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melisa1 [442]
2 years ago
15

Students asked to measure the temperature of a reaction beaker recorded the following temperatures. 103.7°C, 108.4°C, 105.8°C, 1

04.6°C The actual temperature is 105.1°C. Which measurement is the most precise
Physics
1 answer:
zavuch27 [327]2 years ago
3 0

Answer:

The measurement which is the most precise is 104.6 °C.

Explanation:

The measurement which is most precise must be very close to the actual value of the temperature.

Thus, the unit which have less value of the |Δx| (error) must be most precise.

Thus,

Actual value = 105.1 °C

Value = 103.7 °C

<u>|Δx| = 1.4 °C</u>

Value = 108.4 °C

<u>|Δx| = 3.3 °C</u>

Value = 105.8 °C

<u>|Δx| = 0.7 °C</u>

Value = 104.6 °C

<u>|Δx| = 0.5 °C</u>

<u>Thus, The measurement which is the most precise is 104.6 °C.</u>

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Answer:

a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

b. -2.81 m/s²

c. 3.78 s

d. -351.25 N

Explanation:

a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.

For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.

The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².

For the block of ice, v = 15 m/s and Ф = 45°. So,

h = v²sin²Ф/2g

= (15 m/s)²sin²45/(2 × 9.8 m/s²)

= 225 (m/s)²(1/√2)²/19.6 m/s²

= 225 (m/s)²(1/2)/19.6 m/s²

= 112.5 (m/s)²/19.6 m/s²

= 5.74 m

Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

The graph is in the attachment.

b. What is the rate of acceleration while the blocks slow to a stop?

Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m

So, a =  (v² - u²)/2s

substituting the variables into the equation, we have

a =  ((0 m/s)² - (10.61 m/s)²)/2(20 m)

= - 112.57 (m/s)²)/40 m

= -2.81 m/s²

c. How long do the blocks take to slow to a stop?

Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop

So, making t subject of the formula,

t = (v - u)/a

substituting the values of the variables, we have

t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²

= -10.61 m/s/-2.81 m/s²

= 3.78 s

d.  What is the amount of friction between the ice and the snowy ground?

The frictional force, f = net force on block of ice

f = ma where  m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²

f = ma

= 125 kg(2.81 m/s²)

= -351.25 N

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