If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can see

Explanation:

u = 95 m/sec ( Initial speed)

t = 7 sec ( Time of ascent)

According to Equations of Motion :

[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]

Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m

Answer:

332.5 m

Explanation:

At the maximum height, the velocity is 0.

Given:

v₀ = 95 m/s

v = 0 m/s

t = 7 s

Find: Δy

Δy = ½ (v + v₀) t

Δy = ½ (0 m/s + 95 m/s) (7 s)

Δy = 332.5 m

Answer:

K = 227.04 W/m.°C

Explanation:

First we need to find the heat required to melt the ice:

q = m H

where,

q = heat required = ?

m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg

H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg

Therefore,

q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)

q = 2839 J

Now, we find the heat transfer rate through rod:

Q = q/t

where,

t = time = (10 min)(60 s/1 min) = 600 s

Q = Heat Transfer Rate = ?

Therefore,

Q = 2839 J/600 s

Q = 4.73 W

From Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

K = Thermal Conductivity = ?

A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²

L = Length of rod = 60 cm = 0.6 m

ΔT = Difference in temperature = 100°C - 0°C = 100°C

Therefore,

4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m

K = (4.73 W)/(0.0208 m.°C)

K = 227.04 W/m.°C

Answer:

3/4 of 12 = 16

3/4 of 24 = 32

Those are the answers based on how your question sounded

Explanation:

Answer:

27

Explanation:

3/4 of (12 and 24)

of means ×

and means +

therefore,3/4× (12+24)

3/4×(36)

3×9

27

Answer:

 M₁₂ = 1.01 10⁻⁴ H ,   Fem = 3.54 10⁻³ V

Explanation:

The mutual inductance between two systems is

        M₁₂ = N₂ Ф₁₂ / I₁

where N₂ is the number of turns of the inner solenoid N₂ = 21.0, i₁ the current that flows through the outer solenoid I₁ = 35.0 A / s and fi is the flux of the field of coil1 that passes through coil 2

the magnetic field of the coil1 is

   B = μ₀ n I₁ = μ₀ N₁/l   I₁

the flow is

             Φ = B A₂

the area of ​​the second coil is

             A₂ = π d₂ / 4

             Φ = μ₀ N₁ I₁ / L  π d² / 4

we substitute in the first expression

            M₁₂ = N₂ μ₀ N₁ / L    π d² / 4

            M₁₂ = μ₀ N₁ N₂ π d² / 4L

           d = 0.170 cm = 0.00170 m

            L = 4.00 cm = 0.00400 m

let's calculate

            M₁₂ = 4π 10⁻⁷ 6750  21 π 0.0017²/ (4 0.004)

             M₁₂ = π² 0.40966 10⁻⁷ / 0.004

             M₁₂ = 1.01 10⁻⁴ H

The electromotive force is

              Fem = - M dI₁ / dt

              Fem = - 1.01 10⁻⁴ 35.0

              Fem = 3.54 10⁻³ V

Answer:

28.3°C

Explanation:

Using

T(t) = (T(0) - 20)*(e^(-k*t)) + 20

for some positive number k, and some initial temperature T(0).

Boundary conditions:

T(4) = T(0) - 5 _______ (i)

T(8) = T(0) - 7 _______ (ii)

==> solving for T(0) and k :

(i):

(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>

(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5

(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5

5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))

5 = (T(0) - 20) * ( 1 - e^(-k*4) )

(ii):

(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7

(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7

7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))

7 = (T(0) - 20) * (1 - e^(-k*8))

In both results, subsitute x = e^(-4k) and C = (T(0) - 20)

(i): 5 = C * (1 - x)

(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)

Substitute C*(1-x) from (i) into (ii):

(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5

back into (i):

(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3

C = T(0) - 20 ==>

T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3

= 28.3°C

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

Answer:

Explanation:i think this would help u

Answer:

T = 3.14 hours

Explanation:

We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.

We know that the radius of Mars is 3,389.5 km.

So, r = 1,787 + 3,389.5 = 5176.5 km

Using Kepler's law,

[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]

M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]

So,

[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]

or

T = 3.14 hours

So, the orbital period is 3.14 hours

Answer:

The  value is  [tex]B = 0.0043 \ T[/tex]

Explanation:

From the question we are told that

   The  length of the solenoid is  [tex]l = 63.5 = 0.635 \ m[/tex]

    The number of turns is  [tex]N = 960 \ turns[/tex]

    The  current is  [tex]I = 2.28 \ A[/tex]

Generally the magnetic field is mathematically represented as

      [tex]B = \mu _o * n * I[/tex]

Where  n is the number of turn per unit length which is mathematically evaluated as

      [tex]n = \frac{N}{l}[/tex]

     [tex]n = \frac{960}{0.635}[/tex]

     [tex]n = 1512 \ turns /m[/tex]

and  [tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

So  

    [tex]B = 4\pi * 10^{-7} * 1512 * 2.28[/tex]

    [tex]B = 0.0043 \ T[/tex]

Answer:

The energy contained in a single pulse is 90,200 J.

