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2 years ago

A 2.34 kg book is dropped from a height of +3.1 m. (a) What is its acceleration?

Physics

1 answer:

frosja888 [35]2 years ago

8 0

<span> 9.8 m/s^2 is the answer</span>

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The lowest speed will lose

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3 years ago

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The main energy source during exercise of low to moderate intensity is

zhannawk [14.2K]

Answer:

THUS,THE MAJOR SOURCES OF ENERGY DURING EXERCISE ARE CARBOHYDRATES AND FATS.

Explanation:

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2 years ago

A basketball has a mass of 1 kg and is traveling 15 m/s. How fast would a 5 kg bowling ball have to travel to have the same mome

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2 years ago

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Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 51.5 A. The resistanc

Sliva [168]

Answer:

a).Jcu= 436.44 $x10^{3} \frac{A}{m^{2}}$

b).λcu=1.05728 $\frac{kg}{m}$

c).Jal= 274.66 $x10^{3} \frac{A}{m^{2}}$

d).λcu=0.487 $\frac{kg}{m}$

Explanation:

a).

ζcu=1.7 $x10^{-8}$ Ωm

ζal=2.76 $x10^{-8}$ Ωm

$A=\frac{w}{R}$

wcu=ζcu*l

$Acu=\frac{1.7x10^{-8}*1000}{0.144} =1.18x10^{-4} m^{2}$

$J=\frac{I}{Acu}=\frac{51.5A}{1.18x10^{-4}m^{2}} \\J=436.44x10^{3} \frac{A}{m^{2}}$

b).

mass per unit Copper

λcu=Dcu*Acu

λcu=8960 $\frac{kg}{m^{3}}$ *1.18 $x10^{-4} m^{2}$

λcu=1.05728 $\frac{kg}{m}$

c).

wal=ζal*l

$Aal=\frac{2.7x10^{-8}*1000}{0.144} =0.187x10^{-3} m^{2}$

$J=\frac{I}{Aal}=\frac{51.5A}{0.187x10^{-3}m^{2}} \\J=274.66x10^{3} \frac{A}{m^{2}}$

d).

mass per unit Aluminum

λal=Dal*Aal

λal=2600 $\frac{kg}{m^{3}}$ *0.1875 $x10^{-3} m^{2}$

λcu=0.487 $\frac{kg}{m}$

3 0

2 years ago

How many seconds are required to deposit 0.110 grams of magnesium metal from a solution that contains Mg2 ions, if a current of

lakkis [162]

Answer: The time required to deposit such amount of Mg is 886secs.

Explanation: According to Faraday Law of Electrolysis, the mass of a substance deposited is directly proportional to quantity of electricity passed.

He also stated that;

96500C(1Farday) of electricity is required to deposit 1 mole of any metal.

For $Mg^2+$ +2e- ==>Mg

193000C(2Faraday) of electricity is required to deposit 1mole of Magnesium metal (1 mole of Mg=24g)

Which implies;

19300C will liberate 24g

xCwill liberate 0.110g

Where x is the amount of electricity to deposit 0.110g

x = (19300 × 0.110)/24

x = 884.6C

Recall that Q =It

Where Q is the quantity of electricity, I is the current and t is the time taken

884.6= 0.998 × t

t= 884.6/0.998

t= 886.3secs.

Therefore the time require is 886.3s.

8 0

3 years ago