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4 years ago

The distance versus time graph for Object A and Object B are shown. Which object moves with changing speed?

Physics

1 answer:

Nadya [2.5K]4 years ago

6 0

Answer:

<em>Both: Object A and Object B, because different distances are traveled in the same time interval</em>

Explanation:

<u>Constant Speed Motion </u>

An object is said to have constant speed it travels same distances at the same time intervals. For example, an object who travels 4 meters in 2 seconds has a constant speed if it also travels 6 meters in 3 seconds and 10 meters in 5 seconds. The constant speed is 2 m/s.

We can see in the image provided two graphs of distance vs time of two objects. If they were traveling at a constant speed, that graph should be a single line (increasing, decreasing of horizontal). None of the graphs show a line, which indicates the object some times travels more or less distance in a given time interval than in other equal time intervals. Thus, the answer is the last option:

Both: Object A and Object B, because different distances are traveled in the same time interval

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Which interaction between x-ray photons and matter results in total absorption of the incident photon?

Leno4ka [110]

Answer:

Photoelectric effect

Explanation:

In the photoelectric effect, when an x-ray strikes on a metal surface, the energy is completely absorbed by the metal. If the energy would be equal to or more than work function of metal, electron ejects out. The kinetic energy of the electron which is ejected depends on the energy of the incident radiation and work function of the metal.

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4 years ago

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$