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3 years ago

In the rectangle to the left, if

Physics

1 answer:

Margarita [4]3 years ago

7 0

perimeter of a rectangle = 2(L+B)

90=2(L+B)

90/2=L+B

45=L+B

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What makes some chemical bonds more stable than others

Juliette [100K]

it occurs when the energy of the combination has lower energy than the separated atoms

4 0

3 years ago

if the distance between two objects is doubled, the force of gravitation between them becomes __________ of initial value.

dexar [7]

1/4 of initial time :)

8 0

3 years ago

a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed

Juliette [100K]

Mass of the car (m) = 2000 Kg
Initial velocity (u) = 15 m/s
Force (F) = 10000 N
Time (t) = 3 s
Let the acceleration be a.
By using the formula, F = ma, we get,
10000 N = 2000 Kg × a
or, a = 10000 N ÷ 2000 Kg
or, a = 5 m/s^2
Let the final velocity be v.
By using the formula, v = u + at, we get,
v = 15 m/s + 5 m/s^2 × 3 s
or, v = 15 m/s + 15 m/s
or, v = 30 m/s

Answer:

The new speed of the car is 30 m/s.

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is: