A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, rough surface a distance of 4.95 m.

(a) What is the work done (in J) by the 150 N force?

_________J

(b) What is the coefficient of kinetic friction between the block and the surface?
________

Answer:

A

Explanation:

AWNSER:

awnser:

C

explanation:

The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule.  The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.

To compute the Molecular Mass of [tex]$O_{3}[/tex]

Atomic mass of oxygen(O) = 16

As [tex]$O_{3}[/tex] contains 3 atoms,

The molecular mass of [tex]$O_{3}[/tex]

= (16 x 3) = 48g/mol

Hence the mass of 9.11 moles O3

= 9.11 mol x 48g/mol

= 437.28g

= 0.43728kg

Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

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Answer:

Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.

Explanation:

The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.

Answer:

they repel with each other. object of like charges repel while object of opposite charges attracts with each other.

Answer:

a)  A' =  0.345  m,  b)  f = 2,800 Hz

Explanation:

b) The angular velocity of a simple harmonic motion is

        w =[tex]\sqrt{\frac{k}{m} }[/tex]

angular velocity and frequency are related

        w = 2π f

we substitute

        f = 1 /2π   √k/m

indicates that the initial frequency value f = 3.96 Hz

in this case the mass is reduced by half

       m ’= m / 2

we substitute

       f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]

       f = √1/2    (2π √k/m)

       f = 1 /√2  3.96

       f = 2,800 Hz

a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.

When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment

initial instant. Right before the division

        p₀ = (m₁ + m₁) v

final instant. Right after the split

        p_f = m₁ v '

        p₀ = p_f

        (2 m₁) v = m₁ v ’

        v ’= 2v

At this point we can use conservation of energy for the system with only half the block.

Starting point. Where the block divides

         Em₀o = K = ½ m v'²

Final point. Point of maximum elongation

          Em_f = Ke = ½ k A²

how energy is conserved

         Em₀ = Em_f

         ½ m’ v’² = ½ k A’²

we substitute the previous expressions

         ½ m/2 (2v)² = ½ k A’²

         A’² = 2  m v² / k                       (1)

Let's use the conservation of energy with the initial conditions, before dividing the block

          ½ m v2 = ½ k A2

          A² = mv² / k = 5.95 10⁻² m²

we substitute in 1

         A'² = 2 A²

          A ’²=  2 5.95 10⁻²

          A ’²= 11.9 10⁻² m

          A' =  0.345  m

Answer:

When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases. Part of the wave travels faster for longer causing the wave to turn. The wave is slower but the wavelength is shorter meaning frequency remains the same.

Explanation:

Answer:

-3.77 m/s

3.85 m/s

Explanation:

given that

Frequency at stationary = 180 Hz

Beat frequency = 2 Hz

Using Doppler effect, we know that

f' = f[(v ± v0) / (v ± vs)], where

v = speed of sound, 343 m/s

v0 = speed of the observer, 0

vs = speed of light, ?

f = stationary frequency, 180 Hz

f' = stationary ± beat frequency, 180 ± 2

Applying the formula, we have

f' = f[(v ± v0) / (v ± vs)]

182 = 180 [(343 + 0) / (343 + vs)]

182/180 = 343 / 343 + vs

343 + vs = 343 * 180/182

343 + vs = 339.23

vs = 339.23 - 343

vs = -3.77 m/s

Again, using

f' = f[(v ± v0) / (v ± vs)]

178 = 180 [(343 + 0) / (343 + vs)]

178/180 = 343 / 343 + vs

343 + vs = 343 * 180/178

343+ vs = 346.85

vs = 346.85 - 343

vs = 3.85 m/s

Answer:

FRICTION

Explanation:

Friction is a force, the resistance of motion when one object rubs against another.

Frictional force

Explanation:

Its the opposing force against horizontal motion

Vab= E = 20V

because I = 0 and the voltage drop across the resistances R1 and R2 is also 0.

Second circuit:

Vab = 10V (no voltage drop across R1)

Vcd= E2-E1 = 20V

Series connection of voltage sources. But the sources are connected to the contrary and voltage drop across R1 or R2 is 0 V.

