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3 years ago

Why is the SI system a consistent system?

2 answers:

7 0

S.I. System is same for all countries, so if any Scientist uses it, then another Scientist from any place around the world can figure it out, that what is he saying, how exactly is the magnitude of the object. That's why it is a Consistent system

Hope this helps!

ELEN [110]3 years ago

6 0

The SI system or System Internationale is measuring system used around the world with the exception of some previous dominions of the United Kingdoms.

The units used to meaure fields, like distance is measured in meters (m), translate to all languages with shortcase symbols and the actual unit word having no variency.

For example :
The unit for luminous intensity in English is candela (cd)
In French the words for luminous intensity is "Intensité lumineuse," however the unit is still candela (cd)

SI units are also used as the main components of derived units such as Newtons (N), Omhs (Ω)

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a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an

Mademuasel [1]

We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
$T_1 - W_p - W_s + T_2 =0$
where
$W_p = (90kg)(9.81 m/s^2)=883 N$ is the weight of the person
$W_s = (200kg)(9.81 m/s^2)=1962 N$ is the weight of the scaffold
Re-arranging, we can write the equation as
$T_1 = 2845 N-T_2$ (1)

2) Equilibrium of torques:
$T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0$
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using $W_p = 883 N$ and replacing T1 with (1), we find
$2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0$
from which we find
$T_2 = 1128 N$

And then, substituting T2 into (1), we find
$T_1 = 1717 N$

8 0

3 years ago

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. How long was the ball in the air? Rou

Slav-nsk [51]

Answer:

3.6 seconds

Explanation:

Given:

y₀ = y = 0 m

v₀ = 31 sin 35° m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²

0 = 17.78t − 4.9t²

0 = t (17.78 − 4.9t)

t = 0 or 3.63

Rounded to the nearest tenth, the ball lands after 3.6 seconds.

4 0

3 years ago

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0

3 years ago

Read 2 more answers

Two sinusoidal waves are moving through a medium in the positive x-direction, both having amplitudes of 7.00 cm, a wave number o

lys-0071 [83]

Answer:

0.99 m

Explanation:

Parameters given:

Amplitude, A = 7.00cm

Wave number, k = 3.00m^-1

Angular Frequency, ω = 2.50Hz

Period = 6.00 s

Phase, ϕ = π/12 rad

Note: All parameters are the same for both waves except the phase.

Wave 1 has a wave function:

y1(x, t) = Asin(kx - ωt)

y1(x, t) = 7sin(3x - 2.5t)

Wave 2 has a wave function:

y2(x, t) = Asin(kx - ωt + ϕ)

y2(x, t) = 7sin(3x - 2.5t + π/12)

π is in radians.

When Superposition occurs, the new wave is represented by:

y(x, t) = 7sin(3x - 2.5t) + 7sin(3x - 2.5t + π/12)

y(x, t) = 7[sin(3x - 2.5t) + sin(3x - 2.5t + π/12)]

Using trigonometric function:

sin(a) + sin(b) = 2cos[(a - b)/2]sin[(a + b)/2]

Where a = 3x - 2.5t, b = 3x - 2.5t + π/12

We have that:

y(x, t) = (2*7)[cos(π/24)sin(3x - 2.5t + π/24)]

Therefore, when x = 0.53cm and t = 2s,

y(x, t) = (2*7)[cos(π/24)sin{(3*0.53) - (2.5*2)+ π/24}]

y(x, t) = 14 * 0.9914 * 0.0713

y(x, t) = 0.99 m

The height of the resultant wave is 0.99cm

5 0

3 years ago

A pencil has a density of 0.875 g/ml. It has a volume of 4.0 ml. Find the mass

loris [4]

Answer:

To find the mass using density and volume we just multiply them against each other which causes ml to cancel and just leaves us with grams which represents how much the item weights.

$mass=density*volume$

$mass=0.875\frac{g}{ml}*4.0\ ml$

$mass=3.5\ g$

Therefore, our final answer is that our pencil weight 3.5 grams

<u><em>Hope this helps! Let me know if you have any questions</em></u>

3 0

2 years ago