P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism articulated to the ground at its vertex A, while vertex C is attached to the vertical cord fixed to the ground. If the coefficient of friction between the prism and the blocks is 0.4; determine the maximum angle that measures the inclination of the AC face with respect to the horizontal so that the system remains in equilibrium.

P-weight Blocks D And E Are Connected By The Rope Which Passes Through Pulley B And Are Supported By

Answers

Answer 1

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

P-weight Blocks D And E Are Connected By The Rope Which Passes Through Pulley B And Are Supported By

Related Questions

Light with an intensity of 1 kW/m2 falls normally on a surface with an area of 1 cm2 and is completely absorbed. The force of the radiation on the surface is

Answers

Answer:

The force of the radiation on the surface is 3.33 x 10⁻¹⁰ N

Explanation:

Given;

intensity of light, I = 1kw/m² = 1000 W/m²

area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²

Since the light is completely absorbed, the force of the radiation is given by;

F = P/c

where;

c is the speed of light = 3 x 10⁸ m/s

But P = IA

F = IA /c

F = (1000 X 1 X 10⁻⁴) / 3 x 10⁸

F = 3.33 x 10⁻¹⁰ N

Therefore, the force of the radiation on the surface is 3.33 x 10⁻¹⁰ N

The force of radiation will be "3.33 × 10⁻¹⁰ N"

Intensity and Force

According to the question,

Intensity of force, I = 1 kW/m² or,

                               = 1000 W/m²

Area of surface, A = 1 cm² or,

                              = 1 × 10⁻⁴ m²  

Speed of light, c = 3 × 10³ m/s

As we know the relation,

→ F = [tex]\frac{P}{c}[/tex]

or,

  P = IA

or,

  F = [tex]\frac{IA}{c}[/tex]

By substituting the values, we get

     = [tex]\frac{1000\times 1\times 10^{-4}}{3\times 10^3}[/tex]

     = 3.33 × 10⁻¹⁰ N

Thus the response above is correct.

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What does E=mc2 stand for?

Answers

It stands for energy=mass times the speed of light squared.

That mean mass-energy equivalence

A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is

Answers

Answer:

At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage

Explanation;

Because at series connection the battery and resistor have equal voltage

A person is being pulled by gravity with a force of 500 N. What is the force with which the person pulls Earth?
1,000 N
O100 N
500 N
0 250 N

Answers

Answer:

The correct answer is 500 N

Explanation:

This is an exercise in Newton's third law or law of action and reaction

The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.

Which force we call action does not matter, the analysis and conclusions are the same

The correct answer is 500N

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?

Answers

Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.

              Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m

Explanation: A speed of wave on a string under a tension force can be calculated as:

[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]

[tex]F_{T}[/tex] is tension force (N)

μ is linear density (kg/m)

Determining velocity:

[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{0.00874 }[/tex]

[tex]|v| =[/tex] 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]

Δx = 11.5×[tex]10^{-6}[/tex]

With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex]  m.

Doubling Tension:

[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{2.0.00874 }[/tex]

[tex]|v| = \sqrt{0.01568}[/tex]

|v| = 0.1252 m/s

Displacement for same time:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]

[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]

With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m

Megan accelerates her skateboard from 0 m/s to 8 m/s in 2 seconds. What is the magnitude of the acceleration of the skateboard?
O 8 m/s^2
O 16 m/s^2
O 2 m/s^2
O 4 m/s^2​

Answers

Answer:

chk picture for eqn

Explanation:

A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less than the actual distance, the same as the actual distance, or more than the actual distance? Explain.

Answers

Answer:

p = -q  

he distance is equal to the current distance, so the distance does not change

Explanation:

For this exercise we can solve it using the equation of the constructor

            1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies

          1 / p = - 1 / q

           p = -q

Therefore the distance is equal to the current distance, so the distance does not change

An electron has an initial velocity to the south but is observed to curve upward as the result of a magnetic field. This magnetic field must have a component:___________
a) north
b) upwards
c) downwards
d) east
e) west

Answers

Answer:

e) west

Explanation:

According to Lorentz left hand rule, the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb to the south (towards your body), with the palm facing up, then the fingers will point west.

In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v= E/B. The formula is the same for a negatively charged particles.

a. True
b. False

Answers

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

Calculate the electromotive force produced by each of the battery combinations shown in the figure, if the emf of each is 1.5 V.

Answers

Answer:

A) 1.5 V

B) 4.5 V

Explanation:

A) Batteries in parallel have the same voltage as an individual battery.

