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3 years ago

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Select the true statements: check all that apply check all that apply oxidizing agents can convert co into co2. a reducing agent

gets oxidized as it reacts. if something is reduced, it is formally losing electrons.

1 answer:

andrew11 [14]3 years ago

7 0

Answer:

1.) True

2.) True

3.) False

Explanation:

1.) Oxidizing agents can convert co into co2.

This is true

Carbon monoxide burns in the oxygen to produce carbon dioxide

2.) A reducing agent gets oxidized as it reacts.

This is true

Because it loses electrons and is oxidized in a chemical reaction.

3.) If something is reduced, it is formally losing electrons.

This is false.

If something is reduced, it is gaining electrons.

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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

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3 years ago

A transverse wave is traveling from right to left. what direction does the medium vibrate?

lisabon 2012 [21]

<span>In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.</span>

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3 years ago

Read 2 more answers

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

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3 years ago

If the dartboard below is used to model an atom, which subatomic particles would be located at Z? *

Anna11 [10]

I know that protons and neutrons are located at the center of an atom, so the correct answer is D

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3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago