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Drupady [299]
2 years ago
15

Suppose you had used a less sensitive balance for the Archimedes method of getting volume of an object based on the difference i

n mass in water and out of water. Would that change the precisions of your density for that method?. They define sensitivity as the scale to which things are measured and precision as how repeatable something is.
Physics
2 answers:
alisha [4.7K]2 years ago
8 0
<span>If you think about it, changing the scale to which something is measured does not affect the repeatability of the measurement. For instance, if you have a meter stick which was labeled incorrectly, that doesn't affect the fact that every measurement you take of a certain fixed distance will still be the same. Precision does not equal accuracy.</span>
algol132 years ago
6 0

Answer:

There will be no change in the sensitivity of the method.

Explanation:

A less sensitive measuring method is less likely to record the small errors of measurement. The sensitivity of a measuring device depends on the error of measurement that it can accommodate. For example, is the length of a pen was to be measured by the ruler and a vernier caliper, the two measurements would be different. This is because the ruler would give a measurement to the nearest centimeter. However, the vernier caliper would provide a measurement to the nearest 0.1 cm which improves the accuracy and sensitivity of the measurement as well. Thus, the more sensitive an instrument is, the more accurate it is.

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A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0
2 years ago
The speed v of a sound wave traveling in a medium that has bulk modulus b and mass density ρ (mass divided by the volume) is v=b
PilotLPTM [1.2K]

As it is given that Bulk modulus  and density related to velocity of sound

v = \sqrt{\frac{B}{\rho}}

by rearranging the equation we can say

B = \rho * v^2

now we need to find the SI unit of Bulk modulus here

we can find it by plug in the units of density and speed here

B = \frac{kg}{m^3} * (\frac{m}{s})^2

so SI unit will be

B = \frac{kg}{m* s^2}

SO above is the SI unit of bulk Modulus

3 0
2 years ago
A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
thin films of soap sometime display an array of colours. this display is the result of: a) reflection, diffraction, and interfer
kumpel [21]
The array of colors observed when viewing films of soap are caused by the interference, and from the reflections of light from two nearby surfaces (outer surface and inner surface). Among the choices, the closest answer would be B. reflection, refraction, and interference.
7 0
2 years ago
Read 2 more answers
Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿
Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.  

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0
2 years ago
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