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2 years ago

3. Silk fibre is obtained from______

Physics

1 answer:

USPshnik [31]2 years ago

8 0

Answer:

Silk fibre is obtained from <u>silkworms</u>

whereas artificial silk is obtained by

chemical treatment of <u>wood pulp/cellulose.</u>

Explanation:

Artificial silk, or rayon, is made from cellulose, obtained from cotton or wood pulp.

So any one of them in 2nd blank is correct.

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Light traveling from water to a gemstone strikes the surface at an angle of 80.0º and has an angle of refraction of 15.2º . (a)

7nadin3 [17]

Answer:

The the speed of light in the gemstone is $v= 0.599*10^8 m/s$

The unreasonable thing about this is that the speed of ligth in the gemstone is too low

the speed is 19% of the speed of light which is very low

One main unreasonable or inconsistent factor is that the assumption that the differnce between the angle of incidence and angle of refraction is very large

Explanation:

From the question we are told that

The angle of incidence $i = 80^o$

The angle of refraction $r = 15.2^o$

From Snell's law we have ,

$n_1 sin \theta_1 = n_2 sin \theta_2$

Where $n_1$ is the refractive index of the first medium (water) with a constant value of $n_1 = 1..333$

$n_2$ is the refractive index of the second medium (gem stone)

$\theta_1$ is the angle between the beam and perpendicular surface of the first medium

$\theta_2$ is the angle between the beam an the perpendicular surface of the second medium

Making the $n_2$ the subject of the formula

$n_2 = n_1 \frac{sin \theta_1}{sin \theta_2}$

$= (1.333)(\frac{sin (80.0)}{sin 15.2} )$

$= 5.007$

Generally refractive index of a material is mathematically represented

$n = \frac{c}{v}$

Where c is the speed light

v is the speed of light observed in a medium

Making v the subject

$v = \frac{c}{n}$

substituting value for gem stone

$v = \frac{3.0*10^8}{5.007}$

$v= 0.599*10^8 m/s$

7 0

3 years ago

A rocket is launched from Earth (mass ME, radius RE) with velocity v° and reaches radial distance r=6RE with velocity v°/10. Exp

Aliun [14]

The velocity, expressed in terms of ME, RE is given as $V_0 = \sqrt{98.99R_E}$ .

The maximum height that the rocket could reach if launched vertically is H = (¹/₂v₀²)/g.

<h3>Maximum height of the rocket</h3>

The maximum height reached by the rocket can be modeled using conservation of energy as shown below;

P.Ei + K.Ei = K.Ef + P.Ef

$M_EgR_E + \frac{1}{2} M_EV_0^2= \frac{1}{2} M_E(\frac{V_o}{10} )^2+ M_Eg(6R_E)\\\\\frac{1}{2} M_EV_0^2 - \frac{1}{2} M_E(\frac{V_o}{10} )^2 = M_Eg(6R_E) - M_EgR_E\\\\0.495M_EV_0^2= 5gM_ER_E\\\\0.495V_0^2= 5gR_E\\\\V_0 = \sqrt{\frac{5gR_E}{0.495} } \\\\V_0 = \sqrt{98.99R_E}$

<h3>Maximum height when it is launched vertically</h3>

P.E = K.E

mgH = ¹/₂mv²

H = (¹/₂v₀²)/g

Learn more about conservation of energy here: brainly.com/question/166559

5 0

2 years ago

A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$

$v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14 \ m/s$

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

3 0

3 years ago

A 2.0 kg book is lying on a 0.78 m -high table. You pick it up and place it on a bookshelf 2.1 m above the floor.Part ADuring th

zloy xaker [14]

Given data:

* The mass of the book is 2 kg.

* The initial height of the book is 0.78 m.

* The final height of the book is 2.1 m.

Solution:

(A). The work done by the gravity on the book is,

$W=mg(h_f-h_i)$

where m is the mass, g is the acceleration due to gravity, h_i is the initial height and h_f is the final height,

The work is done in moving the object in upward direction where as the gravitational force is acting in the down ward direction. Thus, the value g (acceleration due to gravity) is taken as negative in this case.

Substituting the known values,

$\begin{gathered} W=2\times(-9.8)\times(2.1-0.78) \\ W=-19.6\times1.32 \\ W=-25.9\text{ J} \end{gathered}$

Thus, the work done by the gravitational force is -25.9 J.

(B). The work done by the hand on the moving the book is,

$\begin{gathered} W_1=-W \\ =25.9\text{ J} \end{gathered}$

Thus, the work done by the hand on the book is 25.9 J.

7 0

10 months ago

When water freezes, its volume increases by 9.05% (that is, equation). What force per unit area is water capable of exerting on

PIT_PIT [208]

Answer:

1.991 × 10^(8) N/m²

Explanation:

We are told that its volume increases by 9.05%.

Thus; (ΔV/V_o) = 9.05% = 0.0905

To find the force per unit area which is also pressure, we will use bulk modulus formula;

B = Δp(V_o/ΔV)

Making Δp the subject gives;

Δp = B(ΔV/V_o)

Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²

Thus;

Δp = 2.2 × 10^(9)[0.0905]

Δp = 1.991 × 10^(8) N/m²

3 0

3 years ago