Calculate the work done to raise a charge of 25 coulombs through an emf of 8 volts.

1) 3
2) 200

Answer:

pitch and loudnesss

thanks for your question

Why words are more important than numbers: ... Words on the other hand are harder to manipulate, they also tell you why someone voted a particular way and to improve your delivery and thus your customer satisfaction you need to understand the why's...

#pglubestiehere

Answer:

The automobile is running at speed of 23.806 meters per second.

Explanation:

From Kinematic we remember that acceleration ([tex]a[/tex]) can be defined by this ordinary differential equation in terms of distance:

[tex]a = v\cdot \frac{dv}{ds}[/tex] (1)

Where:

[tex]v[/tex] - Speed of the automobile, measured in meters per second.

[tex]s[/tex] - Distance travelled by the automobile, measured in meters.

If we know that [tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex], then the equation of acceleration is:

[tex]a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)[/tex]

[tex]a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)[/tex]

[tex]a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)[/tex]

But distance covered by the vehicle is defined by the following formula:

[tex]s = \theta \cdot r[/tex] (2)

Where:

[tex]\theta[/tex] - Arc angle, measured in radians.

[tex]r[/tex] - Radius, measured in radians.

Then, we expand (1) by means of this result:

[tex]a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)[/tex]

[tex]a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)[/tex]

[tex]a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r[/tex]

And finally we get the following third order polynomial:

[tex]1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0[/tex] (3)

If we know that [tex]r = 50\,m[/tex], [tex]a = 0.6\cdot g[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the polynomial becomes into this:

[tex]1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0[/tex] (3b)

This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:

[tex]\theta_{1} \approx 4.365\,rad[/tex], [tex]\theta_{2} \approx 0.818+i\,0.441\,rad[/tex], [tex]\theta_{3}\approx 0.818 -i\,0.441\,rad[/tex], [tex]\theta_{4} \approx 1.563\,rad[/tex]

Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid first. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.

The distance travelled by the automobile is: ([tex]r = 50\,m[/tex], [tex]\theta \approx 1.563\,rad[/tex])

[tex]s = (1.563\,rad)\cdot (50\,m)[/tex]

[tex]s = 78.15\,m[/tex]

Lastly, the speed of the automobile at this location is: ([tex]s = 78.15\,m[/tex])

[tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex] (4)

[tex]v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}[/tex]

[tex]v = 23.806\,\frac{m}{s}[/tex]

The automobile is running at speed of 23.806 meters per second.

Answer:

the speed of electron is 5.021 x 10⁶ m/s

the speed of proton is 2733.91 m/s

Explanation:

Given;

magnitude of electric field, E = 554 N/C

charge of the particles, Q = 1.6 x 10⁻¹⁹ C

time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s

The force experienced by each particle is calculated as;

F = EQ

F = (554)(1.6 x 10⁻¹⁹)

F = 8.864 x 10⁻¹⁷ N

The speed of the particles are calculated as;

[tex]F = ma\\\\F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{Ft}{m_e}\\\\v_e = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{9.11 \times \ 10^{-31}}\\\\v_e = 5.021 \times 10^{6} \ m/s[/tex]

[tex]v_p = \frac{Ft}{m_p}\\\\v_p = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{1.673 \times \ 10^{-27}}\\\\v_p = 2733.91 \ m/s[/tex]

Answer:

the diameter of the bright circular spot formed is 0.787 m  

Explanation:

Given that;

Radius of the flat circular mirror = 0.100 m

height of small ight source = 0.920 m

high ceiling = 2.70 m  

now;

Diameter(mirror) = 2×r = 2 × 0.100 = 0.2 m

D(spot) = [Diameter(mirror) × ( 2.70m + 0.920 m)] /  0.920 m

so

D(spot) = 0.2m × 3.62m /  0.920 m

D(spot) = 0.724 m / 0.920 m

D(spot) = 0.787 m  

Therefore, the diameter of the bright circular spot formed is 0.787 m  

Answer:

LAW 1 :  For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.  

---------------------------------------------

LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

-----------------------------------------------

LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

-----------------------------------------------

LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

Explanation:

Answer:

LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.

Answer:  

a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface; [tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0

b) For constant temperature; [tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])

c) The heat transfer in the conducting rod and the cladding material is the same, i.e; [tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] =  [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]

d) The convection surface conduction by cooling fluid will be;

[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )

Explanation:  

Given the data in question;

we write the general form of the heat conduction equation equation in cylindrical coordinates with internal heat generation.

1/r[tex]\frac{d}{dr}[/tex]( kr[tex]\frac{dT}{dr}[/tex] ) + 1/r² [tex]\frac{d}{d\beta }[/tex](  ( k[tex]\frac{dT}{dr}[/tex] ) + [tex]\frac{d}{dz}[/tex]( k[tex]\frac{dT}{dr}[/tex]) + q = 0

where radius of cylinder is r, thermal conductivity of the cylinder is k, and q is heat generated in cylinder.

Now, Assume one dimensional heat conduction

lets substitute the condition for conducting rod with steady state condition.

[tex]k_{y}[/tex]/r [tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{y} }{dr}[/tex] ) + q = 0

Apply the conditions for cladding by substituting 0 for q

[tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{r} }{dr}[/tex] ) = 0

Apply the following boundary conditions;  

a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface;

[tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0

b) For constant temperature

[tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])

c) The heat transfer in the conducting rod and the cladding material is the same, i.e

[tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] =  [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]  

d) The convection surface conduction by cooling fluid will be;

[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )

Answer:

exerts force

Explanation:

The accumulation of excess electric charge on an object is called static electricity. ... An electric field surrounds every electric charge and exerts the force that causes other electric charges to attract or repel. Electric fields are represented by arrows showing the electric field would make a positive charge move.

Answer:

a) k= 3232.30 N / m,  b)  f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

          w = √k/m

the angular velocity is related to the period

          w = 2π / T

we substitute

          T = 2[tex]\pi[/tex]  √m/ k

a) empty car

           k = 4π² m / T²

           k = 4 π² 1310/2 2

           k = 12929.18 N / m

This is the equivalent constant of the short springs

           F1 + F2 + F3 + F4 = k_eq x

           k x + kx + kx + kx = k_eq x

           k_eq = 4 k

           k = k_eq / 4

           k = 12 929.18 / 4

            k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

            m_total = 1360 + 4 70

            m_total = 1640 kg

angular velocity and frequency are related

              w = 2pi f

we substitute

             2 pi f = Ra K / m

in this case the spring constant changes us

             k_eq = 12929.18 N / m

             f = 1 / 2π √ 12929.18 / 1640

             f = π / 2 2.80778

             f = 4,410 Hz

Answer:

2.16 inch

Explanation:

area under water = 66 km²

= 66 x ( 3280.84 x 12 )² inch²

= 1.023 x 10¹¹ sq inch

volume of rain = 9.57 x 10⁸  gallon = 9.57 x 10⁸ x 231 inch³

= 2.21 x 10¹¹ inch³

If depth of rainfall be t

volume of rain = surface area x depth

= 1.023 x 10¹¹ x t

So ,

1.023 x 10¹¹ x t  = 2.21 x 10¹¹

t = 2.16 inch

Answer:

I think the answers March 21

Answer:

Once a year, mean solar time and sidereal time will be the same.

Answer:

i dont know i am right but here Luminous intensity is defined as dI=dΨλ / dΩ, where dΨλ is the luminous flux (light energy flux in watts per m2) emitted within a solid angle dΩ. The light energy flux may be expressed in terms of the incident x-ray energy flux and the x-ray absorption and conversion properties of the scintillator(7,8,9).

Explanation:

Answer:

hello your question has a missing information

The other is Δ-connected with an impedance of (60 - j45) ohm per phase.

answer : A) 5A ∠0° ,

               p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

 B) 193.64 v

C) current at load 1 = 2.236 A , current at load 2 = 4.472 A

 D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )

      Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )

Explanation:

First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution

A) determine the current, real power and reactive power delivered by the sending-end source

current power delivered (Is)  =  5A ∠0°

complex power delivered ( s ) = 3vs Is  

                                                  = 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )

also s = p + jQ  ------ ( 2 )

comparing equation 1 and 2

p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

B) determine Line-to-line voltage at the load

Vload = √3 * 111.8

           = 193.64 v

c) Determine current per phase in each load

[tex]I_{l1} = Vl1 / Zl1[/tex]

     = [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A   hence current at load 1 = 2.236 A

[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]  

     = [tex]\frac{111.8<-10.3}{25<-36.87}[/tex]  = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A

D) Determine the Total three-phase real and reactive powers absorbed by each load

For load 1

3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR

for load 2

3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex]  = 1200 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR

The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.

(a) The impedance per phase of the equivalent Y,

[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]

The phase voltage,

[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]

Total impedance from the input terminals,

[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]

The three-phase complex power supplied  [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]  

P =1800 W and Q = 0 VAR delivered by the sending-end source.

(b) Phase voltage at load terminals will be,  

[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]  

The line voltage magnitude at the load terminal,  

[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]    

(c) The current per phase in the Y-connected load,  

[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} ​}[/tex]

The phase current magnitude,  

[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]

(d) The three-phase complex power absorbed by each load,

[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]

The three-phase complex power absorbed by the line is  

[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]

Since, the sum of load powers and line losses,  

[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]

To know more about voltage,

https://brainly.com/question/2364325

​

Answer:

the work done by the net external force acting on the skier is 3046.12 J.

Explanation:

Given;

initial speed of the water skier, u = 6.3 m/s

final speed of the water skier, v = 10.9 m/s

mass of the water skier, m = 77 kg

The work done by the net external force is calculated as;

W = ΔK.E

[tex]W = \frac{1}{2} m(v^2 - u^2)\\\\W = \frac{1}{2} \times \ 77.0(10.9^2 - 6.3^2)\\\\ W= 3046.12 \ J[/tex]

Therefore, the work done by the net external force acting on the skier is 3046.12 J.

Answer:

Macroscopic energy is energy at a level of system while microscopic energy is energy at the level of atoms and molecules

Explanation:

1. Macroscopic energy is possessed by a system as whole while microscopic energy is possessed by its constituents’ atoms or molecules.  

2. The common form of macroscopic energy is Kinetic and potential energy while the microscopic form of energy are atomic forces due its random, disordered motion and due to intermolecular forces

3. At microscopic level we consider behaviour of every molecule and in macroscopic approach we consider gross or average effects of various molecular infractions

Answer:

the second time there is a gas between you and the star,

Explanation:

When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.

 When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,

The information obtained from the two spectra is the same, the type of atoms that make up the star

Answer:

If she bends forward

Explanation:

because the equilibrium of gravity will not stay the same causing her to move forward

Answer:

the inertia of the crate is (49.67 kg)r²

Explanation:

Given the data in the question;

First; we will use the law of conservation of momentum to determine the mass of the crate;

m₁v₁ - m₂v₂ = 0

given that; m₁ = 0.60 kg and v₂ = 0.058 m/s

we substitute

0.60 × v₁ = m₂ × 0.058 = 0

m₂ = 0.60v₁ / 0.058 ----------- EQU 1

Next, we use the energy conservation relation to find the velocity

According to conservation of energy;

1/2m₁v₁² + 1/2m₂v₂² = 7 J

we substitute

1/2×0.60×v₁² + 1/2×m₂×(0.058)² = 7 J

0.3v₁² + 0.001682m₂ = 7 J ----- EQU 2

substitute value of m₂ form equ 1 into equ 2

0.3v₁² + 0.001682(0.60v₁ / 0.058) = 7 J

0.3v₁² + 0.0174v₁ = 7 J

0.3v₁² + 0.0174v₁ - 7 J = 0

we solve the quadratic equation;

{  x =  [-b±√( b² - 4ac)] / 2a   }

v₁  =  [-0.0174 ±√( 0.0174² - 4×0.3×-7)] / 2×0.3

=  [-0.0174 ±√(8.4003)] / 0.6

= [-0.0174 ± 2.8983 ] / 0.6  

= -4.8595 or 4.8015     but{ v₁ ≠ - }

so v₁ = 4.8015 m/s ≈ 4.802 m/s

next we input value of  v₁ into equation 1

m₂ = (0.60×4.8015) / 0.058

m₂ =  2.8809 / 0.058

m₂ =  49.67 kg

So, the moment of inertia of the crate will be;

I₂ = m₂r²

we substitute value of m₂

I₂ = (49.67 kg)r²

Therefore, the inertia of the crate is (49.67 kg)r²

Answer:

5.8 m/s

Explanation:

Let v = velocity of bike relative to Betty = -12.0 m/s (since the bike is moving away from betty).

u = velocity of ball relative to bike = + 17.8 m/s

and V = velocity of ball relative to Betty.

