A 3.10-mm-long, 430 kgkg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 69.0 kgkg construction worker stands at the far end of the beam.What is the magnitude of the gravitational torque about the point where the beam is bolted into place?

The question is incomplete as it does not have the options which have been provided in the attachment.

Answer:

Option-D

Explanation:

In the given question, the effect of the sunlight on the growth of the plant has been studied. The values provided in the Option-D can be considered correct as the values are measured in the decimal value up to two decimal value.

The values are measured after the first week, second week, and the initial readings. The difference in the values provided in Option-D does not show much difference as well as are up to two decimal places.

Thus, Option-D is the correct answer.  

Answer:

Replacement

Explanation:

in replacements, like ions replace like. in this equation, we can see that Bromine replaced Chlorine. so, the answer is replacement.

Answer:

Single-replacement or replacement

Explanation:

The single-replacement reaction is a + bc -> ac + b, compare them.

NaBr + Cl2 -> 2 NACl + Br.

AB + C -> AC + B

As you can see they are the same ( even though the b is with the a and not with the c. The formula can be switched around a little with the order of b and c ) ((also like ions replace like ions in replacements, which they are in this))

Answer:

The angular velocity is [tex]w_f = 4.503 \ rad/s[/tex]

Explanation:

From the question we are told that

   The mass of the child is  [tex]m_c = 46.2 \ kg[/tex]

    The radius of the merry go round is  [tex]r = 1.9 \ m[/tex]

     The moment of inertia of the merry go round is [tex]I_m = 130.09 \ kg \cdot m^2[/tex]

      The angular velocity of the merry-go round is  [tex]w = 2.4 \ rad/s[/tex]

       The position of the child from the center of the merry-go-round is  [tex]x = 0.779 \ m[/tex]

According to the law of angular momentum conservation

    The initial angular momentum  =  final  angular momentum

So  

       [tex]L_i = L_f[/tex]

=>     [tex]I_i w_i = I_fw_f[/tex]

Now   [tex]I_i[/tex] is the initial moment of inertia of the system which is mathematically represented as

          [tex]I_i = I_m + I_{b_1}[/tex]

Where  [tex]I_{b_i}[/tex] is the initial moment of inertia of the boy which is mathematically evaluated as

      [tex]I_{b_i} = m_c * r[/tex]

substituting values

      [tex]I_{b_i} = 46.2 * 1.9^2[/tex]

      [tex]I_{b_i} = 166.8 \ kg \cdot m^2[/tex]

Thus

   [tex]I_i =130.09 + 166.8[/tex]        

   [tex]I_i = 296.9 \ kg \cdot m^2[/tex]      

Thus  

     [tex]I_i * w_i =L_i= 296.9 * 2.4[/tex]

       [tex]L_i = 712.5 \ kg \cdot m^2/s[/tex]

Now  

     [tex]I_f = I_m + I_{b_f }[/tex]

Where  [tex]I_{b_f}[/tex] is the final  moment of inertia of the boy which is mathematically evaluated as

         [tex]I_{b_f} = m_c * x[/tex]

substituting values

         [tex]I_{b_f} = 46.2 * 0.779^2[/tex]

         [tex]I_{b_f} = 28.03 kg \cdot m^2[/tex]

Thus

      [tex]I_f = 130.09 + 28.03[/tex]

      [tex]I_f = 158.12 \ kg \ m^2[/tex]

Thus

     [tex]L_f = 158.12 * w_f[/tex]

Hence

      [tex]712.5 = 158.12 * w_f[/tex]

       [tex]w_f = 4.503 \ rad/s[/tex]

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = 360 km-m/s

The direction of the change is up /\ .

The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.

Solve the problem ?

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Answer:

Explanation:

Momentum = mass*velocity of a body

For a 4.00 kg ball is moving at 4.00 m/s to the EAST, its momentum = 4*4 = 16kgm/s

For a 6.00 kg ball is moving at 3.00 m/s to the NORTH;

its momentum = 6*3 = 18kgm/s

Total momentum = The resultant of both momentum

Total momentum = √16²+18²

Total momentum = √580

total momentum = 24.1kgm/s

For the direction:

[tex]\theta = tan^{-1} \frac{y}{x}\\\theta = tan^{-1} \frac{18}{16}\\ \theta = tan^{-1} 1.125\\\theta = 48.4^{0}[/tex]

The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST

Answer:

mass 20 times of an amazing and all its motion

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              [tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             [tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             [tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          [tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex]  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         [tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]

                          [tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]

Answer:

Dear user,

Answer to your query is provided below

The angle for which the rope will break θ = 41.8°

Explanation:

Explanation of the same is attached in image

A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. The angle θ for which the rope will break, is 41.81°

The tension is a kind of force which acts on linear objects when subjected to pull.

