Will give brainliest ASAP! Please help (1/10 questions, will mark 5 stars and brainliest for all answers if correct)

Answer:

a)   x = 2.46 m

b)   318.2 N

c)    177.8 N

Explanation:

Need to resolve the tension of the string at end say B.

The vertical upward force at B due to tension is 450 sin 45°.  

Using Principle of Moments, with the pivot at A,

Anti clockwise moments = Clockwise moments

450 sin 45° X 3.2   = 220 X (3.2/2) + (270 X x)  

x = 2.46 m

(b) The horizontal force is only due to the wire's tension, so it is  

450 cos 45° = 318.2 N

(c) total downward forces = 270 + 220 = 496 N  

Total upward forces = 450 sin 45° (at B) + upForce (at A)

Equating, upForce = 496 - 318.2  

= 177.8 N

Answer:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Explanation:

The frequency of an electromagnetic radiation can be given by the following formula:

υ = c/λ

where,

υ = frequency of electromagnetic wave = ?

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = 290 nm to 320 nm

FOR LOWER LIMIT OF FREQUENCY:

λ = 320 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(3.2 x 10⁻⁷ m)

υ = 9.375 x 10¹⁴ Hz

FOR UPPER LIMIT OF FREQUENCY:

λ = 290 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(2.9 x 10⁻⁷ m)

υ = 10.34 x 10¹⁴ Hz

Therefore, the frequency range for UVB radiations is:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Answer

0.2067m or 0.2067m

Explanation;

Let lenght of spring= Lo= 11cm=0.110m

It is hang from a mass of

3.05-kg having a length of L1= 12.40cm= 0.124m

Force required to stretch the spring= Fkx

But weight of mass mg= kx then K= Mg/x

K= 3.05-kg× 9.8)/(0.124m-.110m)

K=2135N

But potential Energy U= 0.5Kx

X=√ 2U/k

√(2*10)/2135

X=0.0967m

The required new length= L2= L0 ±x

=

.110m ± 0.0967m

X= 0.2067m or 0.2067m hence the total lenghth

Answer:

C. The spinning of an object on its axis.

Explanation:

The definition of rotation answers this question.

A good way to remember is that when you rotate, you turn, and when you revolve, you move around.

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]

The induced emf in the shorter coil is calculated as;

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]

The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:

[tex]E=-\frac{\delta\phi}{\delta t}[/tex]

where E is the induced emf,

[tex]\phi[/tex] is the magnetic flux through the coil.

The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:

[tex]\phi = \frac{\mu_oNIA}{L}[/tex]

so;

[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]

where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.

[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]

The emf produced in the coil at the center of the solenoid which has 14 turns will be:

[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]

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Answer:

The longer the cord, the lower the illumination

Explanation:

The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.

Long wires have more electrical resistance than shorter ones.

Let us consider this formula:

Resistance =[tex]\frac{\rho L}{A}[/tex]

From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.

Answer:

Explanation:

volume is same

π r₁² L₁  =π r₂²L₂

L₁ / L₂ = r₂² / r₁²

For resistance the formula is

R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area

R₁ = ρ L₁ / S₁

R₂ = ρ L₂ / S₂

Dividing

R₁ / R₂ = L₁ / L₂ x S₂ / S₁

=  L₁ / L₂ x r₂² / r²₁

= L₁ / L₂ x L₁ / L₂

= L₁²/ L₂²

L₂ = 1.1 L₁ ( GIVEN )

= L₁²/ (1.1L₁)²

1 / 1.21

R₂ / R₁ = 1.21 .

= 1.2

Answer:

Explanation:

A microscope is made up of a convex lens and the nature of the image formed by the object viewed from it is a virtual image. Since the image is virtual, the image distance will be negative and the focal length will be positive (for convex lenses).

Using the lens formula to first calculate the image distance from the lens;

1/f = 1/u + 1/v

f is the focal length = 0.150 cm,

u is the object distance = 0.155 cm

v is the image distance.

Since the image distance is negative,

1/f = 1/u - 1/v

1/0.150 = 1/0.155 - 1/v

1/v = 1/0.155 - 1/0.15

1/v = 6.542 - 6.667

1/v = -0.125

v = 1/-0.125

v = -8 cm

Magnification = image distance/object distance

Mag = 8/0.155

Mag = 51.6

Magnification produces is 51.6

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

Answer:

U_f = (U_o)/2)

Explanation:

The capacitance of a given capacitor is given by the formula;

C = ε_o•A/d

While energy stored in plates capacitor is given as; U_o = Q²/2C

Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;

C' = ε_o•4A/2d

Hence, C' = 2C

so capacitance is now doubled

Thus, the final energy stored between the plates of capacitor is given as;

U_f = Q²/2C'

From earlier, we saw that C' = 2C.

