A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.

Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?

How far does block two travel, d2 in meters, before coming to rest after the collision?

Answer:

A professional basketball game depends on the association presiding over the game. An NBA game lasts for 48 minutes whereas FIBA games take 40 minutes. The total time taken to play for any specialized game is over 2 hours 15 minutes. The time includes the time disruptions like fouls, timeouts, and breaks.

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Complete Question

On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every 2.7 s .

If we assume their arms are each 0.90 m long and their individual masses are 65.0 kg , how hard are they pulling on one another?

Answer:

The force is  [tex]F  = 316.8 \  N[/tex]

Explanation:

From the question we are told that

    The period is  T   =  2.7 s

    The radius of the circle formed by their arms  is  r =  0.90 m

      Their individual  mass is  [tex]m =  65.0 \  kg[/tex]

Generally their angular velocity is mathematically represented as

      [tex]w = \frac{2 \pi}{T}[/tex]

=>    [tex]w = \frac{2 *  3.142 }{2.7}[/tex]

=>  [tex]w =2.327 \ rad/ s [/tex]

Generally the pulling force is mathematically represented as

      [tex]F  = m *  w ^2 *  r[/tex]

=>   [tex]F  = 65 *  2.327^2 *  0.90[/tex]

=>   [tex]F  = 316.8 \  N[/tex]

Answer:

The number of turns in the solenoid is 22366.

Explanation:

The number of turns in the solenoid can be found using the following equation:

[tex] B = \mu_{0} I\frac{N}{L} [/tex]

Where:

B: is the magnetic field = 8.90 T

L: is the solenoid's length = 0.300 m

N: is the number of turns =?

I: is the current = 95 A

μ₀: is the magnetic constant = 4π×10⁻⁷ H/m

By solving equation (1) for N we have:

[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]

Therefore, the number of turns in the solenoid is 22366.

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Answer:

it keeps it growing because if it stays warm and yuh put it by a plant that needs sun yk

Explanation:

A flower of a plant that can stay warm at night as it keeps it growing.

The roots are the underground part of the plant which plays a major role in pulling the water and minerals,  expands within the ground  for better water absorption, anchors which helps in creating better stability.

A stem present above the ground bears leaves, fruits plus flowers.  as it distributes the nutrients and minerals to the leaves, shields the plant and assists in asexual dissemination.

leaf is one of the most major parts of a plant as it contain chlorophyll pigment which assists the plants in preparation for food, the veins allow the flowing of nutrients plus water, it has the parts like petiole, leaf base and lamina help in photosynthesis.

Flower is the most bright and beautiful part of the plant show an important role in making food, used in fertilization process , the basic parts are petals, sepals, stamens, and pistil.

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Answer:

20000

Explanation:

Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000

Answer:

Uniformly accelerated rectilinear motion (MRUA), also known as uniformly varied rectilinear motion (MRUV), is one in which a mobile moves along a straight path being subjected to a constant acceleration.

Explanation:

Answer:

Acceleration = m/s²

Explanation:

T= Newtons compared to the weight W = Newtons for the hanging mass. If the weight of the hanging mass is less than the frictional resistance force acting on the mass on the table, then the acceleration will be zero.

Answer:

The acceleration of the box is 0.67 m/s²

Explanation:

Given that,

Mass of box = 30.0 kg

Horizontal force = 230 N

Friction force = 210 N

We need to calculate the acceleration of the box

Using balance equation

[tex]F-f_{k}=ma[/tex]

[tex]a=\dfrac{F-f_{k}}{m}[/tex]

Where, F = horizontal force

[tex]f_{k}[/tex] =frictional force

m= mass of box

a = acceleration

Put the value into the formula

[tex]a=\dfrac{230-210}{30}[/tex]

[tex]a=0.67\ m/s^2[/tex]

Hence, The acceleration of the box is 0.67 m/s²

Answer:

The appropriate response is "Optical printer ".

Explanation:

Answer:

Dividing the silicon density by 1000 and then multiply it by 1000000.

Explanation:

A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:

[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]

[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]

In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.

Answer:

We are given:

initial velocity (u) = 12 m/s

final velocity (v) = 0 m/s

time taken (t) = 15 seconds

acceleration (a) = a m/s²

Solving for acceleration:

from the first equation of motion

v = u + at

replacing the variables

0 = 12 + (a)(15)

0 = 15a + 12

a = -12 / 15

a = -4 / 5 m/s²

v² - u² = 2 a ∆x

where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆x

∆x = (27 m/s)² / (16 m/s²)

∆x ≈ 45.6 m

The stopping distance of car achieved during the braking is of 45.56 m.

Given data:

The initial speed of car is, u = 27 m/s.

The final speed of car is, v = 0 m/s. (Because car comes to stop finally)

The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].

In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.

Therefore,

[tex]v^{2}=u^{2}+2(-a)s[/tex]

Here, s is the stopping distance.

Solving as,

[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]

Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.

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Answer:

It is si-32

Explanation:

Answer:

silicon-32

Explanation:

just took the quiz and got it right

Answer:

x = 73.71 [m]

Explanation:

In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.

[tex]v_{f }= v_{i}-(a*t)[/tex]

where:

Vf = fnal velocity = 0

Vi = initial velocity = 32.4 [m/s]

t = time = 4,55 [s]

a = acceleration or desacceleration [m/s^2]

0 = 32.4 - (a*4.55)

a = 7.12 [m/s^2]

Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.