The maximum tension Tmax =2W

From the work energy principle,

T₂ = 1/2 mv²

Total energy before and after pushing off

0+mglsinθ = 1/2 mv²

v² = 2gflsinθ..............(1)

From the equilibrium of forces, we have

T= ma +mgsinθ = mv²/l +mgsinθ

2mg = mv²/l +mgsinθ

2g = v²/l +gsinθ

Substitute the value of v² ,we get the expression for θ

θ = sin⁻¹(2/3)

θ =41.81°

Hence, the angle θ for which the rope will break, is 41.81°

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Answer:

a) WT = 137.5 J

b) v2 = 2.34 m/s

Explanation:

a) The total work done on the block is given by the following formula:

[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex]          (1)

Fp: force parallel to the displacement of the block = 150N

Ff: friction force

d: distance = 5.0 m

Then, you first calculate the friction force by using the following relation:

[tex]F_f=\mu_k N=\mu_k Mg[/tex]        (2)

μk: coefficient of kinetic friction = 0.25

M: mass of the block = 50kg

g: gravitational constant = 9.8 m/s^2

Next, you replace the equation (2) into the equation (1) and solve for WT:

[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]

The work done over the block is 137.5 J

b) If the block started from rest, you can use the following equation to calculate the final speed of the block:

[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex]     (3)

WT: total work = 137.5 J

v2: final speed = ?

v1: initial speed of the block = 0m/s

You solve the equation (3) for v2:

[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]

The final speed of the block is 2.34 m/s

Answer:

The correct answer is - option b. 90.6

Explanation:

The scalar x-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle

If you shine a light straight down onto that vector, then the length of its shadow on the x-axis is  -

x-component = 112· cosine(36°)

x-component = 112 · (0.8090)

x-component = 90.60

Thus, The correct answer is - option b. 90.6

Answer:

a)  DECREASE , b) Decreases , c)     DECREASE , d)  the wavelength must increase , e) increasses,

Explanation:

Young's double-slit experience is explained for constructive interference by the expression

          d sin θ = m λ

as in this case, the measured angles are very small,

          tan θ = y / L

         tan θ = sin θ / cos θ = sin θ

          sin θ= y L

        d y / L = m Lam

 we can now examine the statements given

a) if the distance to the screen decreases

        y = m λ / d L

if L decreases and decreases.

The answer is DECREASE

b) if the frequency increases

    the wave speed is

         c = λ f

         λ = c / f

we substitute

          y = (m / d l) c / f

in this case if if the frequency is increased the separation decreases

Decreases

c) If the wavelength decreases

separation decreases

   DECREASE

d) if it is desired that the separation does not change while the separation to the Panamanian decreases the wavelength must increase

      y = (m / d) lam / L

e) if the parcionero between the slits (d) decreases the separation increases

   INCREASES

f) t he gap separation decreases and the distance to the screen decreases so well.

Pattern separation remains constant

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

Answer:

If the initial speed is doubled the time is also doubled

Explanation:

You have that a car with velocity v is decelerated by a constant acceleration a in a time t.

You use the following equation to establish the previous situation:

[tex]v'=v-at[/tex]     (1)

v': final speed of the car  = 0m/s

v: initial speed of the car

From the equation (1) you solve for t and obtain:

[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex]     (2)

To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:

[tex]v'=v_1-at'[/tex]     (3)

v' = 0m/s

v1: new initial speed = 2v

t': new time

You solve the equation (3) for t':

[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]

If the initial speed is doubled the time is also doubled

Answer:

A. Physics has changed the course of the world.

Explanation:

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

[tex]F=qvB[/tex]     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

[tex]F=qvB=ma[/tex]       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

[tex]B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T[/tex]

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

Answer:

[tex]53.1\mu C/m^2[/tex]

Explanation:

We are given that

Electric field,E=[tex]3\times 10^6V/m[/tex]

We have to find the value of maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs.