Answer:

changing the direction in which a force is exerted

Complete question is;

A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 3 ft/s. How fast is the weight rising when the worker has walked:

A) 10 feet

B) 30 feet

Answer:

A) 0.728 ft/s

B) 1.8 ft/s

Explanation:

Let the the position of the worker in ft be denoted by s.

Since he begins to walk away at a constant rate of 3 ft/s, then;

ds/dt = 3 ft/s

Now, the rope will form a triangle, with width "s" and the height 40. Since distance from the connection point to the ceiling = 40 ft

Using pythagoras theorem, we can find the length of the rope on this side of the pulley.

Hence, the length of rope on this side of the pulley = √(s² + 40²)

Meanwhile, on the other side the length will be;

(80) - √(s² + 40²)

Also, height of the weight will be;

h = 40 - ((80) - √(s² + 80²))

h = √(s² + 80²) - 40

Differentiating this, we have;

dh/dt = (ds/dt) × (s/√(s² + 40²))

From earlier, we saw that ds/dt = 3 ft/s

Thus;

dh/dt = 3s/√(s² + 40²)

A) when he has walked 10 ft, it means that s = 10. Thus;

dh/dt = (3 × 10)/√(10² + 40²)

dh/dt = 0.728 ft/s

B) when he has walked 30 ft, it means that s = 30. Thus;

dh/dt = (30 × 3)/√(30² + 40²)

dh/dt = 1.8 ft/s

Answer: We can identify the different types of mirrors without touching them by looking at the image it produces. Look into each mirror, the nature of the image produced will tell you the type of mirror it is.

- A plane mirror will produce an image of the same size as your face.

- A concave mirror will produce a magnified image of your face.

- A convex mirror will produce a diminished image of your face.

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Answer:

a) v_f = 0.898 m / s, b)   v₂ = -6.286 m / s

Explanation:

a) For this exercise we use the conservation of momentum, we define a system formed by the astronaut, her equipment and the expelled gases. We must also define a stationary frame of reference, let's place the system on the platform, so the speed of the subject is v = -3 m / s

Initial instant. Before you start to pass gas

        p₀ = (M + Δm) v

M is the mass of the astronaut M  = 80Kg and Δm the masses of the gases

Final moment. When you expel the gases

        p_f = M (v + Δv) + Δm (v-v_e)

where v_e is the gas velocity v_e = 100 m / s

momentum is conserved

        p₀ = p_f

        M v + Δm v = Mv + M Δv + Δm v -Δm ve

          0 = M Δv - Δm v_e

if we make the very small quantities Δv → dv and Δm → dm, furthermore the quantity of output gas is equal to the decrease in the total mass dm = -dM

         M dv = -v_e dM

         ∫ dv = - v_e ∫ dM / M

We solve, between the lower limits v₀ = v with M = M₀   and the upper  limit v = v_f for M = M_f

         v_f - v₀ = - v_e (ln M_f - Ln M₀)

         v_f - v₀ = v_e ln ([tex]\frac{M_o}{M_f}[/tex])

         v_f = v₀ + v_e ln (\frac{M_o}{M_f})

let's calculate

         v_f = -1.3 + 100 ln (80 + 10 + 2/80 + 10)

          v_f = -1.3 +2.20

          v_f = 0.898 m / s

b) launch the jetpack to increase its speed up to the speed of the platform

  initial instant. Before launching the tanks

        p₀ = (M + m') v_f

final instnte. After launching the tanks

       p_f = M v₁ + m' v₂

indicate that the final velocity of the astronaut is the platform velocity v₁=0 m / s, since the reference system is fixed on it

       p₀ = p_f

       (M+ m) v_f = M v₁ + m v₂2

       v₂ = [tex]\frac{ M ( v_f - v_o) + m' v_f}{m'}[/tex]

        v₂ = [tex]\frac{M}{m}[/tex] (v_f -v₁) + v_f

let's calculate

        v₂ = 80/10 (0.898 - 0) + 0.898

        v₂ = -7.1874 + 0.898

        v₂ = -6.286 m / s

Answer:

      [tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]K

Explanation:

