Let F be the magnitude of the frictional force. This force performs an amount of work W on the bullet such that
W = -Fx
where x is the distance over which F is acting. This is the only force acting on the bullet as it penetrates the tree. The work-energy theorem says the total work performed on a body is equal to the change in that body's kinetic energy, so we have
W = ∆K
-Fx = 0 - 1/2 mv²
where m is the body's mass and v is its speed.
Solve for F and plug in the given information:
F = mv²/(2x)
F = (0.00426 kg) (881 m/s)² / (2 (0.0444 m))
F = 37,234.8 N ≈ 37.2 kN
Momentum = (mass) x (velocity)
Original momentum before the hit =
(0.16 kg) x (38 m/s) this way <==
= 6.08 kg-m/s this way <==
Momentum after the hit =
(0.16) x (44 m/s) that way ==>
= 7.04 kg-m/s that way ==>
Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .
-----------------------------------------------
Change in momentum = impulse.
Impulse = (force) x (time the force lasted)
13.12 kg-m/s = (force) x (0.002 sec)
(13.12 kg-m/s) / (0.002 sec) = Force
6,560 kg-m/s² = 6,560 Newtons = Force
( about 1,475 pounds ! ! ! )
Answer:
Explanation:
<u>Given:</u>
Force = f = 60 N
Mass = m = 12 kg
<u>Required:</u>
Acceleration = a = ?
<u>Formula:</u>
F = ma
<u>Solution:</u>
Rearranging formula
a = F / m
a = 60 / 12
a = 5 ms⁻²
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3><h3>Peace!</h3>
There are two atoms of potassium bonded to one atom of sulfur.
Answer:
Explanation:
fundamental frequency, f = 250 Hz
Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.
the formula for the frequency is given by
.... (1)
Now the length is doubled ans the tension is four times but the mass remains same.
let the frequency is f'
.... (2)
Divide equation (2) by equation (1)
f' = √2 x f
f' = 1.414 x 250
f' = 353.5 Hz
