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Iteru [2.4K]
4 years ago
12

(42) Hydrochloric acid (HCl) and sodium hydroxide (NaOH) will react to form which two compounds?

Physics
1 answer:
xxMikexx [17]4 years ago
5 0

Answer:

D) H20 and NaCI

Explanation:

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what is the difference between cytology and histology

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Explain the flow of energy in the food chain and why stored energy is an important part of it.
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2 years ago
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Darna rolls the 7.05-kg ball down the lane and it hits the 1.52-kg pin head on. The ball was moving at 8.24 m/s before the colli
Amiraneli [1.4K]

Answer:

The velocity of the ball after the collision is 5.39 m/s

Explanation:

Hi there!

To solve this problem, we will use the conservation of momentum: the momentum of the system ball-pin remains the same before and after the collision. The momentum of the system is calculated as follows:

momentum before the collision (initial momentum) = mb · vb1 + mp · vp1

momentum after the collision (final momentum) = mb · vb2 + mp · vp2

Where:

mb = mass of the ball = 7.05 kg

vb1 = velocity of the ball before the collision = 8.24 m/s

mp = mass of the pin = 1.52 kg

vp1 = velocity of the pin before the collision = 0 m/s.

vb2 = velocity of the ball after the collsion = unknown.

vp2 = velocity of the pin after the collision = 13.2 m/s

Since momentum is conserved, then:

initial momentum = final momentum

mb · vb1 + mp · vp1 = mb · vb2 + mp · vp2

Solving for vb2:

mb · vb1 + mp · vp1 -  mp · vp2 = mb · vb2

(mb · vb1 + mp · vp1 -  mp · vp2) / mb = vb2

Since the pin is initially at rest, vp1 = 0:

(mb · vb1  -  mp · vp2) / mb = vb2

(7.05 kg · 8.24 m/s - 1.52 kg · 13.2 m/s) / 7.05 kg = vb2

vb2 = 5.39 m/s

The velocity of the ball after the collision is 5.39 m/s

4 0
3 years ago
An amount of energy is added to ice, raising its temperature from -10°C to -5°C. A larger amount of energy is added to the sam
lara [203]

Answer:

The correct option is;

B) The specific heat of ice is less than that of water.

Explanation:

Here we have

Let the amount of energy added to the ice at -10 C to raise the temperature to -5 C be X J

Let the amount of energy added to the water at 15 C to raise the temperature to 20 C be Y J

We know that the heat required, ΔQ to raise the temperature of a substance is given by

ΔQ = m·c·Δθ

Where:

m = Mass of the substance

c = Specific heat capacity

Δθ = Temperature change

Since the mass of the ice and the water are the same, so also is the change in temperature, (-5 - (-10) = 5 and 20 - 15 = 5) we have

for m₂·c₂·Δθ₂ > m₁·c₁·Δθ₁

Where:

m₁, c₁, Δθ₁, is for the ice and m₂, c₂, Δθ₂ is for the water and

m₁ = m₂

Δθ₁ = Δθ₂

Therefore,

c₂ > c₁ = c₁ < c₂

That is the specific heat capacity of the ice is lesser than the specific heat capacity of the water.

4 0
3 years ago
Read 2 more answers
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