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3 years ago

Convert 350.0mL at 740 K to its new volume at standard temperature

Physics

1 answer:

Lisa [10]3 years ago

7 0

Answer:

129.12 ml

Explanation:

If answer is correct feel free to replay me for explanation.

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The voltage in the lines that carry electric power to homes is typically 2000 V. What is the required ratio of the loops in the

mariarad [96]

Answer:

yurrrrr 221n54

Explanation:

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6 0

3 years ago

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Hellppppp&/&/&/&//&/&/&/&/&& will give brainiest

Komok [63]

Explanation:

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3 0

3 years ago

At the point of fission, a nucleus of ^235 U that has 92 protons is divided into two smaller spheres, each of which has 46 proto

IrinaK [193]

Answer : $F = 3.5\times10^{3}\ N$

Explanation :

Given that

Radius of sphere $r = 5.90\times 10^{-15}\ m$

The distance between the centers of the two spheres is

$r = 2\times 5.90\times 10^{-15}\ m$

The charge of the sphere $q = 46\times1.6\times10^{-19} C$

The magnitude of the repulsive force between the charges pushing them a part is

Using coulomb law

$F = \dfrac {kq_{1}q_{2}}{r^{2}}$

$F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}$

$F = 3501.3\ N$

$F = 3.5\times10^{3}\ N$

Hence, this is the required solution.

3 0

3 years ago

a 20-kg object travelling at 20 m/s collides head on with an 18-kg object travelling at 17 m/s.If they were locked together afte

djverab [1.8K]

Answer:

2.47 m/s

Explanation:

Momentum = Mass X Velocity

If they were locked together, it means its a perfectly inelastic collision. Therefore,

Total momentum before = Total momentum after

Total momentum before = (20 X 20) - (18 X 17)

= 94

Total momentum after = 94

Y = Object speed after collision

94 = (20+18)Y

Y = 2.47368421 m/s

3 0

2 years ago

Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!

krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

$L = 10 Log\frac{I}{I_0}$

$20 = 10 Log{I}{10^{-12}}$

$I_1 = 10^{-10}W/m^2$

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

$L = 10 Log\frac{I}{I_0}$

$50 = 10Log\frac{I}{10^{-12}}$

$I_2 = 10^{-7}W/m^2$

now we know that

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}$

$r_2 = 47 cm$

so now the distance from friend must be 47 cm

8 0

3 years ago