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3 years ago

Một nguồn O phát sóng cơ dao động theo phương trình u = 4cos(4pi t - pi/4). Tốc độ truyền sóng là 1m/s.

Physics

1 answer:

brilliants [131]3 years ago

5 0

Answer:

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ykrjrhrnirjtjjtt

tkrjthr74uu3jt

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Explanation:

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jebfufkr

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The distance from Earth to the star Epsilon Eridani is about 10.5 light years. Which of the following statements is true?

Simora [160]

<span>The correct answer is B. - It would take a ray of light 10.5 light years to travel from Earth to Epsilon Eridani, or vice-versa. Using our current technology it would take far longer than 21.0 years for a space ship from Earth to travel that far - I would have to guess many hundreds of years.</span>

4 0

3 years ago

Read 2 more answers

The function s(t) represents the position of an object at time t moving along a line. Suppose s (2 )equals 146and s (6 )equals

sineoko [7]

Answer:

v(t) = 27 units

Explanation:

The function s(t) represents the position of an object at time t moving along a line such that,

$s(2)=146$

and

$s(6)=254$

We need to find the average velocity of the object over the interval of time [2,6]. The velocity of the object is equal to the total distance divided by time. It is given by :

$v(t)=\dfrac{s(6)-s(2)}{6-2}$

$v(t)=\dfrac{254-146}{6-2}$

v(t) = 27 units

So, the average velocity of the object is 27 units. Hence, this is the required solution.

5 0

3 years ago

What is the net charge of a metal ball if there are 21,749 extra electrons in it?

pickupchik [31]

Answer:

$Q=3.47\times 10^{-15}\ C$

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is $q=1.6\times 10^{-19}\ C$

To find the net charge if there are n number of extra electrons is :

Q = n × q

$Q=21749\times 1.6\times 10^{-19}\ C$

$Q=3.47\times 10^{-15}\ C$

So, the net charge on the metal ball is $3.47\times 10^{-15}\ C$ . Hence, this is the required solution.

6 0

3 years ago

The pages of a book are numbered 1 to 200 and each

never [62]

Answer:

10.4mm

Explanation:

2 pages = 1 leaf

200 pages = 100 leaves

100 × 0.10 = 10 mm thickness

Total thickness = 2(0.20) +10 = 0.4+10 = 10.4mm

6 0

3 years ago

g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located

Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is: P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P /RT ⋯ ⋯( I )

Here

m is the mass of the air,

v is the volume of the air,

P is the atmospheric pressure,

T is the standard temperature at the atmospheric pressure and

R is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101 kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here A is the area of the windmill

Expression to calculate the A is

A = π /4 d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94 k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94 k g / s ( (6m/s)²/2 )

P w = 160144.92 W × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0

3 years ago