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Lubov Fominskaja [6]
2 years ago
8

How does having waters of hydration affect the enthalpy for dissolving?

Physics
1 answer:
poizon [28]2 years ago
4 0

Answer:

it makes it so wet

Explanation:

hope it helps

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D
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And
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3 years ago
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The conservation of momentum is most closely related to
BartSMP [9]

Answer:

<h2>B) Newton's 2nd law</h2>

Explanation:

<h2>From; force= mass × acceleration </h2><h2> f= m×a </h2><h2>where a(acceleration)= velocity/time</h2><h3> force = mv/t</h3><h3>But momentum(p) = Mass × velocity </h3><h2>hence force =p/t </h2><h3>that is Momentum = force × time ( Newton's 2nd law)</h3>
6 0
1 year ago
Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
What are the order of the 7 prefixes you are required to know in order from largest to smallest
vaieri [72.5K]
10^9 giga, 10^6 mega, 10^3 kilo, 10^-3 milli, 10^-6 micro, 10^-9 nano, 10^-12 pico
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A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft
vivado [14]

Their velocity afterwards is 2.88 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, for an isolated system (= no external force), the total momentum must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v

where: in this case:

m_1 = 91.5 kg is the mass of the first player

u_1 = 2.73 m/s is the initial velocity of the first player (choosing east as positive direction)

m_2 = 63.5 kg is the mass of the second player

u_2 = 3.09 m/s is the initial velocity of the second player

v is their combined velocity afterwards

Solving for v, we find:

v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88 m/s

And the sign is positive, so the direction is east.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

7 0
3 years ago
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