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Bogdan [553]
4 years ago
7

When heather grabbed the glass of ice water, the ice cubes were floating at the top. why were the ice cubes floating in the wate

r?
Physics
1 answer:
tiny-mole [99]4 years ago
8 0
The ice cubes were floating in water because they are less dense than liquid water. When water is frozen, a structure that is crystalline is formed that is held by hydrogen bonding. Due to the orientation of these bonds, the moleules would push far away from each other causing it to have a bigger volume and a lower density.
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You see the moon rising, just as the sun is setting. What phase is the moon in?.
Readme [11.4K]

Answer:

full phase

Explanation:

sorry if its wrong

4 0
2 years ago
A bus covers 10 km in 7 minutes. Find the speed of the bus in km/h
Shtirlitz [24]

Answer:

Explanation: so how many minutes are in an hour 60 right, and the bus travels 10km in 7 minutes right so use math the bus travels 14km in 10 minutes so the bus travels 98km in an hour

8 0
3 years ago
A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator
bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

\eta = 1 - \frac{T_2}{T_1}

Where,

T_2 = Temperature of Cold Body

T_1 =Temperature of Hot Body

\eta= Efficiency

According to the statement our values are:

\eta = 0.3

T_2 = 210K

Replacing we have that

0.3 = 1- \frac{210}{T_1}

\frac{210}{T_1} = 0.7

T_1 = \frac{210}{0.7}

T_1 = 300K

Therefore the temperature of the heat source is 300K

6 0
3 years ago
A ball is thrown off a cliff at a speed of 10 m/s in a horizontally direction. The ball reaches the ground 1.5 seconds. If the b
Tems11 [23]
I am pretty sure it is d
5 0
3 years ago
You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between
Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Where,

m_1 = mass of ball

m_2 = mass of the person

u_1 = Velocity of ball before collision

u_2 = Velocity of the person before collision

v_1 = velocity of ball afer collision

v_2= velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

v_1 = v_2

m_1u_1 + m_2u_2 = (m_1+m_2)v_2

Re-arrenge to find v_2,

v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}

Our values are,

m_1= 0.425kg

u_1= 12m/s

m_2= 68.5kg

u_2= 0m/s

Substituting,

v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}

v_2 = 0.074m/s

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0
3 years ago
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