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Nutka1998 [239]

3 years ago

9

El carrito a control remoto de Juanito corre en una pista recta de 40 m de un extremo a otro en 90 segundos; luego de regreso so

bre la pista recorre 15 metros y se ubica en medio en 25 segundos, cual es la rapidez?
y cual es la velocidad?

1 answer:

PSYCHO15rus [73]3 years ago

8 0

Explanation:

hir ceh3co8d ueirfkgrewkfgv ruk gr

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What is a concentration?

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3 years ago

Read 2 more answers

An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit

nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

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3 years ago

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3 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago