PHYSICS COLLEGE

Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids into the empty cubes of an ice tray, and then places the ice tray in the freezer overnight. The next day, she pulls the ice tray out and sets each cube on its own plate. She then waits and watches for them to melt. When the last part of the frozen liquid melts, she records the time.

Answers

Answer 1

Answer:

Answer:

its 45 over 6

Explanation:the answer is in the question

Answer 2

Answer:

Answer: Only the melted cube's shape changed.

Explanation:

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Answers

Answer:

I want to say a because you want to subtract and simplify

You have a small piece of iron at 75 °C and place it into a large container of water at 25 °C. Which of these best explains what will occur over time. A:The iron will cause the water to boil and turn to steam. B:The iron will take the heat from the water and get hotter. C:The water will cool significantly due to its colder temperature. D:Some heat from the iron will move to the water causing both to change temperatures.

Answers

D. Heat energy will be transferred within the system and if left long enough, there will be enough transferred energy to make both of them the same temperature.

D:Some heat from the iron will move to the water causing both to change temperatures.

7) Straws work on the principle of the outside atmospheric pressure pushing the fluid (for example water) up the straw after you have lowered the pressure at the top of the straw (in your mouth). Assuming you could create a perfect vacuum in your mouth, what is the longest vertical straw you could drink water from?

Answers

Answer:

The longest straw will be 10.328 meters long.

Explanation:

The water will rise up to a height pressure due to which will balance the atmospheric pressure.

We know

$P_(atm)=101325N/m^(2)$

Pressure due to water column of height 'h'

$P_(water)=1000* 9.81* h$

Equating both the values we get the value of height 'h' as

$h=(101325)/(1000* 9.81)\n\nh=10.328m$

Why do electrical devices have resistance

Answers

As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answers

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

$V_f^2=V_o^2-2gy$

As the spheres are released v₀ = 0

$V_f^2=2gH$

$V_f=√(2gH)$

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

$V_(10)=\sqrt2gH$

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

$V_h=\sqrt2gH$

Small sphere m₂ and $V_(20)=-\sqrt2gH=-V_h$

Large sphere m₁ and $V_(10)=√(2gH)=V_h$

Before crash

$P_o=m_1V_(10)+m_2V_(20)$

After the crash

$P_f=m_1V_(1f)+m_2V_(2f)$

$P_o=P_f$

$m_1V_(10)+m_2V_(20)=m_1V_(1f)+m_2V_(2f)$

The conservation of kinetic energy

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$

$K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

$K_o=K_f$

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$ $=$ $K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

Let's write the values

$-m_1V_h+m_2V_h=m_1V_(1f)+m_2V_(2f)$

$m_1V_h^2+m_2V_h^2=m_1V_(1f)^2+m_2V_(2f)^2$

The solution to this system of equations is

$m_t=m_1+m_2$

$V_(1f)=((m_1-m_2))/(m_tV_(10)) +(2m_2)/(m_tV_2)$

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_2)$

The large sphere is labeled 1, we are asked for the mass so that $V_(1f)$ = 0, let's clear the equation

$V_(1f)=(m_1-m_2)/(m_tV_(10)) +(2m_2)/(m_tV_(20))$

$0=(m_1-m_2)/(m_tV_h) +(2m_2)/(m_t(-V_h))$

$(m_1-m_2)/(m_tV_h) =(2m_2)/(m_tV_h)$

$(m_1-m_2)=2m_2$

$m_1=3m_2$

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_(20))$

$V_(2f)=(2m_1)/(m_tV_h) +(m_2-m_1)/(m_t(-V_h))$

In addition, we know that m₁ = 3 m₂

$m_t=3m_2+m_2$ mt = 3m2 + m2

$m_t=4m_2$

$V_(2f)=(2* 3m_2)/(4m_2V_h-(m_2-3m_2)4m_2V_h)$

$V_(2f)=(3)/(2) V_h+(1)/(2) V_h$

$V_(2f)=2V_h$

$V_(2f)=2\sqrt{2gh$

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity $V_f$ = 0

$V^2=V_(2f)^2-2gy$

$0=(2√(2gh))^2-2gy$

$y=4H$

Thus

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

To know more about the Laws of collisions follow

brainly.com/question/7538238

Answer:

a) the large sphere has 3 times the mass of the small sphere

b) y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

vf² = vo² - 2g y

As the spheres are released v₀ = 0

vf² = - 2g H

vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

pf = m₁ v₁f + m₂ v₂f

po = pf

m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

Ko = KF

½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

-m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

mt = m₁ + m₂

v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

(m₁-m₂) / mt vh = 2 m₂ / mt vh

(m₁-m₂) = 2m₂

m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

mt = 3m2 + m2

mt= 4m2

v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

v₂f = 3/2 vh +1/2 vh

v₂f = 2 vh

v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

V² = v₂f² - 2 g Y

0 = (2√2gh)² - 2gy

2gy = 4 (2gH)

y = 4H

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?