The Heisenberg uncertainty principle is

          Δx Δp ≥ h' / 2

          h’ =[tex]\frac{h}{2\pi }[/tex]

The kinetic energy of a particle is

          K = ½ m v²

           p = mv

           v = [tex]\frac{p}{m}[/tex]

substitute

           K = [tex]\frac{1}{2} \frac{p^2}{m}[/tex]

from the uncertainty principle,

           Δp = [tex]\frac{h'}{2 \ \Delta x}[/tex]

we substitute

          K = [tex]\frac{1}{2m} ( \frac{h'}{2 \ \Delta x})^2[/tex]

          [tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]

Answer:

The second option- a substance that a wave can travel through.

Explanation:

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Answer:

B. It has allowed for faster transmission of Internet signals.

Explanation:

i took the test on engenuity

Fiber-optic technology has allowed for faster transmission of Internet signals.

The term fiber optics, often known as optical fiber, describes the technique used to transport data via light pulses travelling along a glass or plastic fiber.

Here,

In fiber-optic communications, optical fibres are widely used to send light over longer distances and at higher bandwidths (data transfer rates) than electrical cables. They are most frequently used to convey light between the two ends of the fiber.

The concept of complete total internal reflection governs the operation of optical fibres. When a light beam strikes the interior surface of an optical fiber cable with an incidence angle greater than the critical angle, the incident light beam reflects in the same medium, and the occurrence is repeated.

Hence,

Fiber-optic technology has allowed for faster transmission of Internet signals.

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Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

Answer:

F = 0

Explanation:

The magnetic force is described by two expressions

for a moving charge

          F = q v x B

for a wire with a current

         F = I L xB

bold indicates vectors

let's write this equation in module form

         F = I L B sin θ

where the angle is between the direction of the current and the direction of the magnetic field

In this case they indicate that the cable goes from the South wall to the North wall, so this is the direction of the current

The magnetic field of the Earth goes from the south to the north and in this part it is horizontal

Therefore the current and the magnetic field are parallel, the angle between them is zero

           sin 0 = 0

consequently the magnetic force is zero

            F = 0

Explanation:

Regular reflection: It is the reflection from a smooth surface such that the light rays are evenly parallel to each other and an image is formed. ... Irregular reflection: It is the diffused reflection from uneven surface such that the light rays are not parallel to each other and do not form an image.

I think you need Graph to figure it out

Using Newton's second law and kinematic projectile motion we can find the proton deflection y = 5.43 10⁻⁷ m, in the opposite direction to the electron deflection.

given parameters

to find

For this exercise we will use Newton's second law where the force is electric

            F = ma

            F = q E

where F is the force, q the charge, E the electric field, m the mass and the acceleration of the particle

           q E = m a

           a = q / m E

This acceleration is the direction of the electric field that is perpendicular to the initial velocity (v_i)

Having the acceleration we can use the kinematics relations

If we make the direction of the initial velocity coincide with the x-axis

             v_i = cte

             v_i = x / t

             t = x/ v_i

on the y-axis is in the direction of the electric field

            y = v_{iy}  t + ½ a t²

on this axis the initial velocity is zero

            y = [tex]\frac{1}{2} (\frac{q}{m} E) \ t^2[/tex]

subtitute

            y =            (1)

Electron motion.

Let us propose the expression for the electron situation, the length of the displacement must be the same for electron and proton, suppose that it is x = L

In this case the charge q = -e and the mass m = m_e

its substitute in  equation 1

            y₁ = [tex]\frac{1}{2} \ ( \frac{-e}{m_e} E) \ \frac{x^2}{v_i^2}[/tex]  

where y₁, is the lectron deflection.

Proton motion

Between the proton and the electron we have some relationships

          q_p = -e

          m_ = 1840 m_e

we substitute in the equation  1

         y₂ = ½ e / 1840 me E x² / vi²

         y₂ =

         y₂ = - y₁ / 1840

         y₂ = - 0.001 / 1840

         y₂ = - 5.43 10⁻⁷ m

The negative sign indicates that the deflection of the proton is in the opposite direction to the deflection of the electron.

