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melisa1 [442]
3 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]3 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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A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the s
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Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

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2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

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1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

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a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

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Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

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