PHYSICS COLLEGE

An alternating current is supplied to an electronic component with a rating that the voltage across it can never, even for an instant, exceed 16 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit?A)8 sqrt 2 V

B) 16 sqrt 2 V

C) 256 V

D) 8

Answers

Answer 1

Answer:

Answer:

A) $V_(rms)=8√(2) V$

Explanation:

Maximum voltage = $V_(max)=16 V$

Maximum voltage and rms voltage are related to each other by

$V_(max)=V_(rms) * √(2) \nV_(rms)=(V_(max))/( √(2))\nV_(rms)=(16)/(√(2)) \nV_(rms)=8√(2) V$

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Define reflection of sound?
Two loudspeakers, 4.0 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m.a. What is the frequency of the sound?b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
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The direct sunlight at Earth's surface is about 1050 W/m2 . Compute mass lost by Sun in a thousand years as a fraction of Earth Mass? The lost Mass of sun/in a million yrs = ……?…..% of Earth Mass.

Answers

Answer:

Explanation:

Energy falling on 1 m² surface of earth per second = 1050

Energy in one million years on 1 m²

= 1050 x 60 x 60 x 24 x 365 x 10⁶ = 3.311 x 10¹⁶ J

In order to calculate total energy coming out of the surface of the sun , we shall have to sum up this energy for the while spherical surface of imaginary sphere having radius equal to distance between sun and earth.

Area of this surface = 4π R² = 4 X 3.14 X (149.6 X 10⁹ )²

= 2.8 X 10²³ m²

So total energy coming out of the sun = 2.8 x 10²³ x 3.311 x 10¹⁶

= 9.271 x 10³⁹ J

From the formula

E = mc² { energy mass equivalence formula }

m = E / c² = $(9.271 *10^(39))/(9 * 10^(16))$

1.03 x 10²³ kg

mass of earth = 5.972 x 10²⁴

Answer in percentage of mass of earth

= $(1.03*10^23)/(5.972*10^(24))*100$

= 1.72 %

A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?

Answers

(a) The capacity will be "4006.4 veh/h".

(b) The speed at capacity be "25 mph".

(c) The density will be "299 veh/mi".

Given:

$q = 50k - 0.156 k^2$

At max. flow density,

$(dd)/(dk) =0$

$((dq)/(dt) ) = 50-0.321k =0$

(a)

→ $k = ((50)/(0.312) )$

$= 160.3 \ or \ 160 \ veh/mi$

By substituting the value,

→ $q = 50k-0.156k^2$

$= 50* 160.3-0.156* (160.3)^2$

$= 4006.4 \ veh/h$

(b)

The speed will be:

→ $U = (q)/(k)$

$= (4006.4)/(160.3)$

$= 25 \ mph$

(c)

The density be:

→ $1001.6 = 50k-0.156k^2$

$0.156k^2-50k+1001.6 =0$

$k = 21.5 \ veh/mi \ or \ 299 \ veh/mi$

Thus the responses above are correct.

Find out more information about density here:

brainly.com/question/6838128

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

A mass of 0.14 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.28 m)cos[(8 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass .

Answers

Answer:

The amplitude of oscillation for the oscillating mass is 0.28 m.

Explanation:

Given that,

Mass = 0.14 kg

Equation of simple harmonic motion

$x(t)=(0.28\ m)\cos[(8\ rad/s)t]$ ....(I)

We need to calculate the amplitude

Using general equation of simple harmonic equation

$y=A\omega \cos\omega t$

Compare the equation (I) from general equation

The amplitude is 0.28 m.

Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.

A block, initially at rest, has a mass m and sits on aplane inclined at angle (theta). It slides a distance d before hitting a spring and compresses the spring by a maximum distance of xf. If the coefficient of kinetic friction between the plane and block is uk, then what is the force constant of the spring?

Answers

Answer: k = ma + uk×mgcosθ/ xf

Explanation: The body is placed on a frictionless inclined ramp.

The weight of the object has 2 components, horizontal component (mgsinθ) and vertical component (mgcosθ).

The horizontal component of weight is responsible for making tje object slide down the plane even with no applied force.

So from newton's second law of motion

mgsinθ - uk×R = ma

Where uk = coefficient of kinetic friction.

R = normal reaction = mgcosθ

mgsinθ - uk×mgcosθ = ma

mgsinθ = ma + uk×mgcosθ

mgsinθ is the applied force in this case. This applied force compresses a spring.

According to hooke's law,

F =ke

Where F = ma + uk×mgcosθ, e =xf

F = applied force , e = extension and k = spring constant.

k = F/e

k = ma + uk×mgcosθ/ xf

An element emits light at two nearly equal wavelengths, 577 nm and 579 nm If the light is normally incident on a diffraction grating with 2000 lines/cm., what is the distance between the 3rd order fringes of the two wavelengths on a screen 1 m from the grating?

Answers

Answer:

Explanation:

d = width of slit = 1 / 2000 cm =5 x 10⁻⁶ m

Distance of screen D = 1 m.

wave length λ₁ and λ₂ are 577 x 10⁻⁹ and 579 x 10⁻⁹ m.respectively.

distance of third order bright fringe = 3.5 λ D/d

for 577 nm , this distance = 3.5 x 577 x 10⁻⁹ x 1 /5 x 10⁻⁶

= .403 m = 40.3 cm

For 579 nm , this distance = 3.5 x 579 x 1 / 5 x 10⁻⁶

= 40.5 cm

Distance between these two = 0.2 cm.

What displacement do I have if I travel at 10 m/s E for 10 s? A. 1 m E B. 1 m C. 100 m D. 100 m E Scalar quantities include what 2 things? A. Number and direction B. Numbers and units C. Units and directions D. Size and direction What measures distance in a car? A. Odometer B. Pressure gauge C. Speedometer D. Steering wheel What displacement do I have if I travel 10 m E, then 6 m W, then 12 m E? A. 28 m E B. 16 m E C. 16 m D. 28 m