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meriva
3 years ago
11

Which measurements are equal to 321 decimeters

Physics
1 answer:
vlabodo [156]3 years ago
7 0
You are currently converting Distance and Length units from Centimeters to Feet 321 Centimeters (cm) = 10.5315 Feet (ft) This is a hard one but see if this helps if not let me now and i can try again..
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In the above lightwave the property labeled a determines which characteristic of visible light
kap26 [50]
The light must be either very dim or else non-existent. 
We can't see the light wave or the label.
7 0
3 years ago
Which type of electronic exchange connects buyers with sellers within a specific industry?
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I think the correct answer would be horizontal exchanges or market. It is a type of market wherein a service or a product would meet a need of a very wide range of consumers from different sectors. Hope this answers the question. Have a nice day.
5 0
2 years ago
A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a
Aleks [24]

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

3 0
3 years ago
What is the mechanical advantage of the wheel and axle shown below?
BigorU [14]

The correct answer is A. 32.5

Mechanical advantage is the ratio of force that is input into a machine to the force output.

Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.

MA=R/r where R is the radius of the wheel and  r is the radius of the axle.

Substituting for the values in the question gives:

MA=26cm/0.8cm

    =32.5

6 0
3 years ago
Read 2 more answers
A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
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