HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?

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:

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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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:

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:

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:

0.1 J/m³

Explanation:

We know that

V = k Q / R

We also know that

E = k Q / R²

Joining the two equations together, we have

E = V / R

To solve the question proper, we'd be using the formula

u = 1/2 E• E², substitute for E, we have

u = 1/2 E• (V/R)²

u = 1/2 * 8.85*10^-12 * (9000 / 0.06)²

u = 1/2 * 8.85*10^-12 * 150000²

u = 1/2 * 8.85*10^-12 * 2.25*10^10

u = 1/2 * 0.199125

u = 0.0996

u = 0.1 J/m³

The energy density is 0.1 J/m³

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

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:

I would guess its leaves

Answer:

Leaves

Explanation:

Answer:

Explanation:

Given parameters:

Initial velocity = 0

Acceleration = 11m/s²

Time  = 6s

Unknown:

Final velocity  = ?

Solution:

 From the given parameters, we use one of the appropriate equations of motion to solve this problem.

     V = U + at

V is the final velocity

U is the initial velocity

a is the acceleration due to gravity

t is the time taken

Input the parameters and solve;

     V  = 0 + 11 x6

     V  = 66m/s

The final velocity is 66m/s

1. Ohm's law shows the relationship between:

Formula: voltage = current x resistance

2. The negative charge on the hairbrush will induce a positive charge on your hair. As a result, your hair is going to be attracted to the hairbrush (and repelled by other strands of hair.)

3. V = IR, so if the resistance of the current increases, and the voltage of the current stays the same, there is as a result, going to be less current.

Best of Regards!

Answer:

1. p = 562.3 kg*m/s

2. F = 142.7 N

Explanation:

1. The linear impulse (p) is given by:

[tex] p = mv [/tex]

Where:

m: is the passenger's mass = 73.7 kg

v: is the speed = 7.63 m/s

[tex] p = mv = 73.7 kg*7.63 m/s = 562.3 kg*m/s [/tex]

Hence, the magnitude of the linear impulse experienced by a passenger is 562.3 kg*m/s.

2. The average force can be calculated using the following equation:

[tex] F = \frac{m(v_{f} - v_{0})}{t} = \frac{73.7 kg(7.63 m/s - 0)}{3.94 s} = 142.7 N [/tex]  

Therefore, the average force experienced by the passenger is 142.7 N.

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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:

3

Explanation:

Answer:

I do not understand what you are asking

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:

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]

Answer:

I think it will get colder

Explanation:

Answer:

The water molecules go faster as it gets colder they go slower

Explanation:

trust me thats the answer

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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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 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

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:

(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

Explanation:

Vectors

Given a vector

[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]

We must determine the following:

a) A vector in the same direction as A with double magnitude 2A.

If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:

[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

b) A unit vector in the same direction of A.

The unit vector needs to compute the magnitude of the vector:

[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]

[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]

[tex]\mid A\mid=7[/tex]

The unit vector is:

[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]

[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]

[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.

The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:

[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]

[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

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:

This question is incomplete, the complete question is;

The displacement of a string carrying a traveling sinusoidal wave is given by y(x,t)=ymsin(kx - ωt -φ) .

At time t = 0, the point at x = 0 has velocity v₀ and displacement y₀.

The phase constant φ is given by tanφ =:

A) ωv₀ /y₀    

B) ωv₀ y₀  

C) v₀ /ωy₀  

D) y₀ /ωv₀      

E) ωy₀ /v₀

Answer:

E) ωy₀ /v₀

Explanation:

Given that;

displacement of a wave is; y(x,t) = ym sin (kx - ωt - φ)

we differentiate the given equation with respect to time

d/dt (y(x,t)) = d/dt(ym sin(kx - ωt - φ) )

v(0,0)) = -ym ωcos (k(0) - ω(0) - φ) )

v₀ = -ym ωcos (-φ)  ......... lets leave thisas equ 1

At t = 0, x = 0

the displacement of the wave is

y(0,0) = ym sin (k(0) - ω(0) - φ)

y₀ = ym sin(-φ) ..............let this be equ 2

y₀/v₀ = (ym sin(-φ)) / (-ym ωcos (-φ)) = ( -ym sin(φ)) / (-ym ωcos (φ))

(tanφ)/ω = y₀/v₀

tanφ = y₀ω/v₀

therefore the required value is y₀ω/v₀

option (E).  

Answer:

Transform Boundary

Explanation:

The just slide past each other

Answer:

Transform Boundaries

Explanation:

Answer:

10 seconds

Explanation:

As it starts from rest, then u=0

and by III rd equation of motion: