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Studentka2010 [4]
3 years ago
10

Four people weigh a standard mass of 10.00 g on the same balance. The first person obtains a reading of 12.00 g, the second pers

on reads 9.14 g, the third person reads 10.33 g, and the fourth person reads 12.55 g. These readings suggest the balance is
Precise

Accurate

Neither accurate nor precise

Accurate and precise
Physics
2 answers:
Mars2501 [29]3 years ago
8 0

Answer:

Neither accurate nor precise.

Explanation:

Accuracy is defined as the degree of agreement between the result of a measurement and a true value of the measurand. In other words, it refers to how close the measurements of a measurement system are to the actual value.

Precision is defined as the degree of agreement between the independent results of a measurement, obtained under stipulated conditions. That is, precision refers to how close the measurements of a measurement system are to the true value.

Both are independent of each other.

The correc answer is "Neither accurate nor precise."

The individual measurements are not close to each other and are therefore not accurate. But y are close to the actual value (10 g) and are therefore not exact.

miskamm [114]3 years ago
7 0
The above readings suggest that the balance is Accurate and Precise.<span />
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The drawing shows three objects rotating about a vertical axis. The mass of each object is given in terms of m0, and its perpend
photoshop1234 [79]

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

m₂ = 2m₀

m₃ = m₀

r₁ = r₀

r₂ = 2r₀

r₃ = 3r₀

We apply the formula

I = mr²

then

I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²

I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²

I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²

finally we have

I₁ > I₃ > I₂

7 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
A center-seeking force related to acceleration is _______ force
34kurt
A because centrifugal is to velocity to how slow or fast something is  and centrifugal has expresssed as ac=v2 / r (1)<span />
5 0
3 years ago
Read 2 more answers
Water at the top of a slope has potential energy. true or false
ASHA 777 [7]
<span>The statement is TRUE. Water does have potential energy at the top of a slope. The reason why is that potential energy is energy possessed by a body based on its position relative to a zero point. In this case, water at the top of the slope is at an elevation above ground (zero point). The energy is not kinetic (moving) energy since the water is not moving.</span>
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3 years ago
Read 2 more answers
1. An 800-gram bowling ball is rolling in a straight line toward you. If its momentum is 16 kg•m/sec, how fast is it traveling?
ziro4ka [17]

<u>Answer:</u>

The ball is rolling at a speed of 0.02 meter per second.

<u>Step by step explanation:</u>

We are given that there is a 800 gram bowling ball rolling in a straight line. If its momentum is given to be 16 kg.m/sec, we are to find its velocity.

For this, we will use the formula of momentum.

<em>Momentum = mass × velocity</em>

16 = 800 × velocity

Velocity = 16/800 = 0.02 meter per second

3 0
3 years ago
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