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Troyanec [42]
3 years ago
12

I need to show my work as well but on the computer so, please show work for how you got the answers. Thank you!

Physics
1 answer:
balandron [24]3 years ago
4 0

Answer:

1) 1H_2 + 1Cl_2 => 2HCl

2) 2Al + 6HCl => 2AlCl_3 + 3H_2

3) 1Ca_2Br_2 + 2NaCO_3 => 2CaCO_3 + 2NaBr

4) 3NaOH + 1FeCl_3 => 3NaCl + 1Fe(OH)_3

Explanation:

You might be interested in
¿Quien desarrolló el método físico experimental?
Viefleur [7K]

Answer:

Explanation:

El 15 de febrero de 1564 nacía Galileo Galilei (1564-1642). Galileo Galilei es una de las figuras claves de la historia de la Ciencia, pudiéndosele considerar el primero que aplicó el método científico experimental-matemático.

7 0
3 years ago
PLZ HELP!! WILL MARK BRAINLIEST!!
natali 33 [55]
As waves get closer to the beach they increase in energy
8 0
3 years ago
You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget
aleksklad [387]

Answer:

35.7 m

Explanation:

Let

\mid A\mid=18.5 m

\mid B\mid=41 m

We have to find the distance between Joe's and Karl'e tent.

A_x=Acos\theta

A_y=Asin\theta

Substitute the values then we get

A_x=18.5cos23^{\circ}=17 m

A_y=18.5sin 23^{\circ}=7.2 m

B_x=41cos37.5^{\circ}=32.5 m

B_y=41sin37.5^{\circ}=-24.96 m

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

B=A+C

C=B-A

C_x=B_x-A_x=32.5-17=15.5 m

C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m

\mid C\mid=\sqrt{C^2_x+C^2_y}

\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m

Hence, the distance between Joe's and Karl's tent=35.7 m

6 0
3 years ago
An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0
3 years ago
The tongue weight of a trailer should be what percent of the gross trailer weight rating
mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

-  Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0
3 years ago
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