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To find the initial speed of spaceship one, we need to apply the principle of conservation of momentum. Since the two spaceships have equal masses, their momenta will be equal and opposite. The momentum of spaceship two is given by 150 kg * V2, where V2 is the initial speed of spaceship two. The momentum of spaceship one is given by 150 kg * V1, where V1 is the initial speed of spaceship one. Since they have equal magnitudes, we have 150 kg * V1 = 900 kg * (-V2). Solving for V1 gives V1 = - 6 V2. Since we want the initial speed in magnitude, the initial speed of spaceship one is 6 times the initial speed of spaceship two in magnitude.
the equation for an ideal gas is pv / t = nr n * r is a constant for a closed system p pressure v volume t temperature in kelvin p1 v1 /t1 = p2 v2 /t2 if p1 = p2 v1/t1 = v2/t2 t2= v2/v1 *t1 directly proportional to the change in volume if v1 = v2 the same can be done and you will find that t is directly proportional to change in pressure. generally t is directly proportional to the product of pressure and volume. pv = nr t
The formula to determine acceleration is acceleration = change in velocity / time taken. It can also be written as a = (v2 - v1) / t, where a is acceleration, v1 is the initial velocity, v2 is the final velocity, and t is the time taken.
When the kinetic energy increases by four times, the speed of the body will double. The relationship between kinetic energy and speed is not linear, but rather quadratic. So if the kinetic energy increases by four times (2^2), the speed will increase by two times (2).
One common formula for calculating speed after a collision is the conservation of momentum equation: m1v1 + m2v2 = (m1 + m2)v, where m1 and m2 are the masses of the objects involved, v1 and v2 are their initial velocities, and v is the final velocity after the collision.
v1 is design speed and v2 rotation speed
v1 = initial velocity v2 = final velocity
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Percentage change from V1 = 12 to V2 = 18:[ ((V2 - V1) / |V1|) * 100 ]= ((18 - 12) / |12|) * 100= (6 / 12) * 100= 0.5 * 100= 50% change= 50% increase
Percentage change from V1 = 750 to V2 = 825:[ ((V2 - V1) / |V1|) * 100 ]= ((825 - 750) / |750|) * 100= (75 / 750) * 100= 0.1 * 100= 10% change= 10% increase.
Calculate percentage changefrom V1 = 60 to V2 = 84[ ((V2 - V1) / |V1|) * 100 ]= ((84 - 60) / |60|) * 100= (24 / 60) * 100= 0.4 * 100= 40% change= 40% increase.
Percentage change from V1 = 700 to V2 = 2000:[ ((V2 - V1) / |V1|) * 100 ]= ((2000 - 700) / |700|) * 100= (1300 / 700) * 100= 1.857143 * 100= 185.7143% change= 185.7143% increase
Calculate percentage changefrom V1 = 3624 to V2 = 8236[ ((V2 - V1) / |V1|) * 100 ]= ((8236 - 3624) / |3624|) * 100= (4612 / 3624) * 100= 1.272627 * 100= 127.2627% change= 127.26% increase.
"V1" is the speed at which the flight crew must either continue with the takeoff or abort the takeoff. It is the last point at which aborting is an option.
V1 is a v-speed, in aircraft terminology v-speeds are specific speeds for various operations. Takeoff speed is V2, maximum speed with landing gear extended is VLE, maximum speed with flaps extended is VFE. These speeds vary from model to model & so its important for pilots to familiarize themselves with the v-speeds of any aircraft they intend to fly. V1 is the maximum speed during takeoff at which a pilot can safely stop the aircraft without leaving the runway. This is also the minimum speed that allows the pilot to safely continue (to V2 takeoff) even if a critical engine failure occurs (between V1 and V2). So basically V1 is the point of no return, once you've hit V1, you are committed to the takeoff, even if you lose an engine you're better off flying than you are trying to stop.
V1 is a v-speed, in aircraft terminology v-speeds are specific speeds for various operations. Takeoff speed is V2, maximum speed with landing gear extended is VLE, maximum speed with flaps extended is VFE. These speeds vary from model to model & so its important for pilots to familiarize themselves with the v-speeds of any aircraft they intend to fly. V1 is the maximum speed during takeoff at which a pilot can safely stop the aircraft without leaving the runway. This is also the minimum speed that allows the pilot to safely continue (to V2 takeoff) even if a critical engine failure occurs (between V1 and V2). So basically V1 is the point of no return, once you've hit V1, you are committed to the takeoff, even if you lose an engine you're better off flying than you are trying to stop.
To find the initial speed of spaceship one, we need to apply the principle of conservation of momentum. Since the two spaceships have equal masses, their momenta will be equal and opposite. The momentum of spaceship two is given by 150 kg * V2, where V2 is the initial speed of spaceship two. The momentum of spaceship one is given by 150 kg * V1, where V1 is the initial speed of spaceship one. Since they have equal magnitudes, we have 150 kg * V1 = 900 kg * (-V2). Solving for V1 gives V1 = - 6 V2. Since we want the initial speed in magnitude, the initial speed of spaceship one is 6 times the initial speed of spaceship two in magnitude.