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3 years ago

PLS HELP

Physics

2 answers:

olasank [31]3 years ago

5 0

A, and C hope this helps u out.

RoseWind [281]3 years ago

4 0

Answer:

A and C are correct.

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GIVING BRAINLIEST PLEASE HELP!!

alisha [4.7K]

Answer:

c is the right answer is it

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3 years ago

The average lifetime of a poodle dog is 13.0 years. How fast is it traveling, u, relative to an observer who measures the averag

bogdanovich [222]

Answer:

The speed is 0.97 c.

Explanation:

Given that,

Dilated time t'= 50.0 years

Rest time t = 13.0 years

We need to calculate the speed

Using formula of time dilation

$t'=\dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Where, t' = change in time

t = rest time

v = velocity

c = speed of light

Put the value into the formula

$50.0=\dfrac{13.0}{\sqrt{1-\dfrac{v^2}{(3\times10^{8})^2}}}$

$v^2=\dfrac{(13)^2\times(c)^2-(c)^2\times50^2}{50^2}$

$v^2= 0.9324c^2$

$v=0.97c$

Hence, The speed is 0.97 c.

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3 years ago

Didn't know what type of science it was? But pls helpp

Alla [95]

The answer is the third one down. New evidence may contradict the old evidence of a certain theory.

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3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$

$d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m$

From the continuity equations in pipes we have to

$A_1V_1 = A_2 V_2$

Where,

$A_{1,2}$ = Cross sectional Area at each section

$V_{1,2}$ = Flow Velocity at each section

Then replacing we have,

$(\pi r_1^2) v_1 = (\pi r_2^2) v_2$

$(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2$

$v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1$

From Bernoulli equation we have that the change in the pressure is

$\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)$

$7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)$

$7300= 8919.01 v_1^2$

$v_1 = 0.9m/s$

Therefore the speed of flow in the first tube is 0.9m/s

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4 years ago

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kenny6666 [7]

Answer:

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3 years ago

Read 2 more answers