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3 years ago

8

A runner starts at the start line runs around a 400 m track two times ending up back at the store now what is the runners distan

ce and displacement

1 answer:

Marta_Voda [28]3 years ago

7 0

Answer:

Distance is 800m and displacement is 0m

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An object initially at rest experiences an acceleration of 9.8 m/s2. How much time will it take to achieve a velocity of 58 m/s?

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5.91(approx) seconds just divide velocity by acceleration

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3 years ago

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

$r_1 = 11cm$

$R_2 = 20cm$

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

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3 years ago

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3 years ago

Jonathan has six strings that are the same thickness and are all the same material. He cuts the strings to different lengths and

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3 years ago

Read 2 more answers

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

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Answer:

$f_e = 1.51 cm$

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

$m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}$

$400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}$

$9.6 = \dfrac{16-f_e}{f_e}$

$9.6f_e = 16-f_e$

$10.6 f_e = 16$

$f_e = 1.51 cm$

3 0

3 years ago