Answer:
at distance 18.33 ft no biological effects
Explanation:
given data
distance = 41 ft
radiation = 5 rem
to find out
how close get so no biological effects
solution
we know here that intensity of radiation R is inversely proportional to (radiation)²
so here equation will be
.............1
here I is intensity
so here I1 = 25
because for 0 to 25 there is no detectable effects and for 25 to 100 it will . temporary decrease so
we take 25 because dose is less than 25 so we take that highest value
so
R1 = 18.33 ft
so at distance 18.33 ft no biological effects
I think the answer would be 78 electrons, maybe?
The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;
Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
Learn more here: brainly.com/question/24430414
Answer:
The force exerted by the wood on the bullet is 399.01 N
Explanation:
Given;
mass of bullet, m = 0.0021 kg
initial velocity of the bullet, u = 497 m/s
final velocity of the bullet, v = 0
distance traveled by the bullet, S = 0.65 m
Determine the acceleration of the bullet which is the deceleration.
Apply kinematic equation;
V² = U² + 2aS
0 = 497² - (2 x 0.65)a
0 = 247009 - 1.3a
1.3a = 247009
a = 247009 / 1.3
a = 190006.92 m/s²
Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;
F = ma
F = 0.0021 x 190006.92
F = 399.01 N
Therefore, the force exerted by the wood on the bullet is 399.01 N
Answer:
I=1,2•10³ kg•m/s
Explanation:
v¹=3.5m/s
vf=5m/s
v=5-3.5=1.5m/s
I=p
I=mv=850•1.5=1275 kg•m/s=1,2•10³ kg•m/s
