An average force of magnitude 300 N exerted to compress a spring by 20 cm compute the work done by the force​

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.

Answer:

the results will be the same.it may be

If the same experiment is repeated in different parts of the world by different scientists, the results will be the same.

An experiment is a procedure that is carried out to support or refute a hypothesis, or to determine the efficacy or likelihood of something that has never been tried before. Experiments shed light on cause-and-effect relationships by demonstrating what happens when a specific factor is changed.

Controls are typically included in experiments to minimise the effects of variables other than the single independent variable. This improves the reliability of the results, often by comparing control measurements to the other measurements. Scientific controls are an essential component of the scientific method.

Hence, If the same experiment is repeated in different parts of the world by different scientists, the results will be the same.

Learn more about experiment here:

https://brainly.com/question/11256472

#SPJ2

Answer:

define d first?

you need to list more variables

Answer:

list more valuable unit

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:

86.33m/s^2

Explanation:

Acceleration = Force/Mass

= 25.9/0.3

= 86.33

Answer:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Explanation:

Given

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]

Point: [tex](f(t0), g(t0), h(t0))[/tex]

[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information

Required

Determine the parametric equations

[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]

Differentiate with respect to t

[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]

Let t = 1 (i.e [tex]t0 = 1[/tex])

[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + k[/tex]

To solve for x, y and z, we make use of:

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]

This implies that:

[tex]r'(1)t = xi + yj + zk[/tex]

So, we have:

[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]

[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]

By comparison:

[tex]xi = it[/tex]

Divide by i

[tex]x = t[/tex]

[tex]yj = \frac{1}{3}jt[/tex]

Divide by j

[tex]y = \frac{1}{3}t[/tex]

[tex]zk = kt[/tex]

Divide by k

[tex]z = t[/tex]

Hence, the parametric equations are:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

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

Answer:

4.552m/s

Explanation:

[tex]V=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1} } =\frac{2400*4.33+60*8.88}{2400}=4.552m/s[/tex]

Answer:

4. unbalanced and Accelerating

5. balance and rest

Answer:

Answer is explained in the explanation section below.

Explanation:

In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.

a) You pass through the highest point:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.

And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.

b) You pass through the lowest point of the ride:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.

And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.

Answer: In picture 1, heat is flowing from the liquid to the air. In picture 2, heat is flowing from the air to the liquid

Explanation:

I don't know if I answered correctly, if not I can provide another answer

Answer:

a) E = 0

b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

From which we have;

[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]

E = 0/A = 0

E = 0

b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]

[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]

By Gauss theorem, we have;

[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]

Therefore, we get;

[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]

The electrical field outside the spherical shell

[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]

[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]

Therefore, we have;

[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Answer:

Answer is explained in the explanation section below.

Explanation:

Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.

Solution:

The relation between electric field and the electric potential is:

E = [tex]\frac{dV}{dr}[/tex]

So, making dV the subject, we have:

dV = E x dr

Integrating the above equation, we get.

V = [tex]\int\limits^_ {} \,[/tex]E x dr      Equation 1

Now, in 2-D

E is inversely proportional to the radius r.

E ∝ 1/r

So, we can write: replacing E ∝ 1/r in the equation 1

V ∝  [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr

Which implies that,

V ∝  log (r)

Hence, distance dependence expected for the electric potential =  ln (r)

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:

0.13 m/s

Explanation:

Unfortunately, I don't have an explanation but I guessed the correct answer.

Answer:

A) P(t) = 2651.25 [ 1 - cos2wt ] W

B)  Real power = 999.79 watts

    Reactive power = 2652.86 VA

c) power factor = 0.3526

Explanation:

Given data:

V(t) = 141.4 cos (ωt)

R(t) = 10 Ω

Inductive reactance XL = ωL = 3.77 Ω

Ir(t) = V(t) / R(t) = 14.14

A) Calculate the instantaneous power absorbed by the resistor and by inductor

By resistor :

Pr(t) = V(t) * Ir(t) = 141.4 * 14.14 [tex]cos^{2} wt[/tex] = 1999.396 [tex]cos^{2} wt[/tex]

      hence Pr  = 999.698 (cos2ωt + 1) w

By Inductor :

Pl(t) = V(t) I'L(t) = 141.4 cosωt * 37.5 cos(ωt - 90)  

