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2 years ago

The specific heat of aluminum is 0.90 J/gC . How much heat is given off when 25 grams of aluminum is cooled from 55 C to 25 C?

Physics

1 answer:

sp2606 [1]2 years ago

8 0

Answer:

2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/gºC? c=0.90J/g. 9 (2 sigs.)

Explanation:

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A girl, standing 150m in front of a tall building, fires a shot using a starting pistol. A boy, standing

Tems11 [23]

Answer:

I believe 650 or C

Explanation:

plz mark brainnlyist :)

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2 years ago

In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how t

kakasveta [241]

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

$p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }$

$p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }$

$p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }$

$N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}$

$N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}$

$N = 1300(16) e^{-3}$

N = 1035.57

N = 1036 times

6 0

3 years ago

Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric fiel

ad-work [718]

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

Thus;

E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))

E = 75.31 N/C ≈ 75 N/C

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3 years ago

What is an output force?

Rasek [7]

An output for is the force a person applies to a simple machine.

4 0

3 years ago

Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to

andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3 ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$ ----- ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 = 2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0

2 years ago