Answer:
Distance = 30m
Displacement = 6m W
Explanation:
Given the following:
Movement 1 = 18m W
Movement 2 = 12m E
Diatance is a scalar quantity with only magnitude and no direction. That is, in Calculating the distance moved by the locomotive, the direction of travel or movement of the object is not considered. It only measures the total amount of movement made during the Time of motion.
Therefore, total distance traveled equals :
Movement 1 + movement 2
18m + 12m = 30m
B) Displacement also measures the movement made by an object. However, Displacement is a vector quantity and therefore, considers both magnitude and direction of travel of the object. Therefore, it measures the overall change in position of the object from its starting position.
Therefore, Displacement of the locomotive equals:
18m W - 12m E = 6m E
-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .
-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .
-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²
-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =
64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .
</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.
You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.
Answer:
a switch that flicks the light on
Answer:
the magnitude of Vpg = 493.711 km/h
Explanation:
given data
speed Vpg = 560 km/h
speed Vwg = 80 km/h
solution
we get here magnitude of the plane velocity w.r.t. ground is
we know that the Vpg = Vpw + Vwg .....................1
writing the component of the velocity that is
Vpw = (0 km/h î - 560 km/h j )
Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)
adding these
Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i
Vpg = (42.025 ) î (-491.92 km/h)j
now we take magnitude
the magnitude of Vpg = 
the magnitude of Vpg = 493.711 km/h