So, by Galilean relativity,

V = v + u

V = -12.0 m/s + 17.8 m/s

V = 5.8 m/s

So, the velocity of the ball as measured by Betty is 5.8 m/s

Answer:

Chemical energy

Explanation:

The energy in the torch is stored as chemical energy before the torch gets switch on.

The chemical energy energy in the battery of cell will power the cell and allows it to produce light.

Answer:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9

Answer:

see below :)

Explanation:

Radiometric dating.

Answer:

the correct one is the first,   the refractive index of the two materials must be the same

Explanation:

When a beam of light passes through two materials, it must comply with the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of each medium.

In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,

       θ₁ = θ₂ = θ

substituting

          n₁ = n₂

therefore the refractive index of the two materials must be the same

When reviewing the answers, the correct one is the first

Answer:

Follows are the solution to this question:

Explanation:

Please find the correct question in the attachment file.

Let:

[tex]\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\[/tex]

Calculating the value of  [tex]\overrightarrow{R} \times \overrightarrow{S}:[/tex]

[tex]\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j][/tex]

Calculating the value of [tex]\overrightarrow{R} \cdot (\overrightarrow{R} \times \overrightarrow{S}):[/tex]

[tex]\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])[/tex]

by solving this value it is equal to 0.

Answer:

True.

Explanation:

Welding requires extensive training because welding involves fire and we need to use fire safety measurements. A normal man can't just simply go and weld so a person must require extensive training for welding.

Answer:

The Answer is B)0.2 kg • m/s

Explanation:

I made a 100 on my test. Sorry if I'm late but hope I helped.

Answer:

B. 0.2 kg x m/s

Explanation:

Answer:

[tex]v=0.8m/s[/tex]

Explanation:

From the question we are told that

Distance [tex]d=1.5m/sm/s[/tex]

Time  [tex]t_1=60s[/tex]  

Time  [tex]t_2=90s[/tex]  

Generally the  the equation for the distance traveled is mathematically given as

[tex]d=vt[/tex]

[tex]d=1.5*90[/tex]

[tex]d=138m[/tex]

Generally equation for speed of side walk is mathematically given as

[tex]d=(v+u)t[/tex]

[tex]v=\frac{d}{t}-u[/tex]

[tex]v=\frac{138}{60}-1.5[/tex]

[tex]v=0.8m/s[/tex]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is [tex]A_{x}[/tex] = 0

the y-component of the force at A is [tex]A_{y}[/tex]  = 400 N

the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m

b)

the sum of the momentum about the right end of the beam is;  ∑[tex]M_{R}[/tex]  = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑[tex]M_{L}[/tex] = 0

Vector force acting at A, [tex]F_{A}[/tex] = [tex]A_{x}i[/tex] + [tex]A_{y}j[/tex]

Now, From the image, we have;

a)

∑f = 0

[tex]F_{A}[/tex] - 600j + 200j = 0i + 0j

[tex]A_{x}i[/tex] + [tex]A_{y}j[/tex] - 600j + 200j = 0i + 0j

[tex]A_{x}i[/tex] + ([tex]A_{y}[/tex] - 400)j = 0i + 0j

now by equating i- coefficients'

[tex]A_{x}[/tex] = 0

so, the x-component of the force at A is [tex]A_{x}[/tex] = 0

also by equating j-coefficient

[tex]A_{y}[/tex] - 400 = 0

[tex]A_{y}[/tex]  = 400 N

hence, the y-component of the force at A is [tex]A_{y}[/tex]  = 400 N

we also have;

∑[tex]M_{L}[/tex] = 0

[tex]M_{A}[/tex]  - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

[tex]M_{A}[/tex] - 30 N-m - 228 N-m + 112 Nm = 0

[tex]M_{A}[/tex] - 146 N-m = 0

[tex]M_{A}[/tex] = 146 N-m

Therefore, the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑[tex]M_{R}[/tex] = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)([tex]A_{y}[/tex] ) + [tex]M_{A}[/tex]

∑[tex]M_{R}[/tex] = (108  N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑[tex]M_{R}[/tex]  = 0

Therefore, the sum of the momentum about the right end of the beam is;  ∑[tex]M_{R}[/tex]  = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m