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2 years ago

14

How many oxygen atoms are used to bond with sulfur in the compound "sulfur dioxide"?

2 answers:

kherson [118]2 years ago

7 0

2 oxygen atoms are used to bond with sulfer

Svetllana [295]2 years ago

4 0

"Di" is a prefix meaning two. Dioxide = 2 oxygen

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When considering the gas laws, the kelvin temperature scale must be used. the reason for this is that the kelvin scale is direct

suter [353]

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3 years ago

How many electrons flow through a point in a wire in 7.00 s if there is a constant current of I = 4.35 A?

OLEGan [10]

Answer:

1.90×10²⁰ Electrons

Explanation:

From the question,

Q = It.................... Equation 1

Where Q = charge flowing through the wire, I = current, t = time

Given: I = 4.35 A, t = 7.00 s

Substitute these values into equation 1

Q = 4.35(7.00)

Q = 30.45 C.

But,

1 electron contains 1.6×10⁻¹⁹ C

therefore,

30.45 C = 30.45/1.6×10⁻¹⁹ electrons

= 1.90×10²⁰ Electrons

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2 years ago

A stable, binary ionic compound exists where the cation (A) has a charge twice in magnitude than that of the anion (X). What wou

svetlana [45]

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3 years ago

Read 2 more answers

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

olga nikolaevna [1]

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Total number of oscillation per unit time is called frequency of oscillation. Here, $f=\dfrac{31}{15}=2.06\ Hz$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{15}{4\pi^2\times 2.06^2}$

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

$v_{max}=A\times \omega$

$v_{max}=A\times 2\pi f$

$v_{max}=0.075\times 2\pi \times 2.06$

$v_{max}=0.970\ m/s$

$v_{max}=97\ cm/s$

Hence, this is the required solution.

5 0

2 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

2 years ago