Does Archimedes principle hold good in a vessel in free fall

solve the following system by any method 2x - 6y = 24 ... -5x + 6y = -6 ... A. (0,-6) B. (4,-1) C. (-6,-6) D. (6,-1)

vodka [1.7K]

Using elimination, the answer is C.

5 0

4 years ago

A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflectio

bogdanovich [222]

Answer:

0.79

Explanation:

Using Snell's law, we have that:

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

=> 1.0003 * sin32 = n(2) * sin42

n(2) = (1.0003 * sin32) / sin42

n(2) = 0.79

3 0

3 years ago

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In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago

A cart loaded with bricks has a total mass of 20.4 kg and is pulled at constant speed by a rope. The rope is inclined at 26.1 ◦

Readme [11.4K]

Answer:

Normal force, N = 154.5 N

Explanation:

Given that,

Total mass of the cart, m = 20.4 kg

Angle of inclination, $\theta=26.1^{\circ}$

Distance moved, d = 20.1 m

The coefficient of kinetic friction between ground and cart is 0.6

The value of acceleration due to gravity, $g=9.8\ m/s^2$

Let N is the normal force exerted on the cart by the floor. It is given by :

$N=mg-T\ sin\theta$

$T\ cos\theta=\mu N$

$T\ cos\theta=\mu (mg-T\ sin\theta)$

$T\ cos(26.1)=0.6 (20.4\times 9.8-T\ sin(26.1))$

T = 103.23 N

So, the normal force is :

$N=20.4\times 9.8-103.23\ sin(26.1)$

N = 154.5 N

So, the normal force exerted on the cart by the floor is 154.5 N. Hence, this is the required solution.

4 0

3 years ago

What do astronomers call a system that is composed of more than two stars

andrew-mc [135]

Astronomers call a system with more than two stars a constellation.

3 0

3 years ago

Does Archimedes principle hold good in a vessel in free fall​

Does Archimedes principle hold good in a vessel in free fall