We know that

[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]

Using the formula

[tex]3\times 10^6=\frac{\sigma}{2\times 8.85\times 10^{-12}}[/tex]

[tex]\sigma=3\times 10^6\times 2\times 8.85\times 10^{-12}[/tex]

[tex]\sigma=5.31\times 10^{-5}C/m^2[/tex]

[tex]\sigma=53.1\times 10^{-6}C/m^2=53.1\mu C/m^2[/tex]

[tex]1\mu C=10^{-6} C[/tex]

Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;

* It creates employment

* It keeps the environment clean

* The practice is highly lucrative

* It saves the earth and conserves energy

The weaknesses of the methods of waste management includes;

* The sites are often dangerous

* The process is mostly

* There is a need for global buy-in

* The resultant product had a short life

Answer:

    Q = 735 J

Explanation:

In this exercise we must assume that all the mechanical energy of the system transforms into cemite energy.

Initial energy

        Em₀ = U = m g h

final energy

        [tex]Em_{f}[/tex] = Q

        Em₀ = Em_{f}

        m g h = Q

let's calculate

        Q = 30  9.8  2.5

        Q = 735 J

Answer:

About 6.26m/s

Explanation:

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

Divide both sides by mass:

[tex]gh=\dfrac{1}{2}v^2[/tex]

Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.

[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]

Hope this helps!

The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

Given data:

The height of vertical cliff is, h = 4.0 m.

Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,

Kinetic energy = Gravitational potential energy

[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]

Here,

m is the mass of rock.

v is the speed of rock.

g is the gravitational acceleration.

Solving as,

[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]

Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

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Explanation:

A ball is thrown straight upward and falls back to Earth. It means that it is coming to the initial position. Displacement is given by the difference of final position and initial position. The displacement of the ball will be 0. As a result velocity will be 0.

Acceleration is equal to the rate of change of velocity. So, its acceleration is also equal to 0.

Hence, displacement, velocity and acceleration are zero.

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

Answer:

(a) Yes, it is possible by raising the object to a greater height without acceleration.

Explanation:

The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in  kinetic energy requires a change in velocity.

If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.

If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).

Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised  to a greater height without acceleration.

Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".

Answer:

The ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil.

Explanation:

Oil is more less dense than water. Thus, the molecules that make up the oil are larger than those that that make up water, so they cannot pack as tightly together as the water molecules will do. Hence, they will take up more space per unit area and are we can say they are less dense.

So, we can conclude that the ball filled with water will reach the bottom of the container first this is because oil is less dense than water and so the glass ball filled with oil will be a lot less denser than the one which is filled with water.

Answer:

22.66Nm²/C

Explanation:

Flux through an electric field is expressed as ϕ = EAcosθ

When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.

From the formula above where:

EA = 25N.m^2/C

θ = 25°

ϕ = 25cos 25°

ϕ = 22.66Nm²/C

Answer:

The bottom two lines.

Explanation:

They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.

Answer:

-67 m/s

Explanation:

We are given that

Mass of ball,m=0.4 kg

Initial speed,u=14 m/s

Impulse,I=-32.4 N-s

Time,t=27 ms=[tex]27\times 10^{-3} s[/tex]

We have to find the mass's velocity in the x- direction.

We know that

[tex]Impulse=mv-mu[/tex]

Substitute the values

[tex]-32.4=0.4v-0.4(14)[/tex]

[tex]-32.4+0.4(14)=0.4 v[/tex]

[tex]-26.8=0.4v[/tex]

[tex]v=\frac{-26.8}{0.4}=-67m/s[/tex]

Answer:

c. [tex]-5. 65 \times 10^6 N m^2/C.[/tex]

Explanation:

The calculation of the electric flux through the sides of the box is shown below:-

Negative charge in insulating pitch ball, [tex]q = 50\times 10^{-6}[/tex]

[tex]Permittivity = 8.854 \times 10^{-12} F/m[/tex]

Now, we are placing the values into the formula which is here below:-

[tex]Flux = \frac{Negative\ charge}{Permittivity}[/tex]

[tex]= \frac{50\times 10^{-6}}{8.854 \times 10^{-12}}[/tex]

= [tex]-5. 65 \times 10^6 N m^2/C.[/tex]

Therefore we divided the negative charge by permittivity to reach out the electric flux through the sides of the box.

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]

k = 1.4

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]

Work done is given as;

[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]

inlet velocity is negligible;

[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]

Therefore, the exit velocity is 629.41 m/s

Answer:

D

Explanation:

Answer:

It is D

Explanation: No cap