The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?

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.

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

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.

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

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]

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.

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.

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╭❤～

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

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

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

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

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"

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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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

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.

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]

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.

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

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 ✌️

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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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.

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

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

Answer:

2.5 km

Explanation:

Answer:

2.5 km

Explanation:

Distance = speed x time

So =5 x 0.5

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

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]

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

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

Answer:

genus yds it's the

Explanation:

xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj

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

Answer:

chk picture for eqn

Explanation: