; (b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.

Answer:

658.2 nm

Explanation:

Since the galaxy is moving at relavitistic speed, we use the equation for relativistic Doppler shift of light.

So, the wavelength of light observed on the Earth is λ

λ = λ'([tex]\sqrt{\frac{ 1 + \frac{v}{c} }{1 - \frac{v}{c} } }[/tex])

where  λ' = wavelength of light emitted by galaxy = 656 nm, v = speed of galaxy = 1000 km/s (positive since the galaxy is moving away from the Earth) and c = speed of light = 300000 km/s

So, substituting the values of the variables into the equation, we have

λ = λ'(√[{1 + (v/c)}/(1 - (v/c)]

λ = 656 nm(√[{1 + (1000 km/s/300000 km/s)}/(1 - (1000 km/s/300000 km/s)]

λ = 656 nm(√[{1 + 1/300}/(1 - 1/300]

λ = 656 nm(√[{(300 + 1)/300}/{(300 - 1)/300}]

λ = 656 nm(√[{(301)/300}/{(299)/300}]

λ = 656 nm(√[301/299])

λ = 656 nm(√1.0067)

λ = 656 nm × 1.0033

λ = 658.19 nm

λ ≅ 658.2 nm

So, the wavelength observed on Earth is 658.2 nm

Answer:

a) 70, 95

b) 95-70= 25cc

c) density= mass/volume

102/25

=4.08g/cc

Answer:

Velocity is the rate of change of displacement.

Answer:

5.0cm

Explanation:

To get the height of the image, we will use the magnification formula as shown:

m = Hi/H = v/u

Hi is the image height

H is the object height

v is the image distance

u is the object distance

First, we need to get the image distance v

Using the mirror formula;

1/f = 1/v+ 1/u

1/-6 = 1/v + 1/12

1/v = -1/12 - 1/6

1/v = -1-2/12

1/v = -3/12

v = 12/-3cm

v = -4cm

Next is to get the image height Hi

Using the expression;

Hi/H = v/u

Hi/15 = 4/12

Hi/15 = 1/3

3Hi = 15

Hi = 15/3

Hi = 5.0cm

Hence the image height is 5.0cm

Answer:

D. 5.0 cm

Explanation:

got it correct on the test

Answer:  An effect of continental drift is causing tectonic plates resting upon the convecting mantle to move which results in natural disasters like earthquakes, volcanic eruptions, and more.

Answer:

C

Explanation:

Answer:

A)

Explanation:

it keeps the electric current flowing in one direction

Answer:

Yes but no

Explanation:

Here the kid cant say that he has created s magnetic field as iron fillings do not have magnetic properties

Answer:

a. Energy lost, Q = 16,800 Joules.

b. Power = 28 J/s

c. Time, t = 2357.14 seconds

d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

Explanation:

Given the following data;

a. To find the energy lost by the water as it cools to 0 degree celsius;

Mathematically, heat capacity is given by the formula;

[tex] Q = mcdt [/tex]

Where;

dt = T2 - T1

dt = 20 - 0

dt = 20°C

We know that the specific heat capacity of water is equal to  4200 J/kg°C

Substituting the values into the formula, we have;

[tex] Q = 0.20 * 4200 * 20 [/tex]

Energy lost, Q = 16,800 Joules.

b. To find the average rate at which the water is losing energy in J/s by using the following formula;

[tex] Power = \frac {energy}{time} [/tex]

First of all, we would have to convert the value of time in minutes to seconds.

Conversion:

1 minute = 60 seconds

10 minutes = X seconds

Cross-multiplying, we have;

X = 60 * 10

X = 600 seconds

Substituting the values into the formula, we have;

[tex] Power = \frac {16800}{600} [/tex]

Power = 28 J/s

c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;

We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.

Mathematically, the latent heat of fusion is calculated by using the formula;

Energy, Q = ml = pt

Substituting the values into the formula, we have;

0.20 * 3.3 * 10⁵ = 28 * t

0.20 * 330000 = 28t

66000 = 28t

[tex] t = \frac {66000}{28} [/tex]

Time, t = 2357.14 seconds.

d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

Answer:

A) The population of prey is greater than that of predator

Answer:

Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.

Answer:

Pressure and velocity are inversely proportional to each other because if pressure increase, the velocity decrease to keep the algebraic sum of potential energy, kinetic energy and pressure constant.

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

[tex]v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2[/tex]

[tex]v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2[/tex]

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

[tex]v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

[tex]v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

The final relative velocity of the satellite, [tex]v_f[/tex] = v₁ + v₂

∴ [tex]v_f[/tex] = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, [tex]v_f[/tex] = 0.190 m/s

Answer:

Explanation:

                    The right side of your heart receives oxygen-poor blood from your veins and pumps it to your lungs, where it picks up oxygen and gets rid of carbon dioxide. The left side of your heart receives oxygen-rich blood from your lungs and pumps it through your arteries to the rest of your body.

        #I AM ILLITERATE

Answer:

1. The efficiency of the radiator is 90 %

2. The efficiency of the torch is 65 %

3. The efficiency of the car is 40 %

4. The efficiency of the energy saver is 70 %

5. The efficiency of the speaker is 50 %

Explanation:

Efficiency = (Useful energy out ÷ Total energy in) × 100 J

1. Useful energy = 900 J

The total energy in = 1000 J

The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %

2. Useful energy = 65 J

The total energy in = 100 J

The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %

3. Useful energy = 4,000 J

The total energy in = 10,000 J

The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %

4. Useful energy = 700 J

The total energy in = 1,000 J

The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %

5. Useful energy = 50 J

The total energy in = 100 J

The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %

Answer:

136 m.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 0.80 s

Speed of sound (v) = 340 m/s

Distance (x) =?

