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Sedaia [141]
3 years ago
13

What is the momentum of a 2.0kg ball rolling at 6.0 m/s

Physics
1 answer:
irga5000 [103]3 years ago
4 0
Momentum can be defined by the formula p=m*V (where m is mass and V is velocity) so if we plug in these numbers:

p = 2kg * 6m/s
p = 12 kgm/s
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Please help!
andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

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2 years ago
What is true about solar energy hitting earth?
polet [3.4K]
B. It’s the same roughly at all latitudes
6 0
2 years ago
a. Give an example of a galvanic cell. What kind of reaction occurs in a galvanic cell? b. If one electrode in a galvanic cell i
Burka [1]

Answer:

a) Batteries and fuel cells are examples of galvanic cell

b) Ag-cathode and Zn-anode

c) Cell notation:  Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)

Explanation:

a) A galvanic cell is an electrochemical cell in which chemical energy is converted to electrical energy. The chemical reaction which drives a galvanic cell is a redox reaction i.e. a reduction-oxidation process.

A typical galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs. Batteries and fuel cells are examples of galvanic cells.

b) The nature of the electrode that will serve as an anode or cathode depends on the value of the standard reduction potential (E⁰) of that electrode. The electrode with a higher or more positive the value of E⁰ serves as the cathode and the other will function as an anode.

In the given case, the E⁰ values from the standard reduction potential table are:

E⁰(Zn/Zn2+) = -0.763 V

E°(Ag/Ag+)=+0.799 V

Therefore, Ag will be the cathode and Zn will be the anode

c) In the standard cell notation, the anode half cell is written on the left followed by the salt bridge '||' and finally the cathode half cell to the right.

Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)

4 0
3 years ago
A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie
ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, A=7.4\ cm^2=7.4\times 10^{-4}\ m^2

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, B_1=0.5\ T

Final magnetic field, B_2=3\ T

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_2-B_1}{t}

\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}

\epsilon=1.65\times 10^{-3}\ V

Induced current in the loop of wire is given by :

I=\dfrac{\epsilon}{R}

I=\dfrac{1.65\times 10^{-3}}{2.4}

I=6.87\times 10^{-4}\ A

So, the induced current in the loop of wire over this time is 6.87\times 10^{-4}\ A. Hence, this is the required solution.

7 0
3 years ago
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