An elevator filled with passengers has a mass of 1603 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 2.50 s. Calculate the tension in the cable (in N) supporting the elevator.

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]

or,

    [tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]

On substituting the values, we get

      [tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]

      [tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)

On the Y-axis,

⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]

                        [tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]

                    [tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]

From equation 1, we get

           [tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]

                        [tex]T_1+3T_1=278.4 \ N[/tex]

                                [tex]4T_1=278.4 \ N[/tex]

                                  [tex]T_1=\frac{278.4}{4}[/tex]

                                       [tex]=69.6 \ N[/tex]  

Answer:

69.58

Explanation:

Answer:

The speed of the vehicle is 156 miles per hour.

Explanation:

Let suppose that the Maglev, that is, a vehicle who works on the principle of superconductive magnetic levitation, moves at constant speed. Hence, the speed of the vehicle ([tex]v[/tex]), in miles per hour, is defined by this kinematic model:

[tex]v = \frac{s}{t}[/tex] (1)

Where:

[tex]s[/tex] - Travelled distance, in miles.

[tex]t[/tex] - Time, in hours.

If we know that [tex]s = 624\,mi[/tex] and [tex]t = 4\,h[/tex], then the speed of the vehicle is:

[tex]v = \frac{624\,mi}{4\,h}[/tex]

[tex]v = 156\,\frac{mi}{h}[/tex]

The speed of the vehicle is 156 miles per hour.

Answer:

doubled

Explanation:

When thrust is double so will the pressure I hope this helps

enjoy

Answer:

The sensitivity of the device = 1.234 Hz per km

Explanation:

Given  

Frequency (f) = 10 gHz

Speed of the car = 120 Km/h

As per the doppler’s effect

V = (change in frequency /frequency) *(c/2)

Substituting the given values, we get –  

Change in frequency = {(2*10^9*120)/(3*10^8)} * (1000/3600)

Change in frequency =  37.03 Hz

b) speed of light = wavelength * frequency

3*10^8 = wavelength * 10*10^9

Wavelength = 0.03 m

Sensitivity = change in frequency /wavelength = 37.03/0.03 = 1234 Hz/m

1.234 Hz per km

This question is incomplete, the complete question is;

People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.

a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)  __ mm

Answer:

1) the power in D of the eyes of a woman is 61.7647 D

2) the size in mm of an image of a 8.00 mm object is -1.882 mm

Explanation:

Given the data in the question;

a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,

so

u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m

v = 2.00 cm = ( 2 / 100 )m =  0.02 m

Now, we know that power of lens p = 1 / u + 1 / v

so we substitute

p = ( 1 / 0.085 ) + ( 1 / 0.02 )

p = 11.7647 + 50

p = 61.7647 D

Therefore,  the power in D of the eyes of a woman is 61.7647 D

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)

we know that;

m = -v / u

we substitute

m = -0.02 / 0.085

m = -0.2353

since H₀ = 8.0 mm

H[tex]_i[/tex] = m × H₀

H[tex]_i[/tex] = -0.2353 × 8.0

H[tex]_i[/tex] = -1.882 mm

the size in mm of an image of a 8.00 mm object is -1.882 mm

Answer:

E = 16581.6 J

Explanation:

Given that,

Current, I = 4.9 A

Time for which the current is set up, I = 4.7 min = 282 s

The voltage of the battery, V = 12 V

We need to find how much chemical energy of the battery reduced. Let It is E. We know that,

E = P t

Where

P is power of battery, P = VI

So,

[tex]E=VIt[/tex]

Put all the values,

[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]

So, 16581.6 J of chemical energy of the battery is reduced.

Answer:

a) Ink X is likely to be pure because it only contain 1 spot.

b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.

c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.

The different spots from Y are found at various heights because they represent different compounds.

The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.

We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.

Learn more about chromatography:https://brainly.com/question/26491567?

#SPJ6

Answer:

the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

Explanation:

The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

            F = q E

Solution :

The conditions for the maximum in the Young's experiment is :

d sin θ = m λ,     where m = 0, 1, 2, 3, .....

