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zhuklara [117]
3 years ago
8

CO + 7H2 - C3Hg + 3H20

Physics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

Espanol 7h2 that's c coupon

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Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco
11Alexandr11 [23.1K]

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

5 0
3 years ago
Describe how the gravitational force between 2 objects depends on the mass of the objects and the distance between them. your an
ollegr [7]
The strength of the gravitational force between two objects depends<span> on </span>two<span>factors, </span>mass<span> and </span>distance<span>. the </span>force<span> of gravity the </span>masses<span> exert on each other. If one of the </span>masses<span> is doubled, the </span>force<span> of gravity </span>between<span> the </span>objects<span> is doubled. increases, the </span>force<span> of gravity decreases</span>
3 0
3 years ago
I hope y'all can read this I need help​
Ivanshal [37]

Answer:

its basically on where u live is more hot or cold. Is rain and snow common or uncommon. where is this. is it easier to live or harder to live wherever u live at. list 12 different organisms in the place u live. and one fun fact its basically about what u see where u in the city u live

4 0
3 years ago
Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190
julia-pushkina [17]

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C

8 0
3 years ago
A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

We have the following data:

Speed of the wind from East to West: 6.68 m/s

Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

\theta=53.31\°

6 0
4 years ago
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