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

Which is the BEST definition of refraction? A) Light or sound waves change direction. B) Light or sound waves bounce off a mediu

melisa1 [442]

The answer to this is A. this is because, refraction with a light or sound wave changing its direction involve propagation,(in which propagation is the change in direction of a light or sound wave)

4 0

3 years ago

Read 2 more answers

The breaking car had 10,000 J of kinetic energy before breaking after breaking it had 2000 J of kinetic energy. How much thermal

WINSTONCH [101]

Answer:

8000J

Explanation:

The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.

Kinetic energy loss by car = thermal energy gained by brakes.

∆K.E = ∆T.E ....1

The Kinetic energy loss by car can be expressed as;

∆K.E = K.E1 - K.E2

Initial K.E = K.E1 = 10000J

Final K.E = K.E2 = 2000J

∆K.E= 10000J - 2000J = 8000J

From equation 1,

∆K.E = ∆T.E

∆T.E = 8,000J

thermal energy gain by brakes = 8,000J

8 0

2 years ago

A block slides down a frictionless inclined ramp. If the ramp angle is 17.0° and its length is 30.0 m, find the speed of the blo

shutvik [7]

Answer:

v = 17.15 m/s

Explanation:

given,

angle of ramp = 17.0°

length of ramp(l) = 30 m

height of the ramp =

$h = l sin \theta$

$h = 30 sin 30^0$

h = 15 m

using energy of conservation

$\dfrac{1}{2}mv^2 = mgh$

$\dfrac{1}{2}v^2 = gh$

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8\times 15}$

$v = \sqrt{294}$

v = 17.15 m/s

speed of block reaching at the bottom = v = 17.15 m/s

5 0

3 years ago

Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

3 years ago

You are on a new planet, and are trying to use a simple pendulum to find the gravity experienced on this new planet. You find th

levacccp [35]

To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

$T= 2\pi \sqrt{\frac{L}{g}}$

g = Gravity

L = Length

T = Period

Re-arrange to find the gravity we have

$g = \frac{4\pi^2 L}{T^2}$

Our values are given as

$L = 0.35m\\T = 2s\\$

Replacing we have

$g = \frac{4\pi^2 L}{T^2}$

$g = \frac{4\pi^2 0.35}{2^2}$

$g = 3.45 m/s^2$

Therefore the correct answer is C.

4 0

3 years ago