V = 1.5 V

B) Batteries in series have a voltage equal to the sum of the individual batteries.

V = 1.5 V + 1.5 V + 1.5 V

V = 4.5 V

A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction between the sled and the incline is .45, What is the tension force between the block and the tree

Answers

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

            28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

In which example is kinetic friction most involved? a sled stuck on a snowy hill a bottle of water wedged in a vending machine an explorer unsuccessfully pushing on a massive stone that is blocking the entrance to a cave a volleyball player sliding across the court while diving for the ball

Answers

Answer:

I believe the answer is A volleyball player sliding across the court while diving for the ball.

Explanation:

Kinetic friction is a body moving on the surface experiences a force in the opposite direction of its movement.

Hope this helps! (づ ̄3 ̄)づ╭❤~

Suppose you want a telescope that would allow you to see distinguishing features as small as 3.5 km on the Moon some 384,000 km away. Assume an average wavelength of 550 nm for the light received.Required:What is the minimum diameter mirror on a telescope?

Answers

Explanation:

[tex]\theta=1.22 \frac{\lambda}{D}[/tex]

And, from equation ( 2 ), we get

[tex]\theta=\frac{x}{d}[/tex]

Thus,

[tex]\frac{x}{d} &=1.22 \frac{\lambda}{D}[/tex]

[tex]D &=1.22 \frac{\lambda d}{x}[/tex]

[tex]=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}[/tex]

[tex]=0.0515 \mathrm{m}[/tex]

Thus, the diameter of the telescope's mirror that would allow us to see details as small as is

The magnitude of the magnetic field at point P for a certain electromagnetic wave is 2.12 μT. What is the magnitude of the electric field for that wave at P? (c = 3.0 × 108 m/s)

Answers

Answer:

The electric field is  [tex]E = 636 \ V/m[/tex]

Explanation:

From the question we are told that

     The magnitude of magnetic field is [tex]B = 2.12 \mu T = 2.12*10^{-6} \ T[/tex]

      The value for speed of light is  [tex]c = 3.0 *10^8 \ m/s[/tex]

Generally the magnitude of the electric field at point P is

        [tex]E = B * c[/tex]

substituting values

         [tex]E = 2.12 *10^{-6} * 3.0 *10^{8}[/tex]

         [tex]E = 636 \ V/m[/tex]

The magnitude of electric field for the wave at point P is 636 V/m.

Given data:

The strength of magnetic field at point P is, [tex]B = 2.12 \;\rm \mu T=2.12 \times 10^{-6} \;\rm T[/tex].

The speed of light is, [tex]c = 3.0 \times 10^{8} \;\rm m/s[/tex].

The given problem is based on the concept of electric field and magnetic field. The electromagnetic wave works on the principle of oscillating magnetic field and electric field at the same region. We can find any of the two using the expression,

[tex]E = B \times c[/tex]

here,

E is the strength of electric field.

Solving as,

[tex]E = (2.12 \times 10^{-6}) \times (3 \times 10^{8})\\\\E = 636 \;\rm V/m[/tex]

Thus, we can conclude that the magnitude of electric field for the wave at point P is 636 V/m.

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What is the mass of a rectangular block of
density 2.5 ×10³ k gm³that measures 10cm by 5 cm by 4 cm?
A. 0.002 kg
B. 0.080 kg
C. 0.200 kg
D. 0.500 kg
E. 1.000 kg​

Answers

Answer:

Option (D) : 0.5 kg

Explanation:

[tex]mass = density \times volume[/tex]

[tex]mass = {2500} \times 0.1 \times 0.05 \times 0.04[/tex]

Mass of block = 0.5 kg

the mass of a rectangular block of density 2.5 ×10³ k gm³ that measures 10cm by 5 cm by 4 cm is 0.5 kg.

What is density ?  

Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water)

The Volume of the block is,

V = LBD, where L = length, B = breadth , D = depth of the block.

V = 10 × 5 × 4 = 200 cm³

Density of Block = 2.5 ×10³ kg/m³

Density = Mass / Volume

2.5 ×10³ kg/m³ =  Mass /  200 cm³

2.5 ×10³ kg/m³ × 200 cm³ =  Mass

2.5 ×10³ kg/m³ × 0.2 × 10⁻³ m³ =  Mass

Mass = 0.5 kg

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#SPJ2.

A mass is tired to spring and begins vibration periodically the distance between it's lowest position is 48cm what is the Amplitude of the vibration

Answers

Answer:

The amplitude of vibration of the spring is "24 cm"

The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.

