To determine the initial speed, v0, of the bullet, you need to use the given information and relevant equations from physics, such as the equation for projectile motion. By analyzing the trajectory of the bullet and considering factors like distance traveled and time taken, you can calculate the initial speed of the bullet.
In physics, v0 typically represents the initial velocity of an object at the start of a motion or trajectory. It is used to describe the speed and direction of an object at the beginning of an experiment, calculation, or analysis.
This is a momentum problem.Pbullet inital + Pcan initial = Pbullet final + Pcan final(Mbullet)(Vbullet initial) + (Mcan)(Vcan initial) = (Mbullet)(Vbullet final) + (Mcan)(Vcan final)Then we solve for Velocity can final by inserting known values(0.012kg)(400m/s) + (0.047kg)(0m/s) = (0.012kg)(290m/s) + (0.047kg)(velocity)V= 28.1m/s
No. V=v0 +at is the formula for velocity, the acceleration 'a' can be the same but the initial velocity v0 may be different. If v0 is the same for the two automobiles , the velocity would be the same.
The speed of a bullet coming down, when fired vertically upwards and then falling back under the influence of gravity, will be the same as its speed when it was initially fired upwards. This is because gravity affects both the upward and downward trajectories equally, assuming no other forces are involved. The speed will depend on the initial velocity of the bullet when fired.
The gun and the bullet are already traveling at a high speed. Firing the gun results in the bullet leaving the gun. Its relative velocity to someone standing still would be twice the speed. But its relative velocity to the gun would simply be the normal speed of the bullet. A similar question is the one if your traveling in a car at the speed of light what would happen when you turn on the headlights? No one is really sure. in short, the bullet would go twice the regular speed in a vacuum. I agree with the first bit and have ed a similar question on here. I find the vacuum bit of the above a bit irrelevant. A bullet fired in a vacuum would emerge at the normal speed (around 2,000+mph for a rifle round) but would carry on forever as there would be no air resistance.
In physics, v0 typically represents the initial velocity of an object at the start of a motion or trajectory. It is used to describe the speed and direction of an object at the beginning of an experiment, calculation, or analysis.
This is a momentum problem.Pbullet inital + Pcan initial = Pbullet final + Pcan final(Mbullet)(Vbullet initial) + (Mcan)(Vcan initial) = (Mbullet)(Vbullet final) + (Mcan)(Vcan final)Then we solve for Velocity can final by inserting known values(0.012kg)(400m/s) + (0.047kg)(0m/s) = (0.012kg)(290m/s) + (0.047kg)(velocity)V= 28.1m/s
No. V=v0 +at is the formula for velocity, the acceleration 'a' can be the same but the initial velocity v0 may be different. If v0 is the same for the two automobiles , the velocity would be the same.
That depends on its initial velocity and its acceleration. V1 = V0 + a * t
Her final speed is 14.5 m/s. The kinematics equation v = at + v0 will be useful here. Note that t is the time measured in seconds, a is the acceleration, v0 is the initial velocity, and v is the velocity after t seconds (the final velocity). We are given that v0 = 10 m/s, a = 0.500 m/s2, and t = 9 s. Using the above kinematics equation we get v = (0.500 m/s2)(9 s) + 10 m/s = 14.5 m/s. Since speed = |velocity|, then her final speed = |14.5 m/s| = 14.5 m/s.
The answer depends on what v0 is. And since you have not bothered to define it I cannot provide a more useful answer.
The speed is v and the starting speed is v0.
The speed of a bullet coming down, when fired vertically upwards and then falling back under the influence of gravity, will be the same as its speed when it was initially fired upwards. This is because gravity affects both the upward and downward trajectories equally, assuming no other forces are involved. The speed will depend on the initial velocity of the bullet when fired.
v^2 = v0^2 +2a*height v = speed v0 = starting speed (0 in this case) a = acceleration (9.8 is the acceleration of gravity) the speed of the object won't change based on the mass
The gun and the bullet are already traveling at a high speed. Firing the gun results in the bullet leaving the gun. Its relative velocity to someone standing still would be twice the speed. But its relative velocity to the gun would simply be the normal speed of the bullet. A similar question is the one if your traveling in a car at the speed of light what would happen when you turn on the headlights? No one is really sure. in short, the bullet would go twice the regular speed in a vacuum. I agree with the first bit and have ed a similar question on here. I find the vacuum bit of the above a bit irrelevant. A bullet fired in a vacuum would emerge at the normal speed (around 2,000+mph for a rifle round) but would carry on forever as there would be no air resistance.
If a bullet was fired towards the groung it would be accelerated at a rate of 9.9 meters per second2 due to the force of gravity. The time gravity would have to act and the velocit of the bullet is so great that the extra velocity would be insignificant.
v = v0 + a t v = velocity (m/s) V0 = initial velocity at t = 0 (m/s) a = acceleration (9.81 for earth) t = time (sec)