In conclusion they use Newton's second law and kinematics we can find the proton deflection is y = 5.43 10⁻⁷ m

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Answer:

I think it is transpiration

Answer:

transpiration is the right answer

Answer:

The power generated by the windmill is approximately 1.364 MW

Explanation:

The diameter of the windmill, d = 30 m

The inlet speed of the wind, [tex]V_e[/tex] = 40 mph = 17.88 m/s

The exit stream velocity, [tex]V_i[/tex] = 20 mph = 8.94 m/s

The pressure at the inlet, P₁ = 2 atm

The pressure at the outlet, P₂ = 1 atm

The density of air, ρ = 1.2 kg/m³

The power obtained from the windmill, 'P', is given as follows;

[tex]P =\dfrac{1}{4 \cdot g_c} \cdot \rho \cdot A \cdot (V_i + V_e)\cdot (V_i^2 - V_e^2)[/tex]

Where;

[tex]g_c[/tex] = 1.0 kg/(N·s²)

A = Cross-sectional rea of the the windmill =  π·D²/4 = π×(30 m)²/4 = 706.858347 m²

Plugging in the values, we get;

[tex]P =\dfrac{1}{4 \times 1.0} \times1.2 \times 706.858347 \times (17.88 + 8.94)\cdot (17.88^2 - 8.94^2) = 1363668.19438[/tex]

The power generated by the windmill, P ≈ 1363668.19438 W ≈ 1.364 MW.

Answer:

in oil film        λ = 303.57 10⁻⁹ m

in the water film    λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

             v = λ f

in the void we have

             c = λ₀ f

we divide the two expression

            c / v = λ₀ / λ

the refractive index is

              n = c / v

              n = λ₀ /λ

              λ = λ₀ / n

let's calculate

in oil film

            λ = 425 10⁻⁹ / 1.40

            λ = 303.57 10⁻⁹ m

in the water film

            λ = 425 10⁻⁹ / 1.33

            λ = 319.55 10⁻⁹

those wavelengths are in the ultraviolet

Answer:

8. organelle

Explanation:

9. Epithelial tissue

am i correct?

Answer:

25920 ha-cm

Explanation:

Since water evaporates from the reservoir at a rate of 3 cm in 24 hours, its height changes at a rate of 3 cm/24 × 3600 s = 3 cm/86400s = 3.472 10⁻⁵ cm/s.

Now, the volume loss is dV/dt = dV/dh × -dh/dt

= dV/dt × -3.472 × 10⁻⁵ cm/s

= -3.472 × 10⁻⁵ cm/sdV/dh

The reservoir increases in volume at a rate of 3 m³/s = 3 × 10⁶ cm³/s in 24 hours.

So, the net rate of volume change per unit time of the reservoir is

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt where A = area of vertical walled reservoir and dh/dt = change in height of the reservoir with respect to time

So, 3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt

Since dh/dt = 0 in 24 hours(since the water level remains the same after 24 hours, that is dh = 0)

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = A × 0

3 × 10⁶ cm³/s  - 3.472 × 10⁻⁵ cm/sdV/dh = 0

3.472 × 10⁻⁵ cm/sdV/dh = 3 × 10⁶ cm³/s

dV/dh = 3 × 10⁶ cm³/s ÷ 3.472 × 10⁻⁵ cm/s

dV/dh = 8.64 × 10¹¹ cm²

dV = (8.64 × 10¹¹ cm²)dh

Integrating both sides with V from 0 to V and h from h = 0 to h = 3 cm, we have

∫dV = ∫(8.64 × 10¹¹ cm²)dh

∫dV = (8.64 × 10¹¹ cm²)∫dh

V = (8.64 × 10¹¹ cm²)[h]₀³

V = (8.64 × 10¹¹ cm²)[3 cm - 0 cm]

V = (8.64 × 10¹¹ cm²)(3 cm)

V = 25.92 × 10¹¹ cm³

V = 2.592 × 10¹² cm³

V = 2.592 × 10¹² cm² × 1 cm

Since 1 ha = 10⁸ cm²,

V = 2.592 × 10¹² cm² × 1 ha/10⁸ cm² × 1 cm

V = 2.592 × 10⁴ ha-cm

V = 25920 ha-cm