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Luba_88 [7]
3 years ago
8

I will mark brainlist pls help and. Give right answers

Physics
2 answers:
Vera_Pavlovna [14]3 years ago
7 0

I think its b. Answer:

Explanation:

jeyben [28]3 years ago
5 0
The answer to this question is going to be a
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An object moves 2.5 m. This is an example of a _______.
makvit [3.9K]

The correct answer is A. Distance

Hope this helps

7 0
3 years ago
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The position of an object that is oscillating on a spring is given by the equation x = (17.4 cm) cos[(5.46 s-1)t]. what is the a
nekit [7.7K]

the angular frequency of this motion is 5.46rad /sec

The formula for the angular frequency is = 2π/T. Radians per second are used to express angular frequency. The frequency, f = 1/T, is the period's inverse. The motion's frequency, f = 1/T = ω/2π, determines how many complete oscillations occur in a given amount of time so in this case the It is measured in units of Hertz, (1 Hz = 1/s).

herex

=

(17.4cm)cos[(5.46s− 1)t]

is written in the general form

where we can identify: A=17.4cm and ω

=5.46rad /sec

To learn more about angular frequency :

brainly.com/question/12446100

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8 0
1 year ago
What is the mathematical relationship between voltage, resistance, and current?
Alex73 [517]
The formula for Voltage is V = IR or in words
voltage equals current times the resistance
5 0
3 years ago
Long wavelength corresponds to having _________frequency
Vladimir [108]
Long wavelength corresponds to having lower frequency
4 0
3 years ago
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g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e
rusak2 [61]

Answer:

(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0

Given the function:

p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in [0,\infty).

(2)

\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1

The function p(x) satisfies the conditions for a probability density function.

6 0
3 years ago
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