1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Bogdan [553]
2 years ago
13

A student gathered two boxes of the same size made of different materials: glass and clear plastic. She placed them on a windows

ill in the sun for an hour, and then measured the temperature of the air in each box.
Which statement best describes the purpose of the experiment?


to relate the size of the box to the temperature of air within the box

to relate the amount of time a box is exposed to sunlight to the temperature of air within the box

to relate the type of box material to the mass of the box

to relate the type of box material to the temperature of air within the box
Physics
1 answer:
Naddik [55]2 years ago
3 0

Answer:to relate the size of the box to the temperature of air within the box

Explanation:

You might be interested in
A transformed, visible pattern that holds cues indicating the presence of a particular ethnic group is ____
kogti [31]

Answer:

A transformed, visible pattern on earth that holds cues indicating the presence of a particular ethnic group is cultural landscape.

Explanation:

3 0
3 years ago
Question 23 of 32
marshall27 [118]
Do you still need help?!?!
7 0
2 years ago
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o
garri49 [273]

Answer:

P_{sol}=50.4\ mm.Hg

Explanation:

According to given:

  • molecular mass of glycerin, M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}
  • molecular mass of water, M_w=2+16=18\ g.mol^{-1}
  • ∵Density of water is 0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}
  • ∴mass of water in 316 mL, m_w=316\times 0.992=313.5 g
  • mass of glycerin, m_g=154\ g
  • pressure of mixture, P_x=55.32\ torr= 55.32\ mm.Hg
  • temperature of mixture, T_x=40^{\circ}C

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

n_w=\frac{m_w}{M_w}

n_w=\frac{313.5}{18}

n_w=17.42 moles

<u>moles of glycerin in the given quantity:</u>

n_g=\frac{m_g}{M_g}

n_g=\frac{154}{92}

n_g=1.674 moles

<u>Now the mole fraction of water:</u>

X_w=\frac{n_w}{n_w+n_g}

X_w=\frac{17.42}{17.42+1.674}

X_w=0.912

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

\therefore P_{sol}=X_w\times P_x

\therefore P_{sol}=0.912\times 55.32

P_{sol}=50.4\ mm.Hg

7 0
3 years ago
Why does the chord of an electric heater not glow while the heating element does​
8_murik_8 [283]

Answer:

The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminum is very low so it does not glow.

4 0
3 years ago
A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Maru [420]

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

Hope this Helps!!!

4 0
2 years ago
Other questions:
  • A "synchronous" satellite is put in orbit about a planet to always remain above the same point on the planet’s equator. The plan
    10·1 answer
  • Which of the following is an example of a plasma
    12·1 answer
  • In order to prevent injury in a car crash, it is recommended that you _______. Increase the initial velocity of the collision.
    15·1 answer
  • Discuss the properties of a scalar quantity. give an example of a scalar.
    7·1 answer
  • A wire with a current of 3.40 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at
    15·1 answer
  • Need help on this please
    8·1 answer
  • Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.
    11·1 answer
  • Light of wavelength 480 nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of t
    5·1 answer
  • How much work should be done to lift a 5kg brick to the height of 12m
    5·2 answers
  • What is the speed of a person walking 7mi in 2hours
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!