                        =  5302.5 [tex]sin^2 wt[/tex]

Hence Pl(t) = 5302.5 [tex]sin^2 wt[/tex]   w  =  2651.25 [ 1 - cos2wt ] W

B) calculate the real and reactive power

First we have to determine the power factor

Given that : V(t) = 141.4 cosωt  v ,   Ir(t) = 14.14 cosωt A

IL(t) = 37.5 cos (ωt - 90° )

The phasor representation of the above is :

V = [tex]\frac{141.4}{\sqrt{2} } <0^{0} v[/tex] = 141.4 ∠0° ,  Ir = 10 ∠ 0° , IL = 37. 5 ∠ -90°

Total load current = Ir + IL = 28.35 ∠ -69.35°

power factor = cos -69.35° = 0.3526

Next we will determine the Real and reactive power using the relation below

S = VI = 100 ∠ 0°  * 28.35 ∠ -69.35°

         = 2835 ∠ 69.35°

S = P + jQ = 999.79 + 2652.85 j

Real power  = 999.79 watts

Reactive power = 2652.85 VA

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:

1. cell membrane

2. golgi body

3. mitochondrion

4. cytoplasm

5. nucleolus

6. nucleus

Explanation:

The correct answer to this question is Option A; 6.

Why?

In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.

~Thank you~

Answer:

467 volts

Explanation:

Vs/Vp = Ns/Np

Vs = Ns/Np × Vp

Vs = 350/150 × 200 = 7/3 × 200

Vs = 467 volts

Answer:

Explanation:

The relation between electric field and potential difference is as follows

E = - dV / dr

That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .

Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means

dV / dr = constant

E = constant

Electric field is constant .

So the option which is correct is

" points east and does not vary with position " .

Answer:

v = 7.95 m/s

Explanation:

Given that,

Wavelength of a wave, [tex]\lambda=53\ cm=0.53\ m[/tex]

Frequency of a wave, f = 15 Hz

We need to find the speed of the wave. The speed of a wave is given by :

[tex]v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s[/tex]

So, the wave move with a speed of 7.95 m/s.

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

learn more about electric charge movement here:  https://brainly.com/question/19315467

Answer:

Clara has speed of 80m/min

Explanation:

Clara was jogging at 600 m in 5 minutes. She stopped suddenly which reduced her velocity and then she waited for 10 minutes so that her friends comes near her. She stopped to catch her friend. During this 10 minutes the velocity of Clara is zero. She started to walk again at a slower speed of 80m/min.

Answer:

A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.

(Hope this helps! Btw, I am the first to answer.)

Answer:

1767Kw

Explanation:

Velocity of wind = 10 m/s

diameter of the blades= 60m

ρ= air density = 1.25 kg/m3

Acceleration due to gravity= 9.81 m/s^2

Mechanical energy of the wind can be calculated using the expression below

Energy= (e*m)

= ρ V A e............eqn(1)

Where A= area

ρ= air density

e= wind energy per unit mass of air

e= (v^2)/2..........eqn(2)

If we substitute the values into eqn (2) we have

e= [(10)^2]/2

=50J/Kg

But Area=A= (πd^2)/4

Area= ( π× 60^2)/4

Area=2827.8m^2

If we input substitute the values into eqn (1) we have

Energy= 1.25 ×10 × 50×2827.8

=1767145.7W

We can convert to kilo watt

=1767145.7W/ 1000

= 1767Kw

Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

[tex]V_{A}[/tex] = 1 m³

[tex]T_{A}[/tex] = 10°C = 283 K

[tex]P_{A}[/tex] = 350 kPa

[tex]m_{B}[/tex] = 3 kg

[tex]T_{B}[/tex] = 35°C = 308 K

[tex]P_{B}[/tex] = 150 kPa

Now, lets apply the ideal gas equation;

[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]

[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150

[tex]V_{B}[/tex] = 265.188 / 150  

[tex]V_{B}[/tex] = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, [tex]m_{A}[/tex] =  [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221

[tex]m_{A}[/tex]  = 4.309 kg

Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg

Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex]  = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

[tex]P_{f}[/tex] =  [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]

given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K

we substitute

[tex]P_{f}[/tex] =  ( 7.309 × 0.287 × 293)  / 2.77

[tex]P_{f}[/tex] =  614.6211119 / 2.77

[tex]P_{f}[/tex] =  221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa