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2 years ago

If you are sitting 8.20 m from a

Physics

1 answer:

Elanso [62]2 years ago

8 0

Answer:

Explanation:

There are a lot of assumptions needed for the calculation:

first is the speaker is a point source

second the speaker is 100% efficient in its output

third the sound output propagates as a sphere

Intensity = Power / Surface Area

= 70 / (4*pi*8.20^2)

= 0.082W/m^2

Converting into dB:

= 10*log ( Intensity / (1*10^-12))

= 10*log(0.0828/ (1*10^-12))

= 109 dB

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A block lies on a plane raised an angle θ from the horizontal. Three forces act upon the block: F⃗ w, the force of gravity; F⃗ n

Verdich [7]

Answer:

YFy = 0 = Ffsinθ + Fncosθ - Fw

Explanation:

From the base of the vector Fn, draw a vertical line. the small angle between this line and Fn is also theta. The component of Fn in the vertical direction is Fncos(theta).

Take a moment to picture extreme cases. Sine is 0 at 0 and 1 at 90. Cosine is 1 at 0 and 0 at 90.

Tilt the incline so that the box is on a flat surface. How much of the gravitational force is along the x direction of the floor.

5 0

2 years ago

The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o

blsea [12.9K]

Answer:

$F_f=840N$

Explanation:

From the question we are told that

Weight of fireman $W_f= 87.5kg$

Pole distance $D=4.10m$

Final speed is $V_f 1.75m/s$

Generally the equation for velocity is mathematically represented as

$v^2 = v_0^2 + 2 a d$

Therefore Acceleration a

$a'= v^2 / 2 d$

$a'= 0.21m/s^2$

Generally the equation for Frictional force $F_f$ is mathematically given as

$F_f=m*a$

$F_f=m*(g-0.21)$

$F_f=87.5*(9.81-0.21)$

Therefore

$F_f=840N$

6 0

2 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

2 years ago

Nucleus A decays into the stable nucleus B with a half-life of 22.07 s. At t=0 s there are 1,293 A nuclei and no B nuclei. At wh

Alexeev081 [22]

Answer:

29.38 seconds

Explanation:

Half life, T = 22.07 s

No = 1293

Let N be the number of atoms left after time t

N = 1293 - 779 = 514

By the use of law of radioactivity

$N=N_{0}e^{-\lambda t}$

Where, λ is the decay constant

λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second

so,

$514=1293e^{-0.0314t}$

$2.5155=e^{0.0314t}$

take natural log on both the sides

0.9225 = 0.0314 t

t = 29.38 seconds

5 0

2 years ago

Please answer this question??

Varvara68 [4.7K]

ANSWER: Itna Bada answer Kisi Ko Pata chalega

EXPLANATION : please Manje brainliest karo man Jay he brainiest Karo .

5 0

2 years ago