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Burka [1]
4 years ago
9

During oxidation-reduction reactions, a material loses electrons to the oxidizer, which has a positive charge. Which power sourc

e directs these lost electrons through an electronic device to give it power?
A. An electric outlet
B. An electromagnet
C. An electric motor
D. A battery
Physics
2 answers:
Nikolay [14]4 years ago
3 0

Answer:

D. Battery

During the oxidation-reduction(redox) reaction, there is always flow of electrons from one point to another. The electrons are then converted to power through the battery which converts chemical energy to electrical energy. If there is zero flow of electrons then there will also be zero power.

Tems11 [23]4 years ago
3 0

Answer:

D. A battery.

Explanation:

A battery cell refers to a single anode and cathode separated by electrolyte used to produce a voltage and current. It is typically an electrolytic cell.

An electrolytic cell is an electrochemical cell that drives a non-spontaneous redox reaction through the application of electrical energy. They are often used to decompose chemical compounds, in a process called electrolysis.

The anode which is positive electrode undergoes oxidation i.e loss of electrons while the cathode(negative) undergoes reduction that is, accept electrons.

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A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t
fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

F = m a\\\\a =\frac{F}{m}

We use movies and find lips

\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t

The moment is defined by

\to p = m v

The moment change

\Delta p = m v - m v_0

Let's replace the speeds in this equation

\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0  + F t - m v_0\\\\\Delta p = F t

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0
3 years ago
Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of
notka56 [123]

Answer:37 J

Explanation:

Given

Step :1

Heat added Q=44 J

Work done=-20 J

\Delta E_1=Q+W=44-20=24 J

Step :2

Heat added Q=-61 J

work done W_2

\Delta E_2=Q+W_2

\Delta E_2=61+W_2

\Delta E_1+\Delta E_2=0

as the process is cyclic

44-20-61+W_2=0

W_2=37 J

work done in compression is 37 J

3 0
3 years ago
A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
astraxan [27]

Answer:

A) k=34867.3384\ N.m^{-1}

B) \omega'\approx84\ Hz

Explanation:

Given:

mass of car, m=1380\ kg

A)

frequency of spring oscillation, f=1.6\ Hz

We knkow the formula for spring oscillation frequency:

\omega=2\pi.f

\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f

\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6

k_{eq}=139469.3537\ N.m^{-1}

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

k=\frac{k_{eq}}{4}

k=\frac{139469.3537}{4}

k=34867.3384\ N.m^{-1}

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }

\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }

\omega'\approx84\ Hz

7 0
3 years ago
A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
Why is a virus NOT classified as a living thing?
Shkiper50 [21]

The correct answer would be B. A virus is living BUT   however it does NOT have all of the 7 characteristics of life.

Hope that helped ^^

4 0
3 years ago
Read 2 more answers
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