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dimulka [17.4K]
2 years ago
8

In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) (NH4)2SO4 Ba(

NO3)2 BaSO4 NH4NO3
Physics
2 answers:
Arte-miy333 [17]2 years ago
5 0

Answer:

BaSo4

Explanation:

Mariana [72]2 years ago
4 0
I think it's Barium sulfate, the soild and percipitate
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A ray of light strikes a plane mirror at an angle of incidence 35°. If the mirror is rotated through 10°.
xxTIMURxx [149]

Answer:

1 35°

2 20°

3 50°

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7 0
3 years ago
when a constant force acts upon an object the acceleration of the object varies inversely with its mass. when a certain constant
solong [7]

Explanation:

When a constant force acts upon an object the acceleration of the object varies inversely with its mass.

a\propto \dfrac{1}{m}

or

\dfrac{a_1}{a_2}=\dfrac{m_2}{m_1}

If m₁ = 21 kg, a₁ = 3 m/s², m₂ = 9 kg

We need to find a₂

So,

a_2=\dfrac{m_1a_1}{m_2}\\\\a_2=\dfrac{21\times 3}{9}\\\\a_2=7\ m/s^2

So, if mass is 9 kg, its acceleration is 7 m/s².

8 0
2 years ago
What must happen to the electrons in a material to create an electric current?
zhenek [66]
You have to have electrons that are in the materials conduction band. Then you apply a voltage across the material. 
8 0
3 years ago
Read 2 more answers
What is the change in air pressure a good indicator of a change in temperature be a change in the weather I decrease in humidity
Elza [17]

Answer:

A

Explanation:

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3 0
3 years ago
The cannon on a battleship can fire a shell a maximum distance of 26.0 km.
Alla [95]
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is \theta = 45^{\circ}.
In fact, the laws of motions on both x- and y- directions are
S_x(t)= v_0 cos \theta t
S_y(t)= v_0 \sin \theta t -  \frac{1}{2} gt^2
From the second equation, we get the time t at which the projectile hits the ground, by requiring S_y(t)=0, and we get:
t= \frac{2 v_0 \sin \theta}{g}
And inserting this value into Sx(t), we find
S_x(t) = 2  \frac{v_0^2}{g}  \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)
And this value is maximum when \theta=45^{\circ}, so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
S_x(t)=  \frac{v_0^2}{g} \sin (2\theta)
And by using S_x = 26 km=26000 m, we find the value of the initial velocity v0:
v_0 =  \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s
8 0
3 years ago
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