Explanation:

Given;

power of the laser, P = 82 TW = 82 x 10¹² W

time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s

the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m

The energy contained in a single pulse is calculated as;

E = Pt

where;

P is the power of each laser

t is the time to generate the power

E = (82 x 10¹²)(1.1 x 10⁻⁹)

E = 90,200 J

Therefore, the energy contained in a single pulse is 90,200 J

Answer:

The answer is "45.79 m/s"

Explanation:

Given values:

diameter= 25 cm

w= 3500 rpm

Formula:

[tex]\boxed{v=w \times r} \ \ \ \ \ \ _{where} \ \ \ w = \frac{rad}{s} \ \ \ and \ \ \ r = meters[/tex]

Calculating r:

[tex]r= \frac{diameter}{2}[/tex]

  [tex]=\frac{25}{2}\\\\=12.5 \ cm[/tex]

converting value into meters: [tex]12.5 \times 10^{-2} \ \ meter[/tex]

calculating w:

[tex]w= diameter \times \frac{2\pi}{60}\\[/tex]

   [tex]= 3500 \times \frac{2\times 3.14}{60}\\\\= 3500 \times \frac{2\times 314}{6000}\\\\= 35 \times \frac{314}{30}\\\\= 35 \times \frac{314}{30}\\\\=\frac{10990}{30}\\\\=\frac{1099}{3}\\\\=366.33[/tex]

w= 366.33 [tex]\ \ \frac{rad}{s}[/tex]

Calculating v:

[tex]v= w\times r\\[/tex]

  [tex]= 366.33 \times 12.5 \times 10^{-2}\\\\= 366.33 \times 12.5 \times 10^{-2}\\\\= 4579.125 \times 10^{-2}\\\\\boxed{=45.79 \ \ \frac{m}{s}}[/tex]

Answer:

The fish would appear 42.7 cm on the left side from the front of the bowl.

Explanation:

The fish (object) distance = 11.9 cm, radius of curvature of the bowl = 33 cm. The distance of image of the fish (image distance) can be determined by applying the mirror formula;

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

where f is the focal length of the reflecting surface, u is the object distance and v is the image distance.

But, f = [tex]\frac{radius of curvature}{2}[/tex]

         = [tex]\frac{33}{2}[/tex]

       f = 16.5 cm

Substitute f = 16.5 = [tex]\frac{165}{10}[/tex], and u = 11.9 = [tex]\frac{119}{10}[/tex] in equation 1;

[tex]\frac{10}{165}[/tex] = [tex]\frac{10}{119}[/tex] + [tex]\frac{1}{v}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{10}{165}[/tex] - [tex]\frac{10}{119}[/tex]

  = [tex]\frac{1190 - 1650}{19635}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{-460}{19635}[/tex]

⇒ v  = [tex]\frac{19635}{-460}[/tex]

       = -42.6848

    v = 42.7 cm

The fish would appear 42.7 cm on the left side from the front of the bowl.

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

t=5 secs

Answer:

Vrel= 0.75c

Explanation:

See attached file

Explanation:

F = ma

85 N = m (15 m/s²)

m ≈ 5.7 kg

Answer:

42000N

Explanation:

First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.

1) linear thermal expansion coef brass 19e-6 /K

∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm

Second part involves linear elasticity.

for brass, young's modulus is 15e6 psi or 100 GPa

cross-sectional area of rod is π(0.008)² = 0.0002 m²

F = EA∆L/L

F = (100e9)(0.0002)(0.00387) / (1.85)

F = 42000 or 42 kN

Answer:

1.The temperature of each sample will increase by the same amount

Explanation:

This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.

This is shown by the calculation below

Q = mcΔT

ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.

Since Q, m and c are the same for both substances, thus ΔT will be the same.

So, the temperature of each sample will increase by the same amount

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

[tex]F_{pull}=m_{T}.a[/tex]

[tex]m_{T}=\frac{F_{pull}}{a}[/tex]

[tex]m_{T}=\frac{3615}{1.516}[/tex]

[tex]m_{T}=2384.5[/tex]

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For [tex]m_{1}[/tex]:

The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

[tex]m_{1} = \frac{F_{12}}{a}[/tex]

[tex]m_{1} = \frac{1387}{1.516}[/tex]

[tex]m_{1}[/tex] = 915kg

For [tex]m_{2}[/tex]:

There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]

[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]

[tex]m_{2}[/tex] = 605kg

For [tex]m_{3}[/tex]:

[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]

[tex]m_{3} = 2384.5-1520.0[/tex]

[tex]m_{3}[/tex] = 864.5kg

Answer:

3 meters

Explanation:

Answer:4 times more energy will be striking the childbearing

Explanation:

Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4

Answer:

[tex]v_1 = 301 Hz[/tex]

[tex]v_2 = 601 \ \ Hz[/tex]

[tex]v_3 = 901 \ Hz[/tex]

Explanation:

From the question we are told that

     The  length of the string is  [tex]l = 90 \ cm = 0.9 \ m[/tex]

     The mass of the string is  [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]

     The  distance  from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]

    The tension is [tex]T = 540 \ N[/tex]

