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3 years ago

Indicate whether the statement is true or false. The amount of matter in an object is its weight.

Physics

2 answers:

Zigmanuir [339]3 years ago

6 0

False. the amount of matter in an object is the mass

nadezda [96]3 years ago

4 0

Hey there!

The answer is false( it would be mass)

Hope it helped!

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Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of

Elanso [62]

Answer:

c) The slope is not constant and increases with increasing time.

Explanation:

The equation for the position of this particle (starting from rest is)

$s = at^2/2 = 5t^2/2 = 2.5t^2$

We can take derivative of this with respect to time t to get the equation of slope:

$s' = (2.5t^2)' = 2*2.5t = 5t$

As time t increase, the slope would increases with time as well.

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4 years ago

Convert -13°F into (a) °C (b) kelvin

RideAnS [48]

Answer:

-25ºC

Explanation:

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3 years ago

What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic th

asambeis [7]

Answer:

$\lambda = 2.57 \times 10^{-11} m$

Explanation:

Average velocity of oxygen molecule at given temperature is

$v_{rms} = \sqrt{\frac{3RT}{M}}$

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

$v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}$

$v_{rms} = 483.4 m/s$

now for de Broglie wavelength we know that

$\lambda = \frac{h}{mv}$

$\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}$

$\lambda = 2.57 \times 10^{-11} m$

7 0

3 years ago

Which of the following is one way that an electric motor's rotation speed can be controlled?

malfutka [58]

Depending on which type of motor you're talking about, but the first 3 are true. A stronger magnetic field in a DC motor will slow it down but increase its torque.
The amount of current in the motor will control the magnetic fields and therefore affect the speed (and torque). In an induction motor, the rotational speed is given by $n= \frac{120f}{p}$ where f is the line frequency and p is the number of poles. Thus fewer poles makes it go faster.

5 0

3 years ago

Read 2 more answers

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.