In diving to a depth of 995 m, an elephant seal also moves 839 in due east of hi
ID: 1532894 • Letter: I
Question
In diving to a depth of 995 m, an elephant seal also moves 839 in due east of his starting point. What is the magnitude of the seal's displacement? A spam craft is traveling with a velocity of upsilon_0x = 6870 m/s along the + x direction. Two engines are turned on for a time of 772 s. One engine gives the spacecraft an acceleration in the +x direction of a_x = 1.93 m/s^2. while the other gives it an acceleration in the +y direction of a_y = 7.46 m/s^2. At the end of the firing, what is a) upsilon_x and b) upislon_y? A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t = 0 s, the x components of the puck's initial velocity and acceleration are upsilon_0x = +7.8 m/s and a_x = +7.1 m/s^2. The y components of the puck's initial velocity and acceleration are upsilon_0y = +9.4 m/s and a_y = -4.3 m/s^2. Find (a) the magnitude v and (b) the direction theta of the puck's velocity at a time of t = 0.50 s. Specify the direction. A rocket is fired at a speed of 92.0 m/s from ground level, at an angle of 70.0 degree above the horizontal. The rocket is fired toward an 55.2-m high wall, which is located 40.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut flown and the rocket coasts. By how much does the rocket clear the top of the- wall? Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands four times as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.Explanation / Answer
(1)
let +x be east directiion
The seal displacment is
d= 839 x-995 y
magnitude
d= sqrt ( 839)^2 + (-995)^2 = 1301.51 m
(2)
Since the horizontal motion is completely independent of the vertical motion, we can solve the equations of motion independently in each direction.
For x-direction: vx = v0x + axt = 6870 + 1.93*772 = 8359.96m/s
For the y-direction: vy = 0 + (ay - g) t = (7.46 – 9.81)*772 = -1814.2 m/s
(3)
v_x = v_0x + a t
= ( + 7.8 m/s) + ( 7.1 m/s^2 ) ( 0.5 s)
= 11.35 m/s
v_y = v_0y + a_y t
= 9.4m/s + ( - 4.3 m/s^2) ( 0.6 s)
= 6.82 m/s
magnitude
v = sqrt vx^2 + vy^2 = sqrt ( 11.35)^2 + ( 6.82)^2 = 13.24 m/s
direction
theta = tan^-1 ( 6.82/11.35) = 31 degree
As per guide lines I worked first three problem kindly post reamaining posts in the next post
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