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sattari [20]
3 years ago
12

What describes how much matter is in a object??

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
8 0
Mass and weight are two different things. The unit for mass is a gram. A nickel has the mass of about one gram. Objects<span> that take up space and have mass are called </span>matter<span>.</span>
You might be interested in
What line in the sky is created by our revolution around the Sun?
faust18 [17]

Answer:

Ecliptic

Explanation:

You asked for the term that is used to describe Earth going around the sun

The ecliptic is the path the sun, moon, and planets take across the sky as seen from Earth. It best defines the plane of the Earth's orbit around the sun. The imaginary line can best be visualized in the days just before full moon.

Hope it helped

6 0
3 years ago
dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
3 years ago
The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y,
kap26 [50]

W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} }  ]

<u>Explanation:</u>

The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is

r(t) = (1-t) . (a,b,c) + t . (l, m, n)  where t ∈ |0, 1|

So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is

r(t) = (1 - t) . (3,0,0) + t . (3,2,5)  where t ∈ |0, 1|

r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t)  where t ∈ |0, 1|

r(t) = (3, 2t, 5t)

Given:

F(x, y, z) = \frac{Kr}{|r|^3} \\\\F(x, y, z) = \frac{K}{(x^2 + y^2 + z^2)^3^/^2}  (x, y, z)\\\\F(r(t)) = \frac{K}{(3^2 + (2t)^2 + (5t)^2)^3^/^2}  (3, 2t, 5t)\\\\F(r(t)) = \frac{K}{(9 + 29t^2)^3^/^2} (3, 2t, 5t)

dr = (0, 2, 5) dt

Work = \int\limits^1_0 {F} \, dr

W = \int\limits^1_0 {\frac{K}{(9 + 29t^2)^3^/^2} } (3, 2t, 5t) . (0, 2, 5)\, dt\\ \\    = \int\limits^1_0 {\frac{K ( 4t + 25t)}{(9 + 29t^2)^3^/^2} } \, dt\\\\\\

W = \frac{1}{2}\int\limits^1_0 {\frac{K(29t)}{(9 + 29t^2)^3^/^2} } \, dt \\ \\

Substitute = 9 + 29t² = u, 92tdt = du

Limit changes from 0→1 to 9 → 38

W = \frac{K}{2} \int\limits^3_9 {\frac{du}{u^3^/^2} } \,\\\\

On solving this, we get:

W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} }  ]

Therefore, work done is W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} }  ]

5 0
3 years ago
Plz help me with my career!!!<br> part one...
AlladinOne [14]

Answer:

#1 Yes

Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.

8 0
3 years ago
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