Ask question

Ask question

All categories

3 years ago

15

Un auto parte desde un lugar a las 6:00 de la mañana con una rapidez de 150 Km/hr a través de una carretera recta. a) ¿Qué dista

ncia habrá recorrido a las 10:00 de la mañana? b) ¿A qué hora habrá recorrido 900 Km?

1 answer:

Gwar [14]3 years ago

8 0

Answer:

600 km

6 hours

Explanation:

El tiempo que viaja el automóvil de 6 a 10 de la mañana es $10-6=4\ \text{horas}$

La velocidad del coche es 150 km/hr

La distancia está dada por

$\text{Distancia}=\text{Velocidad}\times \text{Tiempo}$

$150\times 4=600\ \text{km}$

A distancia recorrida a las 10 de la mañana es de 600 km.

Ahora el tiempo que se toma está dado por

$\text{Tiempo}=\dfrac{\text{Distancia}}{\text{Velocidad}}$

$\dfrac{900}{150}=6\ \text{horas}$

El tiempo que se tarda en recorrer 900 km es de 6 horas.

You might be interested in

A snail at position 3 cm moves to position 20 cm in 8 seconds.

natulia [17]

Answer: 17cm.

Explanation:

The equation you're using is:

Δd = df - di

Which means the change in position is equal to the final position minus the starting position. In this case that works out to 20cm - 3cm = 17cm. We're only interested in how much the snail moved, not how long it took to move, so even though they give a time it actually doesn't matter for this question.

4 0

3 years ago

A woman flies 300 kilometers north, 200 kilometers west, and 500 kilometers south. How far is she from her starting point?

maria [59]

She is
$\sqrt{200^2+200^2}= ~282.84$ kilometers away from her starting point

5 0

4 years ago

6 latter word and it has a s and a I and it has mass (9.______________liquids, and gases all have mass.)

Firdavs [7]

Answer:

Solids

Explanation:

Solids, liquids, and gases all have mass.

7 0

2 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

4 years ago

Equations to use: v= λ ∙ f v=d/t

Margarita [4]

b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

$f=\frac{v}{2L}$

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

$v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s$

c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

$v=\frac{d}{t}$

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

$d=vt=(6,000 m/s)(3 s)=18,000 m$

3 0

3 years ago