A bullet is fired with a speed of 670.0 m/s. Wat time is required to strike a target 353 m away mor in detail

Question

What is velocity km/h

Answer 1

R X T = D

What items do you have?

Rate that the bullet travels?
Time? No
Distance to target, yes.

Just plug and solve! T = D/R

For the second part,

You need Km/h, you have m/s, so you just need to convert the units

m to km
s to hr

[m (x km/m) ] / [ s (hr/sec) ]

Hope that helps!

Answer 2

it's not that hard. the problem doesn't say anything about the bullet accelaration nor the weight. so let's assume that that the bullet doesn't accelarate and doesn't affected by gravity. which mean, it's just a simple 1D vector.
the basic equation for movement (1D) without acceleration is:
S = V x T
where: S = distance; T = time; and V = velocity
the unit in the problem uses metric.
S = V x T --> T = S / V = 353 / 670 = 0.53 seconds
if you want to do it with velocity in km/h
convert the unit
353 m = 0.353 Km
670 m/s = (670/1000) km / (1/3600 hour) = 2414 Km/h
T = S / V = 0.353/2414 = 0.00015 hour
= 0.00015 x 3600 sec = 0.53 sec
get it? good luck

Answer 3

It depends on what level of Physics you are talking.

If you are at a very high level of Physics, you need to address the fact that a bullet flight path is parabolic to counter the force of gravity, thus the bullet travels a little farther to get to the 353m mark. Also, the air drag on a bullet causes it to slow down, so you'd be dealing with a projectile that is constantly slowing down (the ballistic coefficient is used in this portion of the calculation).

If you are dealing with simple physics, then the equation is Distance = Velocity * Time.

Time = Distance/Velocity = 353m / (670 m/s).

Time = 0.53 sec.

If you are taking drag and flight path into account...I have no idea...I haven't taken Physics in 8 or 9 years. I don't remember how to deal with changing velocities anymore.

Answer 4

353/ 670 = 0.526 sec

Answer 5

velocity = your speed...so just convert the units

Answer 6

Just step in front of it and see.