If you have a pci express version 2 video card, how many pins will the power connector have if the card requires extra power?

Marina CMI [18]

It's the old-but-important concept: "Speed" depends on who's measuring it and how THEY're moving. Different observers may very well observe different speeds, and they're all correct. There's no such thing as "the REAL speed".

When you were a little kid, did you ever get on a moving escalator or walkway, and when you got to the middle, you turned around and walked the <em>opposite</em> way, so that somebody watching you from the outside would see you not moving at all ?

Say you're on a school bus that's driving 10 mph along the street pavement, and you get up out of your seat and run forward up the aisle at 10 mph. Somebody outside the bus sees you passing them at 20 mph !

If instead, you run toward the BACK of the bus at 10 mph, somebody outside the bus sees you bobbing up and down but not moving forward or backward at all.

All this problem is saying is: The bus is driving along at 15 m/s. A passenger on the bus puts his little baseball down on the floor on its little feet, and it runs backwards down the aisle, toward the back of the bus, at 15 m/s. Somebody is standing outside looking into the bus as it passes by and the little baseball is scurrying toward the back of the bus. How will HE describe the motion of the baseball ?

Have you got it now ?

"Relative to the Earth" just means how fast an object is passing the stores and telephone poles and people standing still ... things that are attached to the Earth. "Relative to the bus" would mean how fast an object is passing by people sitting on the bus.

6 0

3 years ago

Question 1

svetlana [45]

300

Explanation:

100 x 3 =300 simple and easy

7 0

2 years ago

A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele

ludmilkaskok [199]

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for $y$ , and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.

For the x-axis

$v_{x}=v_{0}cos(45\°)=constant$

$x=(v_{0}cos(45\°))t$

For the y-axis

$v_{y}=v_{0}sin(45\°)-gt$

$y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}$

Where $v_{0}$ , is the initial velocity.

8 0

3 years ago

Modern oil tankers weigh over a half-million tons and have lengths of up to one-fourth mile. Such massive ships require a distan

sergiy2304 [10]

(a) $-1.46\cdot 10^{-4} m/s^2$

The average acceleration of the ship is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Here we have:

$u=34 km/h =9.44 m/s$ is the initial velocity

v = 0 is the final velocity

$t=18 min =64800 s$ is the time elapsed

Substituting, we find

$a=\frac{0-9.44 m/s}{64800 s}=-1.46\cdot 10^{-4} m/s^2$

(b) 4.72 m/s

Assuming the acceleration is uniform, the average velocity of the ship is given by:

$v_{avg} = \frac{v+u}{2}$

where

v is the final velocity

u is the initial velocity

Here we have:

v = 0

u = 9.44 m/s

So the average velocity of the ship is

$v_{avg} = \frac{0+9.44 m/s}{2}=4.72 m/s$

6 0

4 years ago

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q

Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0 F=0, a=0

x= 0.050 F= -7.5 N, a= -15 m/s²

x= 0.150 F= 22.5 N, a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values are from the equilibrium position of the spring.

Let's calculate the force

x = 0

F = 0

x = 0.050

F = - 150 0.050

F = - 7.50 N

x = 0.150

F = - 150 0.150

F = - 22.5 N

let's use Newton's second law to find the acceleration

F = m a

a = F / m

x = 0 m

a = 0

x = 0.050 m

a = -7.50 / 0.50

a = - 15 m / s²

x = 0.150 m

a = - 22.5 / 0.50

a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte la fuerza es dada por la fuerza de Hoke,

F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es 1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza

x=0

F= 0

x=0.050

F = - 150 0.050

F= - 7.50 N

x= 0.150

F= - 150 0.150

F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

F = m a

a = F/m

x =0 m

a = 0

x= 0.050 m

a = -7.50/ 0.50

a =- 15 m/s²

x= 0.150 m

a= - 22.5 / 0.50

a= - 45 m/s²

7 0

4 years ago