A testing instrument that's used to measure electrical signals
in a circuit and display them as waveforms on a screen is called
an oscilloscope.
The answer to the question is that before the big bang, the universe was much hotter and more dense than it is now. Letter B.
It is because after the big bag occurred, the universe became cooler and less dense.
a. - does not correspond in the answer because the universe became less dense after the big bang.
c - the universe became cool and less dense after the big bang so being cool and less dense does not correspond to the question.
d - cooler does not answer the question because it only became cooler after the big bang.
Ok, this is a 2d kinematics problem, the falls 14 m part is confusing, I think it means in the x direction, but you don't need it anyway.
If we know it goes 4m into the air, we know d = 4m (height of wall), we also know the acceleration a=-9.8m/s^2 (because gravity) and that the vertical velocity when it just clears the wall will be 0 m/s, which we'll call our final velocity (Vf). Using Vf^2 = Vi^2 +2a*d, we can solve this for Vi and drop Vf because it's zero to get: Vi = sqrt(-2ad), plug in numbers (don't forget a is negative) and you get 8.85 m/s in the vertical direction. The x-direction velocity requires that we solve the y-direction for time, using Vf= Vi + at, we solve for t, getting t= -Vi/a, plug in numbers t= -8.85/-9.8 = 0.9 s. Now we can use the simple v = d/t (because x-direction has no acceleration (a=0)), and plug in the distance to the wall and the time it takes to get there v = (4/.9) = 4.444 m/s, this is the velocity in the x direction, we use Pythagoras' theorem to find the total velocity, Vtotal = sqrt(Vx^2 + Vy^2), so Vtotal = sqrt(8.85^2+4.444^2) = 9.9m/s. Yay physics!
The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.