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erastovalidia [21]

2 years ago

9

From what material did the moon form?How are eras and periods of the geologic time scale named?

1 answer:

Darina [25.2K]2 years ago

7 0

Answer:

this

Explanation:

because of this its this ok

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What happens to a path of a light ray parallel to the principal axis, after it passes through a converging

Simora [160]

Answer: The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. ... All three rays should intersect at exactly the same point.

Explanation: Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses.

3 0

3 years ago

A football player kicks a ball with a mass of 0.42 kg. The average acceleration of the football was 14.8 m/s².

kvasek [131]

D.6.22N. because .42kg * 14.8m/s=6.22 N[meaning newtons}.

6 0

3 years ago

Read 2 more answers

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

A wave with a low frequency generally has a _____.

Artyom0805 [142]

Its C because if it is a low frequency it will not change much so it will be a longer wavelength and the higher the frequency the shorter the wavelength

8 0

3 years ago

Find the magnitude of the sum

shusha [124]

Answer:

$\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m$

Explanation:

<u>Sum of Vectors in the Plane</u>

Given two vectors

$\displaystyle \vec{v_1}\ ,\ \vec{v_2}$

They can be expressed in their rectangular components as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_2}=$

The sum of both vectors can be done by adding individually its components

$\displaystyle \vec{v_1}+\vec{v_2}=$

If the vectors are given as a magnitude and an angle $(M\ ,\ \theta )$ , each component can be found as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_2}=$

The first vector has a magnitude of 3.14 m and an angle of 30°, so

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_1}=$

The second vector has a magnitude of 2.71 m and an angle of -60°, so

$\displaystyle \vec{v_2}=$

$\displaystyle \vec{v_2}=$

The sum of the vectors is

$\displaystyle \vec{v_1}+\vec{v_2}=$

$\displaystyle \vec{v_1}-\vec{v_2}=$

Finally, we compute the magnitude of the sum

$\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{(4.08)^2+(-0.78)^2}$

$\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{17.25}$

$\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m$

3 0

3 years ago