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Elan Coil [88]
3 years ago
7

To ancient peoples, why were planets special?

Physics
1 answer:
coldgirl [10]3 years ago
7 0

Answer:

Planets were like gods.

Explanation:

To the people of many ancient civilizations, the planets were thought to be deities. Our names for the planets are the Roman names for these deities. For example, Mars was the god of war and Venus the goddess of love.

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When can a theory be modified if a new type of technology allows for new observations that raise new questions?
olga nikolaevna [1]

Answer:

id say B, but not sure

Explanation:

4 0
3 years ago
Read 2 more answers
Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen
aleksklad [387]

Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

6 0
3 years ago
I need a science fair project that doesn't need anything on the pic and it can be about any subject thanks.
tigry1 [53]

you could make a self propelled car all you need is cardboard, wheels, and a balloons or rubber bands

  

7 0
3 years ago
Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?
makkiz [27]
I think it is B. 100 W
7 0
3 years ago
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