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3 years ago

Accuracy is the correctness of a measurement. true or false.

Physics

2 answers:

Over [174]3 years ago

8 0

Answer:

True

Explanation:

Accuracy indicates how close to the true measurement you actually are.

Let us understand this with an example.

If a rod produced by a manufacturer has a specification of 1 meter length and after measuring it you also get 1 m length then the measurement is highly accurate.

There is another term called precision which is used in the same sense as accuracy called precision but they do not have the same meaning. Precision refers to how close in the measurements you are while taking multiple readings.

mash [69]3 years ago

6 0

It is true i think so

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A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W

Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)$

$V^{2} = 24.5 m/s$

$V = \sqrt{24.5} \ m/s$

$V = 4.95 \ m/s$

V = ± 4.95 m/s

V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)$

$0 = U^{2} - 16.072 m/s$

$U^{2} = 16.072 m/s$

$U = \sqrt{16.072} \ m/s$

U = ± 4.01 m/s

U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

$F = \frac{mv - mu}{t}$

$F.t = m(v - u)$

⇒ Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,

∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )

F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )

F.t = 0.120 kg × 8.96 m/s

Impulse = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

6 0

3 years ago

A student has a small piece of steel.

Simora [160]

Answer:

There are different ways to investigate density. In this required practical activity, it is important to:

record the mass accurately