Now using the following equation:

[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]

where:

xo = initial distance = 0

x = final distance [m]

Therefore we have:

x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)

x = 73.71 [m]

Answer: The options are not given.

Here are the options.

a) There is an additional force lifting up on you.

(b) At the top you continue going straight and the seat moves out from under you.

(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (

d) Both b and c are correct.

(e) a, b, and c are correct.

The correct option Is D.

B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.

Explanation:

At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.

That is the scenarion that occur...

Answer:

Folding

Explanation:

Answer:

Tolerance: [tex]\pm 10\%[/tex]

Explanation:

Resistor Color Codes

Resistor Color Coding uses colored bands to quickly identify the resistive value or resistors and its percentage of tolerance.

Since the question does not provide a specific color table, we'll use the table attached below.

The colors of the resistor shown in the question are:

First band: orange

Second band: blue

Third band: brown

Fourth band: silver

The colors relate to the following numbers respectively:

3, 6, 10Ω, [tex]\pm 1\%[/tex]

The first two colors form the number 36

The third color is the multiplier: 36*10Ω = 360Ω

And the fourth color is the tolerance or the possible variation of the resistance [tex]\pm 1\%[/tex]

Resistance: 360Ω

Tolerance: [tex]\pm 10\%[/tex]

We are given:

the initial height of the object (h) = 100 m

initial velocity (u) = 0 m/s

we will let the value of g = 10 m/s/s

Speed of the object just before hitting the ground:

From the third equation of motion:

v² - u² = 2ah     (where v is the final velocity)

replacing the variables, we get:

v² - (0)² = 2(10)(100)

v² = 2000

v = 10√20 = 44.7 m/s

Therefore, the speed of the object just before hitting the ground is 44.7 m/s

Answer:

10 seconds

Explanation:

As it starts from rest, then u=0

and by III rd equation of motion:

Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

Answer:

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

Explanation:

For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.

Let's start with the projectile launch

as the body leaves the vertical its velocity must be horizontal

         x = v₀ₓ t

         y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero

         t = √ 2 y₀ / g

we substitute

        x = vox √2y₀ / g

        v₀ₓ = √(g / 2y₀)     x

In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)

       v₀ₓ = √(g /(2 (H -L))    D

this is the minimum speed to cross the well.

Now let's use conservation of energy

starting point. On the platform

      [tex]Em_{o}[/tex] = U = m g H

final point. At the bottom of the swing

      Em_{f} = K + U = 1 / 2m v² + m g (H -L)

as there is no friction the mechanical energy is conserved

        Em_{o} = Em_{f}

       m g H = 1 / 2m v² + m g (H -L)

        v = √ (2gL)

let's write our two equations

the minimum speed to cross the well

       v₀ₓ = √ (g /(2 (H -L))    D

the speed at the bottom of the oscillatory motion

       v = √ (2g L)

we analyze the extreme cases

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well

      V₀ₓ = v

      g / (2 (H -L) D² = 2g L

       4 L (H- L) = D²

        4 H L - 4 L2 - D² = 0

        L² - H L - D² / 4 = 0

let's solve the quadratic equation

      L = [H ± √ (H2-D2)] / 2

we assume that H> D

       L = ½ H [1 + - RA (1 - (D / H) 2)]

The two values ​​of La give the range of values ​​for which the two speeds are equal

A) The person lands in the moat if the rope's length is very short because :

B) The person lands in the moat if the rope length is similar to the height of the platform because :

Following the assumptions;

size of the person is much smaller than L and H

D = horizontal distance

Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed  and  The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.

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Answer:

Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name.

Explanation:

Answer:

An element's period and group

Answer:

At constant velocity, his weight equals the force of friction. In other words, there is no net force. If however, he loosens his grip and decreases the friction force, he will accelerate downward.

Explanation:

This question involves the concepts of mass, acceleration due to gravity, weight, Newton's Gravitational Law, gravitational force, and graphs.

Graph "A" re[resents the best relationship between acceleration due to gravity, and mass for objects.

The relationship between the mass of earth and the acceleration due to gravity can be found by equating the weight of the object and the gravitational force, from Newton's Gravitational Law on it:

[tex]Weigth = Gravitational\ Force\\\\mg = \frac{GmM}{r^2}\\\\g = \frac{GM}{r^2} ---------- eqn(1)[/tex]

where,

g = acceleration due to gravity

G = universal gravitational constant

M = mass of Earth

r = radius of Earth

Hence, it is clear from equation (1), that mass of the Earth and the acceleration due to gravity have a direct relationship with each other. Therefore the graph between them will be a straight line, which is Graph A.

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The attached picture illustrates Newton's Law of Gravitation.

Answer:

the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

Explanation:

Answer:

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Explanation:

The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:

[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]

[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.

[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.

And we expand the equation above by definitions of elastic potential energy and kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the block, measured in kilograms.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.

[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.

If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:

[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]

[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Answer:

a: 19.6 meters b: 9.8meters per second

Explanation:

speed and object falls is 9.8 meters per second. 9.8 meter times 2 is 19.8 meters.

Answer:

I do not understand what you are asking