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alekssr [168]
2 years ago
13

Importance of international bureau of weights and measures Inthe world​

Physics
1 answer:
vaieri [72.5K]2 years ago
4 0

Answer:

The international bureau of weights and measures is an international organization to bring about the unification of different measurement systems to make or establish international standards and prototypes, etc.

If my answer helped, kindly mark me as the Brainliest!!

Thank You!!

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You run<br> completely around a 400m track in<br> 80s. What was your average velocity?
dalvyx [7]

Answer:

V=?

S=400m

t=80s

V=S/t

V=400/80

V=5m/sec

6 0
3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the
xeze [42]
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is 
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
6 0
3 years ago
Which statement is a hypothesis?
shusha [124]
I believe the answer should be D
8 0
3 years ago
Read 2 more answers
A communications channel has a bandwidth of 4,000 hz and a signal-to-noise ratio (snr of 30 db. what is the maximum possible dat
levacccp [35]
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as 

C = B × log₂(1 + S/N)

where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.

Since the given SN ratio is in decibels, we must first express it as a ratio with no units as

SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000

Now that we have S/N, we can solve for its capacity (in bits per second) as 

C = 4000 × log₂(1 + 1000) 
C = 39868.91 bps

Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.

Answer: 40 kbps

 
4 0
2 years ago
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