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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

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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At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0

3 years ago

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out

blondinia [14]

Answer:

$\theta = 20.98 degree$

Explanation:

As we know that the speed of the sound is given as

$v = 332 + 0.6 t$

now at t = 273 k = 0 degree

$v = 332 m/s$

so we have

$a sin\theta = N\lambda$

$a sin\theta = N(\frac{v_1}{f})$

now when temperature is changed to 313 K we have

$t = 313 - 273 = 40 degree$

now we have

$v = 332 + (0.6)(40)$

$v_2 = 356 m/s$

$a sin\theta' = N(\frac{v_2}{f})$

now from two equations we have

$\frac{sin19.5}{sin\theta} = \frac{332}{356}$

so we have

$sin\theta = 0.358$

$\theta = 20.98 degree$

7 0

2 years ago

Which of these is an experiment generally regarded as being first carried out by James Joule?

kipiarov [429]

The correct answer for this question is this one: "measuring the temperature increase of water from doing work stirring it." This experiment is generally regarded as being first carried out by James Joule is this one, <span>measuring the temperature increase of water from doing work stirring it.</span>

5 0

3 years ago

The electric field between square the plates of a parallel-plate capacitor has magnitude E. The potential across the plates is m

Helga [31]

Answer:

The magnitude of the electric field between the plates is half its initial value.

Explanation:

We know the electric field E = V/d where V = voltage applied and d = separation between plates.

Since V is constant and V = Ed,

So, E₁d₁ = E₂d₂ where E₁ = initial electric field at separation d₁, d₁ = initial separation of plates, E₂ = final electric field at separation d₂ and d₂ = final separation of plates.

So, E₂ = E₁d₁/d₂

Now, the distance between the plates is twice their original separation. Thus, d₂ = 2d₁

So, E₂ = E₁d₁/2d₁ = E₁/2

So, E₂ = E₁/2

Thus, the magnitude of the electric field between the plates is half its initial value.

5 0

3 years ago

A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

$F_{G}=F_{D}$

$mg=\frac{1}{2}C \rho Av_{T}^{2}$

$v_{T}=\sqrt{\frac{2mg}{\rho CA} }$

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

$v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 9.09

The 95kg skydiver's terminal velocity is

$v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

$t=\frac{4750}{9.09}$

t = 522.5

95 kg at 12.3 m/s:

$t=\frac{4750}{12.3}$

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0

2 years ago