Thus;

U_f = Q²/2(2C)

U_f = Q²/4C

Rearranging, we have;

U_f = (1/2)(Q²/2C)

From earlier, U_o = Q²/2C

Hence,

U_f = (1/2)(U_o)

Or

U_f = (U_o/2)

Answer:

In order to stay hydrated during warm-up(s) drink 8oz of water 20-30 mintues before you start exercising or during your warm-up(s), make sure you drink 7 to 10 oz of water every 10 to 20 minutes during exercise, and drink 8oz of water no more than 30 minutes after you exercise.

In order to stay hydrated during scheduled activity(s) drink 17 to 20 oz of water 2 to 3 hours before you start to exercise, like said before drink 8 oz of water 20 to 30 minutes before you start exercising or during your warm-up(s), drink 7 to 10 oz of water every 10 to 20 minutes during exercise, also said before drink 8 ounces of water no more than 30 minutes after you exercise.

Answer: My scheduled activity was one hour of softball practice. I play catcher, so my thighs and knees take a lot of abuse from kneeling and standing. The lunges were excellent at preparing my thighs for softball. The high knees exercise and arm pumping didn’t feed into softball too well. I suppose that they might help me with base running.

Explanation: EDMENTUM

Answer:

Explanation:

Range is defined as the distance covered in the horizontal direction. In projectile, range is expressed as x = vt where;

x is the range

v is the velocity of the runner

t is the time taken

Before we can get the range though, we need to find the time taken t using the relationship S = ut + 1/2gt²

if u = 0

S = 1/2gt²

2S = gt²

t² = 2S/g

t = √2S/g

t = √2(141)/9.8

t = √282/9.8

t = 5.36secs

The range x = 52.4*5.36

x =280.86 m

Hence, the coyort lies approximately 280.86 m from the base of the cliff

Answer:

P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.

Explanation:

The force, F on a conductor of length, L in a magnetic field of magnetic field strength, B and current, I flowing through it is given by

F = BIL.

Now, if the conductor has a velocity, v, the energy dissipated by this force is P

P = Fv = BIL × v = BILv.

Now, we know that the induced e.m.f due to the motion of the conductor is given by ε = BLv.

From P above, P = BILv = I(BLv)

substituting ε = BLv into P, we have

P = Iε

Thus, P is the electrical energy dissipated due to the motion of the conductor.

Now since P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.

Answer:

The expected year is 2017.

Explanation:

Total years that the millionaire to live = 15 years

Travel away from the earth at  = 0.8 c

This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:

[tex]T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years[/tex]

Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017

Answer:

Using ohms law

The current is found from Ohm's Law.

I = V /R = E /R = Bxy /Rt.

Answer:

On that line segment between the two charges, at approximately [tex]0.7\; \rm m[/tex] away from the smaller charge (the one with a magnitude of [tex]5 \times 10^{-19}\; \rm C[/tex],) and approximately [tex]1.3\; \rm m[/tex] from the larger charge (the one with a magnitude of [tex]20 \times 10^{-19}\; \rm C[/tex].)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let [tex]k[/tex] denote the Coulomb constant, and let [tex]q[/tex] denote the size of a point charge. At a distance of [tex]r[/tex] away from the charge, the electric field due to this point charge will be:

[tex]\displaystyle E = \frac{k\, q}{r^2}[/tex].

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.  

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] and [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex]. Assume that the electric field is zero at [tex]r[/tex] meters to the right of the [tex]5\times 10^{-19}\; \rm C[/tex] point charge. That would be [tex](2 - r)[/tex] meters to the left of the [tex]20 \times 10^{-19}\; \rm C[/tex] point charge. (Since this point should be between the two point charges, [tex]0 < r < 2[/tex].)

The electric field due to [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] would have a magnitude of:

[tex]\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}[/tex].

The electric field due to [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex] would have a magnitude of:

[tex]\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}[/tex].

Note that at all point in this section, the two electric fields [tex]E_1[/tex] and [tex]E_2[/tex] will be acting in opposite directions. At the point where the two electric fields balance each other precisely, [tex]| E_1 | = | E_2 |[/tex]. That's where the actual electric field is zero.

