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2 years ago

•. What is called the error due to the procedure and used apparatuses?

Physics

1 answer:

sashaice [31]2 years ago

6 0

Answer:

$c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}$

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A 55.0-g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum? Express

andrew11 [14]

Answer:

Final temperature of the aluminum = 41.8 °C

Explanation:

We have the equation for energy

E = mcΔT

Here m = 55 g = 0.055 kg

ΔT = T - 27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

E = mcΔT

725 = 0.055 x 921.096 x (T - 27.5)

T - 27.5 = 14.31

T = 41.81 ° C = 41.8 °C

Final temperature of the aluminum = 41.8 °C

6 0

3 years ago

In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction

Sphinxa [80]

Answer: $\frac{V^{2}}{r}$

Explanation:

According to Newton's 2nd Law of motion the force $F$ is proportional to the mass $Fm$ and acceleration $a$ :

$F=m.a$ (1)

On the other hand, the equation for the Centripetal force is:

$F=\frac{mV^{2}}{r}$ (2)

Where:

$V$ is the velocity

$r$ is the radius of the circular motion

Making (1) and (2) equal:

$m.a=\frac{mV^{2}}{r}$ (3)

Hence:

$a=\frac{V^{2}}{r}$ This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

3 0

3 years ago

Read 2 more answers

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

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3 years ago

What factors affect the speed of water waves

Aneli [31]

Hey there,

Your question states: What factors affect the speed of water waves
Let's get one thing out the way, (wavelength) does $NOT$ affect the the speed of water. If anything, it would be how high the wavelength's are. The higher the wavelengths are, the more that it would affect the speed, because there very high, but if it were to go longer on the width side, that would increase the speed, but that's not the case. Your correct answer would be (higher wavelength).

Hope this really helps you.

6 0

3 years ago

Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109

r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

$E=\Delta mc^2\\\Delta m=\frac{E}{c^2}$ (1)