The distance of the player from the wall can be obtained as follow:

v = 2x / t

340 = 2x / 0.80

Cross multiply

2x = 340 × 0.80

2x = 272

Divide both side by 2

x = 272 / 2

x = 136 m

Thus, the distance of the player from the wall is 136 m

Answer:

The number of turns in the solenoid is 19506.

Explanation:

Length of solenoid, L = 0.28 m

Magnetic field, B = 7 T

Current, I = 80 A

Let the number of turns is N.

The formula used to find the magnetic field is

[tex]B = \frac{\mu 0N I}{L}\\\\7 = \frac{4 \times 3.14\times 10^{-7}\times N\times 80}{0.28}\\\\N = 19506[/tex]

Answer:

given,

load = 400 N

effort = 100 N

load distance = 20 cm

we know that ,

E*Ed = L*Ld

=100 N* Ed = 400N * 20 cm

=100N * Ed = 8000N/cm

= Ed =( 8000N/cm ) / 100N

= Ed = 80 cm

b. soln.

given,

load = 600 N

load distance = 2 m

effort = 300 N

effort distance = ?

we know that ,

= E *Ed = L * Ld

= 300N * Ed = 600N * 2 m

= 300N * Ed = 1200N/m

=Ed =( 1200N/m ) / 300 N

= Ed = 4 m

C. soln.

given,

load = 600 N

load distance =20 cm

effort = 200 N

effort distance = ?

M.A = ?

V.R = ?

Efficiency = ?

we know that ,

= E *Ed = L *Ld

= 200N * Ed = 600 N * 20 cm

=200 N *Ed = 12000 N/cm

=Ed = ( 12000 N/cm) / 200 N

= Ed = 60 cm

Also,

M.A = load / effort

=600 N / 200 N

= 3

V.R = Ed/ Ld

= 60 cm / 20 cm

= 4

efficiency = ( M.A / V.R ) 100 %

= ( 3 / 4 ) 100%

= 75 %

d. soln.

given,

load = 600 N

load distance = 0.5 m

effort distance = 2.5 m

effort = ?

we know that ,

= E * Ed = L * Ld

= E * 2.5 m = 600 N * 0.5 m

= E * 2.5 m = 300 N / m

= E = ( 300 N / m ) / 2.5 m

= E = 120 N

here's the answer to your question

Answer:

Answer:

Higher-level human resources related to engineering are those who:

Medium Level engineers will normally comprise the following:

Cheers

Answer:

SI unit is an internationally accepted system and have the same value all over the world .

I don't know the importance sotry

Answer:

SI unit is an internationally accepted system that have same value all over the world.

Answer:

The mass of platinum iridium rod whose diameter and height are equal an kept at international bureau of weight and measurement in Paris of France is called as one standard kg.

Answer:

Explanation:

As seen, batteries have chemical energy in them stored which when used in any electronics produces electricity.

Thus, we can conclude that a battery converts chemical energy into electric energy.

Answer:V₁=300ml

T₁=27°C

V₂=?

T₂= -3°C

as we know

V₁T₁=V₂T₂

By putting values in formula

300ml×27°C=V₂×(-3°C)

300ml×27°C/-3°C=V₂

8100ml/-3=V₂

-2700ml=V₂

or V₂=  -2700ml

Answer:

θ = 66º

Explanation:

This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball

           a = v² / r

the radius of the circle is

          sin θ = r / L

          r = L sin θ

we substitute

          a = v² /L sin θ

now let's write Newton's second law

vertical axis

            T_y -W = 0

             T_y = W

radial axis

            Tₓ = m a                 (1)

let's use trigonometry for the components of the string tension

             cos θ = T_y / T

             sin θ = Tₓ / T

             Tₓ = T sin θ

we substitute in 1

            T sin θ = [tex]\frac{m \ v^2}{L \ sin \theta}[/tex]

             T L sin² θ = m v²

we write our system of equations

             T cos θ  = m g

             T L sin ² tea = m v²

we divide the two equations

             L [tex]\frac{sin^2 \theta}{cos \theta}[/tex] = v² / g

             (1 -cos²)/ cos θ  = [tex]\frac{v^2 }{g \ L}[/tex]

             1 - cos² θ  =  [tex]\frac{4.75^2}{9.81 \ 2.37}[/tex]   cos θ

             cos² θ + 0.97044 cos θ -1   = 0

we change variable    cos  θ = x

             x² + 0.97044 x - 1 =0

             x= [tex]\frac{-0.97 \pm \sqrt{0.97^2 - 4 1} }{2}[/tex]

           since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m

           T sin θ = [tex]\frac{m \ v^2}{ r}[/tex]

           T cos θ  = m g

resolved

           tan θ =  [tex]\frac{v^2}{ r g}[/tex]

           θ = tan⁻¹ ( 4.75²/ 1 9.81)

           θ = 66º

Answer:

It is the inability of the body to change by itself its state of rest or uniform motion or direction. Types of Inertia- It is of three types-(1)Inertia of rest (2) Inertia of motion(3) Inertia of direction. (1) Inertia of rest - It is the inability of the body to change by itself its state if rest.

[tex]\\ \sf\longmapsto P.E=mgh[/tex]

[tex]\\ \sf\longmapsto P.E=150(2)(10)[/tex]

[tex]\\ \sf\longmapsto P.E=3000J[/tex]

In 2nd case

[tex]\\ \sf\longmapsto P.E=200(2)(10)[/tex]

[tex]\\ \sf\longmapsto P.E=4000J[/tex]

We can observe that