[tex]| E_1 | = | E_2 |[/tex] means that [tex]\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}[/tex].

Simplify this expression and solve for [tex]r[/tex]:

[tex]\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0[/tex].

[tex]\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0[/tex].

Either [tex]r = -2[/tex] or [tex]\displaystyle r = \frac{2}{3}\approx 0.67[/tex] will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, [tex]0 < r < 2[/tex]. Therefore, [tex](-2)[/tex] isn't a valid value for [tex]r[/tex] in this context.

As a result, the electric field is zero at the point approximately [tex]0.67\; \rm m[/tex] away the [tex]5\times 10^{-19}\; \rm C[/tex] charge, and approximately [tex]2 - 0.67 \approx 1.3\; \rm m[/tex] away from the [tex]20 \times 10^{-19}\; \rm C[/tex] charge.

Answer:

    f= 4,186  10²  Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

             w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El  momento de inercia de indican que giran un eje que pasa                 por enronqueces

           I= ½ M R2  

reduzcamos las cantidades al sistema SI

         R= 1,4 cm = 0,014  m

         M= 430 g = 0,430 kg

substituimos

           w= √ (2 k/M R2)

calculemos  

           w = RA ( 2 370 / (0,430  0,014 2)

           w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

            w =2pi f

            f= w/2π

            f= 2,963 10³/ (2π)

            f= 4,186  10²  Hz

Answer:

A) 3 interactions

B)  2.78 m/s

C)  2.78 m/s

D) 0 m/s

Explanation:

Given Data :

plastic buttons ( 2 ) = 2.0 g each

charge on each plastic buttons = +65 nC

A) The interactions that have potential energy can be expressed as

V net = v1 + v2 + v3

V net = 0 + [tex]\frac{kq1q2}{r12} + \frac{kq1q3}{r13} + \frac{kq2q3}{r23}[/tex]

hence the total interactions is (3)

B) The final speed of the left 2.0 g plastic button

U = [tex]\frac{9*10^9* 10^-18}{0.02} [ 65*250 + 65*250 + (65*65)/2 ][/tex]

U = 0.0155 J  

also U = [tex]2( 1/2 mv^2 )[/tex]

      U = mv^2 = 0.0155 J

therefore ;  v = [tex]\sqrt{0.0155/(2*10^-3)}[/tex] = 2.78 m/s

C) final speed of the right 2.0 g plastic button = Final speed of the right 2.0 g plastic button

v = 2.78 m/s

D) The final speed of the 5.0 g plastic button

= 0 m/s

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

[tex]F = \frac{k|q_1||q_2|}{r^2}[/tex]

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

[tex]F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N[/tex]

Therefore, the mutual force between the two point charges is 319.64 N

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

A)  The height of the balloon at the end of 40 sec is 820 feet.

B) The average velocity of the balloon is 30 ft/sec.

C) The velocity of the balloon at the end of 30 sec is

Given :

Part A)

The height of the balloon after 40 secs is :

h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

Part B)

The average velocity of the balloon is  :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0 sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

When at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon = 30 ft/sec

The average velocity of the balloon  is  30 ft/sec.

Part C)

The velocity of the balloon at the end of 30 sec is :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec.

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Answer:

The average induced emf in the loop is 0.3 V

Explanation:

Given:

d = 22 cm

Magnetic field  is  B = 1.5 T

Change in time  Δt = 0.20 sec.

Radius of loop  = r = d/2 = 11 x [tex]10^{-2}[/tex] m

According to the faraday's law,  Induced emf is given by   e = - (ΔФ / Δt)

where Ф = magnetic flux

Ф = BAcos0   (in this occasion, Ф=0)

where area A = pi * r²

note that we have to neglect negative sign due to its lenz law...so

       B * pi * r

e = ----------------

          Δt

       1.5 ( 3.1416) 11 x [tex]10^{-2}[/tex] )²

e =  ------------------------------------

                        0.20

e = 0.3 V

Therefore, the average induced emf in the loop is 0.3 V

Given that,

The electric field is given by,

[tex]\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}[/tex]

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of [tex](\vec{E}\times\vec{B})[/tex] vector is the direction of propagation of the wave .

