2 years ago

The force of gravity acting upon an object is the objects mass. True or false

Physics

1 answer:

rosijanka [135]2 years ago

4 0

False , because mass is the amount of matter contains

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A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875

mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

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2 years ago

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What scientific principle causes the different layers of the earth

Vedmedyk [2.9K]

Answer: The stratification of earth into its geologic layers was brought about by th e formation of the earth iron core.

Explanation:

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2 years ago

Which of these is an example of a mechanical wave? A. the light of the Sun B. a car stuck in traffic C. the shaking from an eart

sesenic [268]

A Explanation: I took this test before

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2 years ago

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

mihalych1998 [28]

The radius of a nucleus of hydrogen is approximately $r_{n1}=1\cdot 10^{-15}m$ , while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: $r_{e1}=5.3 \cdot 10^{-11}m$

The radius of a dime is approximately $r_{n2} = 9\cdot 10^{-3}m$ : if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion
$r_{n1}:r_{e1}=r_{n2}:r_{e2}$
from which we find
$r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m$

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.

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3 years ago

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

Helga [31]

Answer:

2.59 m

Explanation:

Coordinates of origin = (0, 0)

Coordinates of Point p where the fly reach = (2.3 m, 1.2 m)

Use the distance formula of coordinates to find the distance between the origin and the point P.

$d=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$

$d=\sqrt{\left ( 2.3- 0 \right )^{2}+\left ( 1.2-0 \right )^{2}}$

d = 2.59 m

Thus, the distance between the origin and the point P is 2.59 m.

5 0

3 years ago