicular to the surface of the disc through its center. We can integrate or look up the result from standard tables

I = ½m(R² + r²)

then use the parallel axis theorem to shift the position of the axis

I = ½m(R² + r²) + md²

where d is the distance of the shift. In this case d = R

I = ½m(R² + r²) + mR²

I = m(1.5R² + 0.5r²)

If we select a mass of say 20 grams

I = 0.020(1.5(0.060²) + 0.5(0.0075²))

I = 0.0001085625 kg•m²

I think static is the correct answer

Answer:

Explanation:

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

A 100 N create is being pulled at a constant velocity by a rope a 30 degrees to the horizontal as depicted in the diagram given in question the force of friction Ff = 100μcos30 / (cos30 + ½μ).

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

the force of friction Ff, is 100μcos30 / (cos30 + ½μ).

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This temperature would be 170° F on the Fahrenheit scale. Hence, option (C) is correct.

The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points and thermometric substances.

The most popular scales are the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the Fahrenheit scale (°F), and the Kelvin scale (K), with the latter being mostly used for scientific purposes.

the relation between Kelvin scale and Fahrenheit scale is given by:

(F - 32)/180 = (K - 273)/100

F - 32 = (350 - 273)(9/5)

F = 32 + (350 - 273)(9/5)

F = 170

Hence,  this temperature would be 170° F on the Fahrenheit scale.

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Answer:

626m

Explanation:

Answer:

The distance traveled before takeoff is 1720 m

Explanation:

a = + 3.2 m/s²

t = 32.8 s

Vᵢ = 0 m/s

d = ?

Now,

d = Vᵢ × t + 0.5 × a × t²

d = (0 m/s) × (32.8 s) + 0.5 × (3.20 m/s²) × (32.8 s)²

d = 1720 m

Thus, The distance traveled before takeoff is 1720 m

-TheUnknownScientist 72

Answer:

Explanation:

s = s₀ + v₀t + ½at²

if the throw point is origin and UP the positive direction and ignoring air resistance.

s = 0 + (-5)(7) + ½(-10)(7²)

s = 0 - 35 - 245

s = - 280 m

Answer:

$13.31

Explanation:

We know that the bill comes to $93.17 and that 7 people will split the bill equally

We can just use the equation

bill = $93.17/7

bill = $13.31

Answer:

Explanation:

a) The spring force will equal the weight.

b) If up is positive

kx - mg = 0

mg = kx              kx = 25 N

c) m = kx/g = 25/10 = 2.5 kg

do your work in class you would know  kids

Answer:

36 C= 96.8 F

373 K= 99.85

Explanation:

C to F: (36 x 1.8) + 32

         = 64.8 +32

         = 96.8 F

K to C: C= K- 273.15

           C= 373-273.15

           C= 99.85

____

= 36°C

=( 36 × 9/5 ) + 32

=(36 ÷ 5 × 9) + 32

=(7,2 × 9) + 32

= 64,8 + 32

______

______

= 373 K

= 373 - 273

[tex] \boxed { \sf semoga \: membantu \: :v }[/tex]

Answer:

Explanation:

v = √(25² + 8.5²) = 26.40549... = 26 m/s

θ = arctan(8.5/25) = 18.77803... = 19° S of E

Hi there!

We can begin by calculating the acceleration of the block and the wheel using the following equation:

d = vit + 1/2at², where initial velocity = 0 m/s

d = 1/2at²

2d/t² = a

2(1.5)/2² = 0.75 m/s²

Now, we can do a summation of torques:

∑τ =  rT

Rewrite using Newton's 2nd Law for rotation:

Iα = rT

Convert α to a using the relationship α = a/r:

I(a/r) = rT

Ia = r²T

I = r²T/a

Plug in the values:

I = (0.40²)(20)/(0.75) = 4.267 kgm²

Hi there!

We can begin by finding the total time taken for the ball to reach the net using the equation:

dₓ = vₓt

12.9 = 44.7t

12.9/44.7 = t = 0.289 s

Now, we can use the following equation to solve for displacement in the Y direction:

d = y₀ + vit + 1/2at²

There is no initial vertical velocity, so:

d = y₀ + 1/2at²

Plug in known values:

d = 1.28 + 1/2(-9.8)(0.289²)

d = 0.87m

Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.

Answer:

L=mvr

α=0

v=ωr

L=mωr

2

2

1

mv

2

=k

L=Iω

k=

2

1

Iω

2

ω

1

=2ω

k

1

=1/2k

∴I

1

=

8

1

I

L

1

=

8

1

I×2ω

=

4

1

Iω

=L/4

Physical motion that occurs when an item rotates or spins on an axis is known as “rotatory motion,” sometimes known as “rotational motion” or “circular motion.”

Angular displacement per unit of time is measured by angular frequency, also referred to as radial or circular frequency (). Therefore, it has degrees (or radians) per second as its units. 1 Hz = 6.28 rad/sec, thus. 1 radian equals 57.3° because 2 radians equal 360°.

The frequency of rotation is the rate at which an item rotates around an axis, sometimes referred to as rotational speed or rate of rotation.

However, the linear velocity also depends on how far the particles are from the axis of rotation. Because every particle distances are different, so too will the linear velocities for each particle.

Therefore, All the particles in a rigid body's rotational motion move in the same direction in a given interval. Therefore, all the particles' angular velocities will be the same.

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Answer:

heating prosedure takes place the opposite of condensation

Answer:

Explanation:

I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero

If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.

Call the traction force at the rear wheels F

The normal force on the front wheels will be zero, so no moment generated by the front wheels.

Summing moments about our chosen point to zero

1600(9.8)[2.6 / 2] - F[0.7] = 0

F = 291,200

this force will create an acceleration of

a = F/m

a = 291200/1600

a = 182 m/s²

which is about 18.6 times gravity acceleration

Answer:

Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself

Answer:

20 m/s at -35°

Explanation:

Ignoring air resistance, the initial vertical velocity will be reversed and the initial horizontal velocity will remain constant.

Answer:

10.83 Amperes

Explanation:

if   A ⇒ current

W = VA

1300 = 120 x A

1300 / 120 = A

10.83 = A

The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2

Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.

So, E = E'

1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'

where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.

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

1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'

1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2

mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2

mR²ω²/4 + 1/2mv² = mgR/2

R²ω²/4 = gR/2 + 1/2v²

R²ω²/4 = (gR + v²)/2

ω² = 2(gR + v²)/R²

ω² = √[2(gR + v²)/R²]

ω = √[2(gR + v²)]/R

Since angular momentum L = Iω, the rolling disk's initial angular momentum is

L = 1/2mR² ×√[2(gR + v²)]/R

L = mR√[2(gR + v²)]/2

the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2

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