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What is the magnitude of the net force acting on this object? And what direction?

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:

D

Explanation:

Unbalanced forces move stuff. Gravity would only increase/decrease movement if the object was already in motion.

Answer:

b

Explanation:

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.

Learn more about obstacle course : https://brainly.com/question/241926

Answer:

10 seconds

Explanation:

As it starts from rest, then u=0

and by III rd equation of motion:

Answer:

Folding

Explanation:

This question is incomplete

Complete Question

m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.

a) if the system is released from rest what will be its acceleration

Answer:

0.7 m/s²

Explanation:

The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.

(a) if the system is released from rest what will be its acceleration

g = acceleration due to gravity = 9.81 m/s²

Coefficient of Kinetic Friction = μk = 0.30

m1 = 10kg

m2 = 4.0kg

The formula to solve question a is given as:

a = acceleration at rest

m2g- μk m1g = (m1+ m2) a

Making a the subject of the formula:

a = (m2g- μk×m1g )/(m1+ m2)

a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)

a = 0.7 m/s²

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

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]

This question is incomplete, the complete question is;

For this force system the equivalent system at P is ___________

A) FRP = 40 lb (along +x-dir.) and MRP = +60 ft.lb

B) FRP = 0 lb and MRP = +30 ft.lb

C) FRP 30 lb (along +y-dir.) and MRP  = -30 ft.lb

D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb

Answer:

D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb

Explanation:

From the figure in the image i uploaded along this answer;

FRP = ( 40 lb i + 30 lb j ) + [30 lb (-j)]

Where i  and j are the unit vectors along X & Y axis respectively.

So, FRP = 40 lb i

that is,  FRP = 40 lb along +X direction

MRP = [ 30 lb x ( 1 ' + 1' ) ] +( -30 lb x 1 ' )

= (30 lb x 2 ' )- 30 lb ft

= 60 lb ft - 30 lb ft

= 30 lb ft

Therefore option(D) is correct    

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:

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:

ggggggggggggggggggggggggggggg

Explanation:

Answer:

The volume of the sample of gold is

16.51 [tex]cm^{3}[/tex]

Explanation:

The formula for density is:

D= [tex]\frac{M}{V}[/tex].

where:

D is density,

M is mass, and

V is volume.

Rearrange the density formula to isolate volume.

V= [tex]\frac{M}{D}[/tex]

V= [tex]\frac{318.97g Au}{19.32g cm^{3}}[/tex]

V= 318.97∅ ×  [tex]\frac{1 cm^{3} Au}{19.32g cm^{3} }[/tex]← Multiply by the multiplicative inverse of the density.

V= 16.51 cm³ Au.

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

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:

Transform Boundary

Explanation:

The just slide past each other

Answer:

Transform Boundaries

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:

θ = 33.0705°

Explanation:

The angle between the two vectors is given by the formula;

Cos θ = (F1 • F2)/(|F1| × |F2|)

We are given;

F1 = 8.92i + 17.37j

F2 = 12.44i + 7.11j

Thus;

Cos θ = [(8.92i + 17.37j) • (12.44i + 7.11j)]/[√(8.92² + 17.37²) × √(12.44² + 7.11²)]

Cos θ = (110.9648 + 123.5007)/(19.5265 × 14.3285)

Cos θ = 0.8380

θ = cos^(-1) 0.8380

θ = 33.0705°

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:

I would guess its leaves

Answer:

Leaves

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:

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.

I hope it helps you!

Answer:

3

Explanation:

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:

B 5.0 A .

Explanation:

Hello.

In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:

[tex]I=\frac{Q}{t}[/tex]

Then, for the given data, we obtain:

[tex]I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A[/tex]

Therefore, answer is B 5.0 A .

Best regards!

9514 1404 393

Answer:

  Not Necessarily

Explanation:

If the object is changing speed or direction, then it is accelerating. If it is maintaining the same speed and direction, it is not accelerating.

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:

The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).

Answer:

The value  is    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]

Explanation:

From the question we are told that

  The atom fraction of metal A at point G is [tex] A  =  0.30 \ m[/tex]

   The atom fraction of metal  A at a distance 5000nm from G is  [tex]A_2 = 0.35[/tex]

   The number of atoms per [tex]m^3[/tex] is    [tex]N_h =  9 * 10^{28}[/tex]

    The diffusion coefficient is  [tex]D =   2* 10^{-14 } m^2/s[/tex]

Generally of the concentration of atoms of metal A at G is  

       [tex] N_A = A * N_h [/tex]

=>    [tex] N_A =  0.3  * 9 * 10^{28}[/tex]

=>     [tex] N_A =   2.7 * 10^{28} 2.7 atoms/m^3[/tex]

Generally of the concentration of atoms of metal A at a distance 5000nm from G is  

       [tex]D =  0.35 *9 * 10^{28}[/tex]

=>     [tex]D =  3.15 * 10^{28} \  atoms / m^3[/tex]

The concentration gradient is mathematically represented as

   [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]

=> [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]  

=>   [tex]\frac{dN_A}{dx}  = 9 * 10^{20} / m^4[/tex]  

Generally the flux of the atoms per unit  area according to Fick's Law  is mathematically represented as

       [tex]J =  -D* \frac{d N_A}{dx}[/tex]

=>    [tex]J =  -2* 10^{-14 * 9 * 10^{20} [/tex]

=>    [tex] J =  18*10^{6}\   atoms\ crossing\ /m^2 s  [/tex]

Generally if the cross-section area is [tex] a  =  1 cm^2 =  10^{-4} \  m^2[/tex]

Generally the number of atom crossing the above area  per second is mathematically is  

      [tex]H  =  18*10^{6}    *  10^{-4} [/tex]

=>    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]