The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,

d sin θ =  λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]

Given : d = 100 λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]

[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]

 [tex]$=0.573^\circ$[/tex]

  = 0.01 rad

Answer:

 K = 7.56 10⁻⁴ J

Explanation:

The rotational kinetic energy is

        K = ½ I w²

They ask us for the kinetic energy of the wheel, which can be approximated as a thin ring, its moment of inertia is

         I = M r²

the linear speed of the gerbils is equal to the linear speed of the wheel. The linear and rotational variables are related

          v = w r

          w = v / r

   we substitute

         K = ½ (M r²) v² / r²

         K = ½ M v²

let's calculate

         K = ½ 5 10⁻³ 0.55²

         K = 7.56 10⁻⁴ J

Answer:

a = -0.4m/s²

Explanation:

v_f = v_I + (a)(t)

a(t) = v_f-v_I

a = (v_f-v_I)/t

a = (0m/s-12m/s)/30s

a = -0.4m/s²

Answer:

I dont. understand the question, maybe insert the picture?

The answer for this question is D

c. a natural process

It is a natural process

SOLVED DOWN BELOW

Explanation:

In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law

I = 25/200

I= .125 amps or 125 ma

__________

R= r1 * r2 / r1 +r2

R= 100 * 100 / 100 + 100

R= 10000 / 200

R= 50 ohms

Answer: b force

Explanation:

yes because the world comin g up with more technique

Answer:

spirit animal?

Explanation:

Answer:

F = 312 N

Explanation:

Given that,

The mass of a crate, m = 40 kg

Acceleration of the crate, a = 2 m/s²

As the carte is falling downward, the net force exerted by the rope on the carte is given by :

F = m(g-a)

Put all the values,

F = 40(9.8-2)

F = 312 N

Hence, the required force exerted by the rope on the crate is equal to 312 N.

Answer: [tex]14.64\ N[/tex]

Explanation:

Given

Inclination of ramp is [tex]\theta=15^{\circ}[/tex]

Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal

Weight of cart [tex]W=40\ lb[/tex]

from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp

Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position

[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]

Answer:

The resulting force on the first object is 800 N.

Explanation:

Given;

force exerted on one of the objects, F₁ = 400 N

let the first charge = q₁

let the second charge = q₂

The force of repulsion between the objects is calculated using Coulomb's law;

[tex]F =\frac{kq_1q_2}{r^2} \\\\\frac{k}{r^2} = \frac{F}{q_1q_2} = \frac{F_1}{q_1\times2q_2} \\\\F_1 = \frac{F(q_1\times2q_2)}{q_1q_2} \\\\F_1 = 2F\\\\F_1 = 2(400 \ N)\\\\F_1 = 800 \ N[/tex]

Therefore, the resulting force on the first object is 800 N.

Answer:

The correct answer is "[tex]0.5\times 10^{-6} \ m[/tex]".

Explanation:

Given:

[tex]\frac{\lambda D}{d} =2.8\times 10^{-3}[/tex]

[tex]d = 0.5\times 10^{-3}[/tex]

[tex]D = 2.80[/tex]

Now,

The wavelength will be:

⇒ [tex]\lambda = 2.8\times 10^{-3}\times \frac{d}{D}[/tex]

By putting the values, we get

⇒    [tex]=\frac{2.8\times 10^{-3}\times 0.5\times 10^{-3}}{2.8}[/tex]

⇒    [tex]=\frac{1.4\times 10^{-6}}{2.8}[/tex]

⇒    [tex]=0.5\times 10^{-6} \ m[/tex]  

Answer: See explanation

Explanation:

The evolutionary stages for the formation of planets from earliest to latest will be:

1. Dust keeps matter inside the disk cool enough for planet formation to start

2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.

3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.

4. Planetesimals begin to accrete, forming protoplanets.

5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

Answer:

Explanation:

From the given information:

Let's assume that the missing function is:

s(t) = t³ - 6t², t ≥ 0

From part (b), we are to find the given  required terms when time t = 2

So; from the function s(t) =  t³ - 6t², t ≥ 0

[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]

[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]

[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]

[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]

[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]

[tex]acceleration\ a(t) = 6t - 12[/tex]

At time t = 2

The position; S(2) = (2)² - 6(2)²

S(2) = 8 - 6(4)

S(2) = 8 - 24

S(2) = - 16 ft

v(2) = 3(2)² - 12 (2)

v(2) = 3(4) - 24

v(2) = 12 - 24

v(2) = - 12 ft/s

speed = |v(2)|

|v(2)|  = |(-12)|

|v(2)| = 12 ft/s

acceleration = 6t - 12

acceleration = 6(2) - 12

acceleration =  12 - 12

acceleration =  0 ft/s²

By Newton's second law, the net force on the object is

∑ F = m a

∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N

Let f be the unknown force. Then

∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f

=>   f = (-2.0 i - 12.0 j ) N