From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.

Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.

Amplitude = 24 cm

A double-slit experiment is performed with light of wavelength 620 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?

Answers

Answer:

1.34 mm

Explanation:

A double slit experiment is conducted with a light which has a wavelength of 620 nm

The fringes are separated 2.3 mm apart

The light is changed to a wavelength length of 360 nm

Let x represent the fringe spacing as a result of the change in wavelength

Therefore,the fringe spacing can be calculated as follows

2.3mm/x= 620nm/360nm

Multiply both sides

x × 620= 2.3×360

620x= 828

x= 828/620

x= 1.34 mm

3 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

Explanation:

[tex]qV = \frac{1}{2}mv^2[/tex]

Multiply both sides by 2 and then divide by m to get

[tex]\dfrac{2qV}{m} = v^2[/tex]

Take the square root of both sides to get

[tex]v = \sqrt{\dfrac{2qV}{m}}[/tex]

What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×10-34 Js.)

Answers

Answer:

9.82 × [tex]10^{-35}[/tex] Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = [tex]\frac{h}{mv}[/tex]

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = [tex]\frac{h}{mv}[/tex]

  = [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]

 = [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]

 = 9.8222 × [tex]10^{-35}[/tex]

The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.

Which of the units of the following physical quantities are derived

Answers

Answer:

You have not provided options but so long as the option is none of these, it's the right answer:

Electric current, Thermodynamic temperature,mass,time amount of substance,length, luminous intensity

When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are generally not more than a few diameters of the Sun or Moon and invariably the innermost ring is blue. The explanation for these phenomena involves:_________
A) reflection
B) refraction
C) interference
D) diffraction
E) Doppler effect

Answers

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

The same force is applied to two hoops. The hoops have the same mass, but the larger hoop has twice the radius. How are the angular accelerations of the hoops related

Answers

Answer:

The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].

Explanation:

Net force on the hoop is given by;

[tex]F_{net} = ma[/tex]

where;

a is linear acceleration

m is the mass

Net torque on the hoop is given by;

[tex]\tau_{net} =I\alpha[/tex]

where;

I is moment of inertia

α is the angular acceleration

But, τ = Fr

[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]

[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]

let the angular acceleration of the smaller hoop = α₁

let the radius of the smaller hoop = r₁

then, the radius of the larger loop, r₂ = 2r₁

let the angular acceleration of the larger hoop = α₂

[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]

Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]

The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would you walk(km)?

Answers

Answer:

2.5 km

Explanation:

Answer:

2.5 km

Explanation:

Distance = speed x time

So =5 x 0.5

1. Rank the transformers on the basis of their rms secondary voltage. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
2. 100 A of rms current is incident on the primary side of each transformer. Rank the transformers on the basis of their rms secondary current. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns

Answers

Answer:

1. Transformer 3> Transformer 1 and 2 > Transformer 4

2. Transformer 2,3,5 > Transformer 1 > Transformer 4

Explanation:

From;

Vs/Vp = Ns/Np

Where;

Vp= voltage in primary coil

Vs= voltage in secondary coil

Ns= number of turns in secondary coil

Np= number of turns in primary coil.

Vs= Ns×Vp/Np

Vs= 480 ×2000/4000

Vs= 240 V

Vs= 480 ×1000/2000

Vs=240 V

Vs= 120 × 2000/500

Vs= 480 V

Vs= 240 × 500/1000

Vs= 120 V

2. Ns/Np= Ip/Is

Is= Np×Ip/Ns

Is= 1000 × 100/2000

Is= 50 A

Is= 2000 × 100/1000

Is= 200 A

Is= 1000 × 100/500

Is= 200 A

Is= 500 × 100/2000

Is= 25 A

Is= 4000 × 100/2000

Is= 200 A

You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 38.5∘ from the ground with an initial speed of 27.5 m/s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.

Answers

The ball's horizontal position in the air is

[tex]x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t[/tex]

It hits the wall when [tex]x=17.5\,\mathrm m[/tex], which happens at

[tex]17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s[/tex]

Meanwhile, the ball's vertical position is

[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2[/tex]

where [tex]g[/tex] is the acceleration due to gravity, 9.80 m/s^2.

At the time the ball hits the wall, its vertical position (relative to its initial position) is

[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}[/tex]

c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until friction causes it to settle back to the equilibrium position. Ignoring friction, what maximum height will it reach above the surface

Answers

Answer:

y = 20.99 V / A

there is no friction    y = 20.99 h

Explanation:

Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy

when the block of ice is submerged it is subjected to two forces its weight  hydrostatic thrust

         

              F_net= ∑F = B-W

the expression stop pushing is

              B = ρ_water g V_ice

where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice

We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³

             W = ρ-ice g V

              F_net = (ρ_water - ρ_ ice) g V

this is the net force directed upwards, we can find the potential energy with the expression

            F = -dU / dy

            ΔU = - ∫ F dy

            ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy

            ΔU = - (ρ_water - ρ_ ice) g A y² / 2

we evaluate between the limits y = 0,  U = 0, that is, the potential energy is zero at the surface

             U_ice = (ρ_water - ρ_ ice) g A y² / 2

now we can use the conservation of mechanical energy

starting point. Ice depth point

             Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2

final point. Highest point of the block

             [tex]Em_{f}[/tex] = U = m g y

as there is no friction, energy is conserved

            Em₀ = Em_{f}

            (ρ_water - ρ_ ice) g A y² / 2 = mg y

let's write the weight of the block as a function of its density

            ρ_ice = m / V

            m = ρ_ice V

we substitute

             (ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y

              y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A

let's substitute the values

             y = 0.913 / (1 - 0.913) 2 V / A

             y = 20.99 V / A

This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction

                V / A = h

where h is the height of the block

                 y = 20.99 h

A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is in the xy plane. The vertical force required to lift this slab is 60 kN (F 60 kN). Find the tensions in cables DA, DB and DC (show all your workings that you do to find these)

Answers

Answer:

Fad = 28.8 kN

Fbd = 16.4 kN

Fcd = 28.1 kN

Explanation:

First, find the length of each cable.

AD = √((2 m)² + (0.5 m)² + (2.5 m)²)

AD = √10.5 m

AD ≈ 3.24 m

BD = √((1.5 m)² + (1 m)² + (2.5 m)²)

BD = √9.5 m

BD ≈ 3.08 m

CD = √((1 m)² + (1 m)² + (2.5 m)²)

CD = √8.25 m

CD ≈ 2.87 m

Next, use similar triangles to find the x, y, and z components of each tension force.

Fadx = 2/3.24 Fad = 0.617 Fad

Fady = 0.5/3.24 Fad = 0.154 Fad

Fadz = 2.5/3.24 Fad = 0.772 Fad

Fbdx = 1.5/3.08 Fbd = 0.487 Fbd

Fbdy = 1/3.08 Fbd = 0.324 Fbd

Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd

Fcdx = 1/2.87 Fcd = 0.348 Fcd

Fcdy = 1/2.87 Fcd = 0.348 Fcd

Fcdz = 2.5/2.87 Fcd = 0.870 Fcd

Now sum the forces in the x, y, and z directions:

∑Fx = ma

-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0

∑Fy = ma

-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0

∑Fz = ma

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.

-0.463 Fad + 0.811 Fbd = 0

0.811 Fbd = 0.463 Fad

Fbd = 0.571 Fad

Substitute into either of the first two equations:

-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0

-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0

-0.339 Fad + 0.348 Fcd = 0

0.348 Fcd = 0.339 Fad

Fcd = 0.975 Fad

Now substituting into the third equation:

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0

60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0

60 kN − 2.083 Fad = 0

Fad = 28.8 kN

Solving for the other two tension forces:

Fbd = 0.571 Fad = 16.4 kN

Fcd = 0.975 Fad = 28.1 kN

Answer:

Tensions of:

DA = 28.81 KN

DB = 16.45 KN

DC = 28.07 KN

Explanation:

see attached

hi guys!!! i have no more points, can someone nice guess all of these for me? :)
1.What happens to the ocean water before the precipitation part of the water cycle
2.During which stage of the water cycle does water from the ocean form clouds?
3.what is a runoff??
4.Which statement about oceans is incorrect? A.Evaporation occurs when water is warmed by the sun. B.Most evaporation and precipitation occur over the ocean. C.97 percent of Earth's water is fresh water from the ocean. D.Water leaves the ocean by the process of evaporation
5.How does most ocean water return to the ocean in the water cycle

tysm to u who answers :)

Answers

1. The ocean water collects back in the ocean.

2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.

3. an excessive amount of water flowing from downslope along earths surface

4. A.Evaporation occurs when water is warmed by the sun.

5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.

Hope this answer correct ✌️

the molecule of magnet are independent _____________​

Answers

Answer:

The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.

What is the direction of the net gravitational force on the mass at the origin due to the other two masses?

Answers

Answer:

genus yds it's the

Explanation:

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