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MrRissso [65]
3 years ago
12

Suppose you were producing zinc chloride by the reaction, Zn + 2 HC1 ---> ZnC12 + H2 If you started with 4 molecules of zinc,

you would need _____molecules of hydrogen chloride and you would produce _____ molecules of zinc chloride.
Physics
1 answer:
harkovskaia [24]3 years ago
8 0
2 molecules of zinc chloride and 2 molecules of hydrogen chloride
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On a cross-country trip, a couple drives 500 mi in 10 h on the first day, 380 mi in 8.0 h on the second day, and 600 mi in 15 h
Zepler [3.9K]

We know that the average speed is simply the ratio of the total distance travelled over the total duration of the trip.

total distance = 500 mi + 380 mi + 600 mi

total distance = 1,480 mi

 

total time = 10 h + 8 h + 15 h

total time = 33 h

 

So the average speed is therefore:

average speed = 1,480 mi / 33 h

<span>average speed = 44.85 mi / h</span>

8 0
3 years ago
In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.
pickupchik [31]
In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because of friction. 
AMA (Actual mechanical advantage) is found by dividing output force by effort force. The actual mechanical advantage will always be less than the ideal mechanical advantage. The ideal mechanical advantage assumes perfect efficiency which doesn't account for friction, while actual mechanical advantage does. Therefore; the IMA is always greater than the actual mechanical advantage because all machines must overcome friction. 
5 0
3 years ago
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
Pluto has been reassigned and is now a dwarf planet. Why did scientists think this reassignment was necessary? If you were a sci
kap26 [50]

Answer:

Explanation:

we humans have our own ify classification for celestial objects, most people are saddened that pluto is not a planet anymore altho it hasn't changed at all.

scientist say that if an object is going to be considered a planet it must fill in these three checkboxes:

You must be spherical, you must orbit a star, and you must have already cleared your path or debris.

Pluto fills in the first two boxes but it does orbit in the keyperbelt and there are 5 other objects just like it. this is why pluto has been dubbed a dwarf planet.

6 0
3 years ago
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a. A lifted parcel of air will be colder (heavier) that the air surrounding it. Because of this fact, the lifted parcel will ten
kvv77 [185]
I think the answer is A because it’s a better explanation
5 0
3 years ago
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