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3 years ago

AlBr3 + K2SO4 → KBr + Al2(SO4)3

Physics

2 answers:

trasher [3.6K]3 years ago

8 0

Answer: D

Explanation:

STALIN [3.7K]3 years ago

6 0

On the left, you have

1 Al
3 Br
2 K
1 S
4 O

while on the right, you have

1 K
1 Br
2 Al
3 S
12 O

To get a balanced equation, you'll need 6 KBr on the right, which would require 2 AlBr3 and 3 K2SO4 on the left, so you end up with

$2\,\mathrm{AlBr}_3+3\,\mathrm K_2\mathrm{SO}_4\to6\,\mathrm{KBr}+\mathrm{Al}_2(\mathrm{SO}_4)_3$

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A satellite with mass 6000 kg is orbiting the planet at 2500 km above the planet's

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By the law of universal gravitation, the gravitational force F between the satellite (mass m) and planet (mass M) is

F = G M m / R ²

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Solve for M :

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2 years ago

Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you

Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

= 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

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As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t

v = u+at

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t = 27.2 min

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3 years ago

Two objects with masses of 2.75 kg and 4.45 kg are connected by a light string that passes over a light frictionless pulley to f

Llana [10]

Answer:

Explanation:

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T = 2.75(9.81 + 2.32) = 33.3 N

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2 years ago

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Mass of the bug= m

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$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

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$m(0.65)l=\frac{13}{10}ml^2 \omega$

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$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

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Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

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$l=2.45cm$

7 0

3 years ago