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kicyunya [14]
3 years ago
15

Why do planets not travel in a straight path?

Physics
1 answer:
Elis [28]3 years ago
6 0

Answer:

Because the gravitational attraction of the Sun hold them in motion around it

Explanation:

For an object travelling in a straight path at constant velocity, the net force acting on the object must be zero.

The planets in the Solar System, however, do not experience a zero net force: in fact, the Sun exerts a gravitational attraction on them, whose magnitude is given by

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the mass of the Sun

m is the mass of the planet

r is the average distance between the Sun and the planet

Due to the presence of this force, the Sun makes the planets 'deviating' from their straight path, forcing them to following an elliptical path around the Sun.

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3 years ago
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En la siguiente expresión matemáticas w=mg el peso w con relación a se relaciona con la masa m en una proporción
s2008m [1.1K]

Answer:

a) Directamente proporcional

Explanation:

El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.

Matemáticamente, el peso de un objeto viene dado por la fórmula;

Peso = mg

Donde;

m es la masa del objeto.

g es la aceleración debida a la gravedad.

De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.

Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.

4 0
3 years ago
The law of inertia applies to objects
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Answer:

<em>at</em><em> </em><em>rest</em><em> </em><em>and</em><em> </em><em>in</em><em> </em><em>motion</em>

Explanation:

<em>The</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>inertia</em><em> </em><em>applies</em><em> </em><em>to</em><em> </em><em>objects</em><em> </em><em>at</em><em> </em><em>rest</em><em> </em><em>and</em><em> </em><em>in</em><em> </em><em>motion</em>

6 0
3 years ago
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The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to
LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

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Power  ,P = F .v

F=force

v=velocity

v= 100 km/h

v=\dfrac{1000}{3600}\times 100\ m/s

v=27.77 m/s

75 x 1000  = F x 27.77

F= \dfrac{75000}{27.77}\ N

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

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time t given as

v= u + a t

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3 0
3 years ago
A spring is used as part of a lift system and follows Hooke's law. If the spring is
salantis [7]

Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

F = force (N)

x = extension length (m)

k = constant of proportionality (N/m)

According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

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Using F = kx, we find the proportionality constant

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F = kx

x = F/k

x = 15500/14705.8

x = 1.05m or 105cm

6 0
3 years ago
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