Answer:
B. From the light due to the explosion
Explanation:
Space is basically a vacuum, meaning it is void of matter or to be more accurate, the matter are very far apart. Now sound and vibrations need matter for them to travel through. Sound is also a type of vibration, and they can only travel through a medium (made up of matter) for them to move along. Since the astonaut is in space, then he won't feel vibrations or hear the sound of the expolosion..
On the other hand, light is an EM wave or an electromagnetic wave. Electromagnetic waves can travel through a vacuum. So the answer would be B.
When the net forces equal 0 Newtons, they are a balanced force.
Answer:
Dx = - 74.82 [m]
Explanation:
In order to solve this problem we must decompose the horizontal and vertical axes using the trigonometric functions of the sine and cosine.
For the x component, we must use the cosine of the angle.
Dx = 101*cos(42.2) = 74.82 [m]
Now we must appreciate that the x component is going to the left therefore we have a negative component in the x-direction.
Dx = - 74.82 [m]
Answer:
A). Enough to fly to the first point of intended landing and to fly after that for 45 minutes at normal cruising speed
Explanation:
<u>Here are Fuel requirements for flight in VFR conditions</u>
No person may begin a flight in an airplane under VFR conditions unless there is enough fuel to fly to the first point of intended landing and, assuming normal cruising speed -
- During the day, to fly after that for at least 30 minutes; or
- At night, to fly after that for at least 45 minutes.
Answer:
Velocity = 20.3 [m/s]
Explanation:
This is a typical problem of energy conservation, where potential energy is converted to kinetic energy. We must first find the potential energy. In this way, we will choose as a reference point or point where the potential energy is zero when the carrriage rolls down 21 [m] from the top of the hill.
![E_{p} =m*g*h\\ where:\\m = mass = 25[kg]\\g = gravity = 9.81 [m/s^2]\\h = elevation = 21 [m]\\E_{p} =potential energy [J]\\E_{p} =25*9.81*21=5150[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5C%20where%3A%5C%5Cm%20%3D%20mass%20%3D%2025%5Bkg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%3D%2021%20%5Bm%5D%5C%5CE_%7Bp%7D%20%3Dpotential%20energy%20%5BJ%5D%5C%5CE_%7Bp%7D%20%3D25%2A9.81%2A21%3D5150%5BJ%5D)
Now this will be the same energy transformed into kinetic energy, therefore:
![E_{p}=E_{k} = 5150[J]\\E_{k} =0.5*m*v^{2} \\where:\\v=velocity [m/s]\\v=\sqrt{\frac{E_{k}}{0.5*25} } \\v=20.3[m/s]](https://tex.z-dn.net/?f=E_%7Bp%7D%3DE_%7Bk%7D%20%3D%205150%5BJ%5D%5C%5CE_%7Bk%7D%20%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv%3Dvelocity%20%5Bm%2Fs%5D%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2A25%7D%20%7D%20%5C%5Cv%3D20.3%5Bm%2Fs%5D)