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dlinn [17]
4 years ago
8

Cuál es el tiempo empleado por una moto que alcanza una velocidad final de 100km/h partiendo del reposo y lo hace a una acelerac

ión constante de 2km/h 2 ?
Physics
1 answer:
victus00 [196]4 years ago
6 0

Answer:

50 horas

Explanation:

Datos:

velocidad inicial: v_{i}=0km/h (inicia del reposo)

velocidad final: v_f=100km/h

aceleración: a=2km/h^2

Usando la formula de la velocidad final con aceleracion constante:

v_f=v_i+at

donde t es el tiempo.

De la formula anterior despejamos para el tiempo t:

t=\frac{v_f-v_i}{a}

sustituimos los valores:

t=\frac{100km/h-0km/h}{2km/h^2}\\ \\t=\frac{100km/h}{2km/h^2}\\ \\t=50h

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Answer:

5 x 10⁻⁷N

Explanation:

Given parameters:

Mass of object 1  = 100kg

Mass of object 2 = 300kg

Distance  = 2m

Unknown:

Force of gravitational attraction between the objects  = ?

Solution:

From Newton's law of universal gravitation we derive an expression:

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G is the universal gravitation constant = 6.67 x 10⁻¹¹

m is the mass

r is the distance between the bodies

 Now insert the parameters and solve;

     Fg  = 6.67 x 10⁻¹¹ x \frac{100 x 300}{2^{2} }   = 5 x 10⁻⁷N

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3 years ago
A Stegosaurus hits' a wall with a force of
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A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q
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Read 2 more answers
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
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