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daser333 [38]
2 years ago
5

A 12.0-g bullet is fi red horizontally into a 100-g wooden block initially at rest on a horizontal surface. After impact, the bl

ock slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact?
Physics
1 answer:
allochka39001 [22]2 years ago
3 0

Answer:

v_1 = 91.3 m/s

Explanation:

By energy conservation and work energy theorem we can say that after bullet hits the block, it will move on the rough floor and comes to rest

so here work done by frictional force = change in kinetic energy

so we know that

W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

-\mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)

-0.650(9.81)(7.5) = \frac{1}{2}(0 - v_i^2)

47.8 = \frac{1}{2}v^2

v = 9.78 m/s

now by momentum conservation we have

m_1 v_1 = (m_1 + m_2) v

12 v_1 = (100 + 12) 9.78

v_1 = 91.3 m/s

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Please awnser and show the ways​
topjm [15]

Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

  • The density of the substance in Crown A;

Density = mass ÷ volume = \frac{1930}{100} = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

  • The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043  g/cm³ ≈ 10.5 g/cm³  (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

  • The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to \frac{29.8}{2} = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

  • An empty masuring cylinder has a mass of 500 g.
  • Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.
  • The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷  volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

\frac{0.02 * 1}{0.001} =  20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

\frac{0.02 * 1,000,000}{1} = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

  • A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
  • in a displacement can
  • The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; \frac{20 * 1}{1} = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density ×  volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0
2 years ago
Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em
mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

6 0
3 years ago
What is lenz law?<br>What is galvanometer?​
yulyashka [42]

Answer:

Lenz's law, in electromagnetism, statement that an induced electric current flows in a direction such that the current opposes the change that induced it. This law was deduced in 1834 by the Russian physicist Heinrich Friedrich Emil Lenz (1804–65).

6 0
1 year ago
A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?
Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒  Density = \frac{Mass}{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

On substituting the values, we get

⇒               =\frac{6.3 \ g}{19.3 \ g/mL}

⇒               =0.32 \ mL

7 0
2 years ago
If a chemical reaction has a AH value of -125 kJ, it would be considered what?
egoroff_w [7]

exothermic

Explanation:

negative means it's losing energy

4 0
2 years ago
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