Answer:
1.a) 1 kJ
1.b) 4 kJ
ratio 1:4
1.c) 4 times as before
2.a) 3.33 m/s2
Explanation:
1.a) bicycle's velocity =Displacement/time
=100/20 m/s
=5 m/s
bicycler's KE =1/2 *mass*(velocity)^2
=1/2*80*5^2
=1000 J = 1 kJ
1.b) bicycle's new velocity =200/20 m/s
=10 m/s
bicycler's new KE =1/2*80*10^2
=4000 J = 4 kJ
Ratio= KE 1 :KE new
= 1 :4
1.c) when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it
ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times
2.a) car acceleration = (20-0)/6 m/s2
= 3.33 m/s2
( 3 yr) · (186,282.397 mile/s) · (86,400 s/day) · (365 day/yr)
= (3 · 186,282.397 · 86,400 · 365) mile
= 1.762380502 x 10¹³ miles
= 1.8 x 10¹³ miles (rounded to the nearest trillion miles)
So first things first its c. because when two plates collide it causes a earthquake so u can rule those out and of course no not volcano so that's it u only have c left as your answer choice.
:/
That's "<em><u>insolation</u></em>" ... not "insulation".
'Insolation' is simply the intensity of solar radiation over some area.
If 200 kW of radiation is shining on 300 m² of area, then the insolation is
(200 kW) / (300 m²) = <em>(666 and 2/3) watt/m²</em> .
Note that this is the intensity of the <em><u>incident</u></em> radiation. It doesn't say anything
about how much soaks in or how much bounces off.
Wait !
I just looked back at the choices, and realized that I didn't answer the question
at all. I have no idea what "1 sun" means. Forgive me. I have stolen your
points, and I am filled with remorse.
Wait again !
I found it, through literally several seconds of online research.
1 sun = 1 kW/m².
So 2/3 of a kW per m² = 2/3 of 1 sun
That's between 0.5 sun and 1.0 sun.
I feel better now, and plus, I learned something.
Answer:
t = 1.42 s and d = 35.5 m
Explanation:
Given that,
Velocity of a roadrunner is 25 m/s
A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.
We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.
Let d is the distance traveled. So,
d = vt
d = 25 m/s × 1.42 s
d = 35.5 m
