Answer:
m1 = 20g (= 0.02 kg)
Mass of pistol, m2 = 2 kg
Initial velocity of the bullet (u1) and pistol (u2) = 0
Final velocity of the bullet, v1 = +150m s-1
Let v be the recoil velocity of the pistol.
Total momentum of the pistol and bullet after it is fired is
= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)
= (3 + 2v) kg m s-1
Total momentum after the fire = Total momentum before the fire
3 + 2v = 0
→v = -1.5 m/s
Answer:
0.0173 V
Explanation:
PARAMETERS GIVEN:
Inductance, L = 2 mH = 0.002 H
Time interval, dt = 0.15 s
Change in current, dI = 1.5 - 0.2 = 1.3 A
The magnitude of the induced EMF is given as:
EMF = | -L*dI/dt |
EMF = (0.002 * 1.3) / 0.15
EMF = 0.0173 V
<span> we suppose that x be distance of man from spot light is
so 12-x is distance from man to wall
we will draw first triangle ABC, where
a= spotlight (on ground)
b= man (feet)
c= man (head)
we know that AB = x and BC = 2
by extend line AB to the wall at point D and extend line AC to the wall at point E
AD = 12 (distance from spotlight to wall)
DE = s (length of shadow)
Now triangles ABC and ADE are similar
Therefore DE/AD = BC/AB
s / 12 = 2 / x
s = 24 / x
Differentiate both sides with respect to t
ds/dt = -24/x² dx/dt
Man is walking toward building at speed of 16 ms
dx/dt = 16
Man is 4 m from the building/wall
12 - x = 4
x = 8
Find ds/dt when x = 8 and dx/dt = 16
ds/dt = -24/x² dx/dt
ds/dt = -24/(64) * 16
ds/dt = -6
So length of shadow is decreasing at rate of 6 m/s
hope it helps
</span>
Answer:
no, yo no lo conosco...porque preguntas?
Explanation:
