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4 years ago

2. Descnbe a in which a set of measurements precise but not accurate. Be sure to explain what about measurement

Physics

1 answer:

jek_recluse [69]4 years ago

6 0

Answer:

Precision

Explanation:

You can be very precise but inaccurate, as described . You can also be accurate but imprecise. For example, if on average, your measurements for a given substance are close to the known value, but the measurements are far from each other, then you have accuracy without precision. Precision is independent of accuracy. You can be very precise but inaccurate, as described above. You can also be accurate but imprecise. For example, if on average, your measurements for a given substance are close to the known value, but the measurements are far from each other, then you have accuracy without precision. Precision refers to the closeness of two or more measurements to each other. Using the example above, if you weigh a given substance five times, and get 3.2 kg each time, then your measurement is very precise.

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Energy can not be created or destroyed but can change from one form to another.

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3 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

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Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

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3 years ago

I got this information for a lab but I don't know how to do the hypothesis and the conclusion please can you guys help me with i

pochemuha

Answer:

A hypothesis is what you think will happen.

A conclusion is the results of an experiment summarized.

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8 0

3 years ago

Anyone know how to do this?

MrMuchimi

The voltage from one side of the battery all the way around to the other side of the battery is 12v .

If 4 of those volts show up across the circle-thing, then the rest of the 12v ... 8v ... Must show up across the set of parallel rectangles.

To get that answer, I subtracted the 4 from the 12.

Just like it says in choice-C.

7 0

3 years ago

The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70

MrRa [10]

Answer:

a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

C = $\epsilon_o \frac{A}{d}$

C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

C = $\frac{Q}{\Delta V}$

Q₀ = C ΔV

Q₀ = 2.62 10⁻¹² 8.70

Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

C₁ = 1.04 10⁻¹² F

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ΔV₁ = Q₀ / C₁

ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

ΔV₁ = 21.9 V

b) initial stored energy

U₀ = $\frac{Q_o}{ 2C}$

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U₀ = 99.2 10⁻¹² J

c) final stored energy

U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)

U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

W = U_f - U₀

W = (249.2 - 99.2) 10⁻¹²

W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0

3 years ago