Generally the frequency is mathematically represented as

        [tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]

Where n is and integer that defines that overtones

i.e  n =   1 is for fundamental frequency

      n =  2   first overtone

       n =3   second overtone

Also  [tex]\mu[/tex] is the linear density of the string which is mathematically represented as

           [tex]\mu = \frac{m_s}{l}[/tex]

=>        [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]

=>       [tex]\mu = 0.003889 \ kg/m[/tex]

So for   n = 1

     [tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v_1 = 301 \ Hz[/tex]

So for  n =  2

     [tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v_2 = 601 \ Hz[/tex]

So for  n =  3

     [tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]

     [tex]v =901 \ Hz[/tex]

Answer:

The distance is  [tex]D = 45000 \ m[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 3.0 *10^{-4 } \ s[/tex]

Generally the speed of the radar is equal to the speed of light and this has a value  

      [tex]c = 3.0*10^{8} \ m /s[/tex]

Now the distance covered by the to and fro movement of the radar is mathematically evaluated as

      [tex]d = c * t[/tex]

=>    [tex]d = 3.0*10^{8} * 3.0*10^{-4}[/tex]

=>  [tex]d = 90000 \ m[/tex]

Therefore the distance between the radar antenna and the object is  

      [tex]D = \frac{d}{2}[/tex]

       [tex]D = \frac{ 90000}{2}[/tex]

      [tex]D = 45000 \ m[/tex]

The distance between the radar antenna and the object will be 45000 m.

A radar antenna is a device that sends out radio waves and listens for their reflections. The ability of an antenna to identify the exact direction in which an item is placed determines its performance.

The given data in the problem is;

t is the time=  3.0 x 10⁻⁴

d is the distance between the radar antenna and the object=?

c is the peed of light=3×10⁸ m/sec

The radar's speed is usually equal to the speed of light, and this has a value. The distance covered by the radars to and fro movement is now calculated mathematically as

[tex]\rm d= c \times t \\\\ \rm d= 3.0 \times 10^8 \times 3.0 \times 10^{-4} \\\\ d=90000 \ m[/tex]

As a result, the radar antenna's distance from the target is

[tex]\rm D=\frac{d}{2} \\\\ \rm D=\frac{90000}{2} \\\\ \rm D=\ 45000 \ m[/tex]

Hence the distance between the radar antenna and the object will be 45000 m.

To learn more about the radar antena refer to the link;

https://brainly.com/question/24067190

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU  is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU  - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

Answer:

space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              [tex]D = \frac{1.22 \lambda }{d }[/tex]

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

Answer:

The wavelength is  [tex]\lambda = 1805 nm[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]

     The  distance of the screen is  D  =  3.0  m

     The  fringe width is  [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]

Generally the fringe width for a bright fringe  is mathematically represented as

          [tex]y = \frac{ \lambda * D }{d }[/tex]  

=>     [tex]d = \frac{ \lambda * D }{ y }[/tex]

=>     [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]

=>     [tex]d = 0.000373 \ m[/tex]

Generally the fringe width for a dark fringe  is mathematically represented as

      [tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

Here  m = 0  for  first order dark fringe

   So  

         [tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

looking at which we see that   [tex]y_d = y[/tex]

         [tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]

=>    [tex]\lambda = 1805 *10^{-9} \ m[/tex]

=>    [tex]\lambda = 1805 nm[/tex]

Answer:

N₁ = 1800 turns

So, the side of the transformer that plugs into the outlet has 1800 turns.

Explanation:

The transformer turns ratio is given by the following equation:

V₁/V₂ = N₁/N₂

where,

V₁ = Voltage of outlet = 120 V

V₂ = Device Voltage = 3 V

N₁ = No. of turns on outlet side = ?

N₂ = No. of turns on side of device = 45

Therefore,

120 V/3 V = N₁/45

N₁ = (40)(45)

N₁ = 1800 turns

So, the side of the transformer that plugs into the outlet has 1800 turns.

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Answer:

The position on the y-axis where the magnetic field is zero is at y = 10 cm

Explanation:

The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by

B = μ₀i/2πR

Now, let y be the point where the magnetic fields of both wires are equal.

So, the magnetic field due to wire 1 carrying current i₁ = 73 A is

B₁ = μ₀i₁/2π(y - 3) and

the magnetic field due to wire 2 carrying current i₂ = 31 A is

B₂ = μ₀i₂/2π(13 - y)

At the point where the magnetic field is zero, B₁ = B₂. So,

μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)

cancelling out μ₀ and 2π, we have

i₁/(x - y) = i₂/(13 - y)

cross-multiplying, we have

(13 - y)i₁ = (y - 3)i₂

Substituting the values of i₁ and i₂, we have

(13 - y)73 = (y - 3)31

949 - 73y = 31y - 93

Collecting like terms, we have

949 + 93 = 73y + 31y

1042 = 104y

dividing through by 104, we have

y = 1042/104

y = 10.02 cm

y ≅ 10 cm

So, the position on the y-axis where the magnetic field is zero is at y = 10 cm