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3 years ago

Answer the question please

Physics

2 answers:

Juli2301 [7.4K]3 years ago

8 0

The property that she is testing is what is known as the boiling point. This describes the temperature (given a pressure) that a liquid begins to transition into the gas phase. By monitoring the production of bubbles (gas), she is investigating the boiling points of the two liquids. Hope this helps!

Nataliya [291]3 years ago

3 0

Answer:

B. solubility I think lmk if its wrong

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Label these parts on the wave below: Amplitude, Wavelength, Crest, Trough, Rest Position

Leviafan [203]

Answer:

Wavelength is the distance between from one crest to another crest or from one trough to another trough. The amplitude is the distance from the midpoint to the crest or trough. Crest is the highest point of the or a wave. Tough is the lowest point of the or a wave. Rest position is the position where it lies on the midpoint line.

Explanation:

I need a diagram to label these parts.

5 0

3 years ago

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri

miv72 [106K]

Answer:

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

T = m_A a

Axis y

N- W_A = 0

Body B

Vertical axis

W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

T = m_A a

W_B –T = M_B a

We solve this system of equations

m_B g = (m_A + m_B) a

a = m_B / (m_A + m_B) g

In this initial case

m_A = M

m_B = M

a = M / (1 + 1) M g

a = ½ g

Let's find the tension

T = m_A a

T = M ½ g

T = ½ M g

Now we change the mass of the second block

m_B = 2M

a = 2M / (1 + 2) M g

a = 2/3 g

We seek tension for this case

T’= m_A a

T’= M 2/3 g

Let's look for the relationship between the tensions of the two cases

T’/ T = 2/3 M g / (½ M g)

T’/ T = 4/3

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

4 0

3 years ago

A 28-kg beginning roller skater is standing in front of a wall. By pushing against the wall, she is propelled backward with a ve

Scorpion4ik [409]

Answer:

33.6 Ns backward.

Explanation:

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = change in momentum

I = mΔv................................. Equation 1

Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.

Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)

Substituting into equation 1

I = 28(-1.2)

I = -33.6 Ns

Thus the impulse = 33.6 Ns backward.

3 0

3 years ago

A cart rolling down an incline for 2.0 seconds has a beginning speed of 3.0m/s, what Is the final speed or volocity?

Kryger [21]

Answer:

1.5 seconds

Explanation:

V=S/T

V=3/2

V=1.5 seconds

8 0

3 years ago

6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final

Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy = ¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

= ¹/₂ x (9.43)²(25+6.8)

= 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

4 0

3 years ago