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NemiM [27]
3 years ago
11

Section of two thermometers are shown. The thermometer on the left is labeled A and the thermometer on the right is labeled B. T

hermometer labeled A has markings forty degrees, forty five degrees, and fifty degrees at equal intervals. The thermometer labeled B has markings forty degrees, forty five degrees, and fifty degrees at equal intervals and there are four small markings equally spaced in between the numbers.Which statement explains which thermometer is more appropriate to measure the temperature of a liquid at 43.6 °C?
a. Thermometer A, because it measures temperature more accurately than B.

b. Thermometer B, because it measures temperature more accurately than A.

c. Thermometer A, because it measures temperature more precisely than B.

d. Thermometer B, because it measures temperature more precisely than A.
Physics
2 answers:
jolli1 [7]3 years ago
8 0

Answer : The correct option is, (b) Thermometer B, because it measures temperature more accurately than A.

Explanation :

Accuracy : It is defined as the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 50 kg and one person weighed 48 kg and another person weighed 55 kg. Then, the weight measured by first person is more accurate.

Precision : It is defined as the closeness of two or more measurements to each other.

For Example: If you weigh a given substance five times and you get 1.8 kg each time. Then the measurement is said to be precise.

Level of precision is determined by the maximum number of decimal places.

As per question, the thermometer B has markings 40°, 45° and 50° at equal intervals and there are four small markings equally spaced in between the numbers and thermometer B has markings 40°, 45° and 50° at equal intervals but there is no small markings equally spaced in between the numbers. That means, thermometer B is measures temperature more accurately than thermometer A.

Hence, correct option is, (b) Thermometer B, because it measures temperature more accurately than A.

Verizon [17]3 years ago
4 0

Answer:

the answer is B

Explanation:

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¡Hellow!

For this problem, lets recabe information:

v (Velocity) = 10 m/s

a (Aceleration) = 1 m/s²

t (Time) = 1/4 min = 25 s

d (Distance) = ?

v' (Final velocity) = ?

First, for calculate distance, lets applicate formula:

                                        \boxed{\boxed{\text{d = Vo * t + (a * t}^{2})\text{ * 0,5}   } }

Lets replace according we information and let's resolve it:

d = 10 m/s * 25 s + (1 m/s² * (25 s)²) * 0,5

d = 250 m + (625 m) * 0,5

d = 2,5 m + 312,5 m

d = 314 meters.

Now, for calculate final speed, lets applicate formula:

                                                 \boxed{\boxed{\text{v' = v + a * t}   } }

Lets replace according we information and let's resolve it:

v' = 10 m/s + 1 m/s² * 25 s

v' = 10 m/s + 25 m/s

v' = 35 m/s

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Answer: 3.

Explanation:

The correct answer is a higher amplitude and lower frequency. Since an opera singer is lowering his pitch it means that he is creating higher amplitude and because he is raising his voice for a song with that higher amplitude he is creating lower frequency.

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A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

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