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arsen [322]
3 years ago
15

Which boxes represent pure substances?

Physics
2 answers:
anygoal [31]3 years ago
7 0
D only because the rest are all compounds and even though for c they are not connected they are still included to be compounded
Mademuasel [1]3 years ago
3 0
A, B and D can I have Brainlyest thx
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A 0.0427 kg racquet-ball is moving
Gwar [14]

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball (m_1) = 0.00427 kg

Velocity of racket-ball before collision (v_{1i}) = 22.3 m/s

Velocity of racket-ball after collision with box (v_{1f}) = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision v_{2i} = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision v_{2f}= 1.53 m/s

To find the mas of the box m_2.

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

p_i=p_f

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

We can plugin the given value to find m_2

(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)

0.9522+0=-0.4911+1.53m_2

Adding both sides by 0.4911

0.9522+0.4911=-0.4911+0.4911+1.53m_2

1.4433=1.53m_2

Dividing both sides by 1.53.

\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}

0.9433=m_2

∴ m_2=0.9433 kg

Mass of the box = 0.9433 kg (Answer)

4 0
3 years ago
A hot air balloon contains 85,000 moles of air. To what temperature must the
Oksi-84 [34.3K]

Answer:

B is the best answer for this question

8 0
3 years ago
State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati
juin [17]
F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)
6 0
3 years ago
An applied force varies with position according to F = k1 x n − k2, where n = 3, k1 = 3.6 N/m3 , and k2 = 76 N. How much work is
aniked [119]

Answer:

The work done is 205 kJ.

Explanation:

Hi there!

Work can be calculated using the following equation:

W = F · Δx

Where:

W = work

F = applied force

Δx = displacement

In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:

W = ∫ F · dx  

F = 3.6 N/m³ · x³ - 76 N

W = ∫ (3.6 x³ - 76)dx

W = 0.9 x⁴ - 76x

Evaluating from xi to xf:

W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m

W = 205 kJ

8 0
3 years ago
An egg drops from a nest at a height of 2.3 m. Which equation can you use to calculate the impact speed of the egg?
igomit [66]

Answer:

V2^2=V1^2+2a/\d

Explanation:

3 0
3 years ago
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