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2 years ago

Why is the moon’s surface cratered, but the Earth’s is not

1 answer:

8 0

The comets hit the moon's surface because there is no atmosphere on the moon to protect it. The earth has an atmosphere so it is protected.

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Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f

ruslelena [56]

Answer:

Moc = -613.25 [lb*in]

Explanation:

Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.

Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.

$M_{o}=r_{A/O} * F_{AB}$ (producto cruz)

Necesitamos identificar los puntos:

O (0,0,0) [in]

A (12,0,0) [in]

B (0, 24,8) [in]

C (12,24,0) [in]

$r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k$

El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].

$F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]$

De esta manera realizando el producto cruz tenemos

$M_{O}=r_{A/O} * F_{AB}$

$M_{O}=0i-685.7j+2057.1k [Lb*in]$

Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.

$OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j +0k$

El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC

$M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]$

7 0

2 years ago

A spaceship traveling to Alpha Centauri at 0.75c sends a message home to Earth saying they are at the halfway point.

Lina20 [59]

Answer:

(a) 24.87 GHz (b) 3.55 GHz

Explanation:

We have given the speed of the spaceship =0.75 c, here c is the speed of the light

(a) In the first case the ship is moving towards the alpha century

So the frequency is given by $f'=f\times \sqrt{\frac{c+0.75c}{c-0.75c}}=9.4\times \sqrt{\frac{c+0.75c}{c-0.75c}}=24.87GHz$

(b) In this case the ship is moving away from the alpha century

So frequency is given by $f'=f\times \sqrt{\frac{c-0.75c}{c+0.75c}}=9.4\times \sqrt{\frac{c-0.75c}{c+0.75c}}=3.55GHz$

5 0

2 years ago

How does gravity affect speed

mixas84 [53]

Answer:

When objects fall to the ground, gravity causes them to accelerate. ... Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2, so by 1 second after an object starts falling, its velocity is 9.8 m/s

Explanation:

3 0

2 years ago

Read 2 more answers

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

The position of the object at t = 0 is x = -14.0 m

The velocity at t = 0 s is $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

$a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=> $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=> $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At t =0 s and $v_{0}x = 7.10 m/s$

=> $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=> $K_1 = 7.10$

$\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=> $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=> $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At t =0 s and x = -14.0 m

$-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=> $K_2 = -14$

$X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At t = 10.0 s

$X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=> $X = 38.3 \ m$

5 0

2 years ago

What is the difference between industrialized countries and developing countries

Anvisha [2.4K]

Answer: Developed nations are generally categorized as countries that are more industrialized and have higher per capita income levels. ... Developing nations are generally categorized as countries that are less industrialized and have lower per capita income levels.

Explanation: this is the source that i used to find the answer:

https://study.com/academy/lesson/economic-inequality-differences-in-developed-and-developing-nations.html

6 0

2 years ago