Direction of magnetic field = [tex]\hat{j}[/tex]

[tex]B=B_{0}\sin(kx-\omega t)\hat{k}[/tex]

We need to calculate the poynting vector

Using formula of poynting

[tex]\vec{S}=\dfrac{E\times B}{\mu_{0}}[/tex]

Put the value into the formula

[tex]\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}[/tex]

[tex]\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]

Hence, The poynting vector is [tex]\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]

Answer:

The refractive index of fluid 2 is 1.78

Explanation:

Refractive index , n = real depth/apparent depth

For the first fluid, n = 1.37 and apparent depth = 9.00 cm.

The real depth of the container is thus

real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm

To find the refractive index of fluid index of fluid 2, we use the relation  

Refractive index , n = real depth/apparent depth.

Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.

So, n = 12.33 cm/6.86 cm = 1.78

So the refractive index of fluid 2 is 1.78

Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8

Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,

Then,

Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]

Real depth = Index of refraction x apparent depth

Since the same container is used, we can make an assumption that;

real depth of fluid 1 = real depth of fluid 2

That is,

1.37 x 9 = n x 6.86

Where n = Index of refraction for the second fluid.

make n the subject of formula

n = 12.33 / 6.86

n = 1.79

Therefore, the index of refraction of the second fluid is 1.8 approximately.

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Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.

N/m

Explanation:

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. The focus point is close to the curved  mirror than the centre of curvature.

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. All incident light striking the surface all converges at a point on the central axis known as the focus.

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Answer:

931.00ft-lb

Explanation:

Pls see attached file

The work done in moving the object from x = 1 ft to x = 18 ft is 935  ft-lb.

Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.

Given that force = 6x - 2 pounds.

So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]

= [ 3x² - 2x]¹⁸₁

= 3(18² - 1² ) - 2(18-1) ft-lb

= 935  ft-lb.

Hence, the work done is  935  ft-lb.

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Answer:

1.03A

Explanation:

For computing the magnitude of the current in the circuit we need to do the following calculations

LCR circuit impedance

[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]

= 148.7Ω

Now the phase angle is

[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]

Now the rms current flowing in the circuit is

[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]

= 0.98 A

The current flowing in the circuit is

[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]

= 1.39 A

And, finally, the current across the generator is

[tex]I'= I cos \phi[/tex]

[tex]= (1.39) cos 42.3^{\circ}[/tex]

= 1.03A

Hence, the magnitude of the circuit current is 1.03A

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Answer:

The only suitable candidate is steel.

Explanation:

We are given;

Initial length;L_o = 120 mm = 0.12 m

Diameter;d_o = 15 mm = 0.015 m

Radius;r = 0.015/2 m = 0.0075 m

Applied force;F = 35000 N

Allowable Diameter reduction;Δd = 1.2 x 10^(-2) mm = 0.012 mm

Formula for Poisson’s ratio is;

v = ε_x/ε_z

Where;

ε_x is lateral strain = = Δd/d_o

ε_z is axial strain = σ/E

Thus;

v = (Δd/d_o)/(σ/E)

Making Δd the subject, we have;

Δd = (vσd_o)/E

Where σ is force per unit area

σ = F/A = 35000/πr² = 35000/(π0.0075)² = 198 × 10^(6) Pa

Now, let's find Δd for each of the metals given;

For Titanium alloy, E = 70 GPa = 70 × 10^(9) Pa and Poisson’s ratio (v) = 0.33

Thus;

Δd_ti = (0.33 × 198 × 10^(6) × 0.015)/70 × 10^(9) = 0.000014m = 0.014 mm

Similarly, for steel;

Δd_st = (0.36 × 198 × 10^(6) × 0.015)/(105 × 10^(9)) = 0.0000102 m = 0.0102 mm

For magnesium;

Δd_mn = (0.27 × 198 × 10^(6) × 0.015)/(205 × 10^(9)) = 0.0000039 m = 0.0039 mm

For aluminum;

Δd_al = (0.2 × 198 × 10^(6) × 0.015)/(45 × 10^(9)) = 0.0000132 m = 0.0132 mm

Since they must not experience a diameter reduction of more than 0.012 mm, then the only alloy that has a reduction diameter less than 0.012 is steel which has 0.0102.

Thus,the only possible candidate is stee.

Answer:

100 degrees Celsius

Explanation:

Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.

At 100 degrees Celsius water begin to boil and turns to steam.

The melting point for water is 0 degrees C (32 degrees F). The boiling point of water varies with atmospheric pressure. At lower pressure or higher altitudes, the boiling point is lower. At sea level, pure water boils at 212 °F (100°C).

If the temperature is much above 212°F, the water will boil. That means that it won't just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. If the water has very few dust flecks etc.

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