First we write the kinematic equations
a
v = a * t + vo
r = (1/2) at ^ 2 + vo * t + ro
We have then that:
(10.4 - t) = time that they run at their maximum speed
For Laura:
d = (1/2) at ^ 2 + (at) (10.4 - t)
100 m = (1/2) a (1.96) ^ 2 + [(1.96) a] (8.44)
100 = 1.9208a + 16.5424a
100 = 18.4632a
a = 100 / 18.4632 = 5.42 m / s ^ 2
For Healen:
100 = (1/2) a (3.11) ^ 2 + [(3.11) a] (7.29)
100 = 4.83605a + 22.6719a
100 = 27.50795a
a = 100 / 27.50795
a = 3.64 m / s ^ 2
Answer:
the acceleration of each sprinter is
Laura: 5.42 m / s ^ 2
Healen 3.64 m / s ^ 2
Answer:
its like game controlls the 1st doesnt move R L R R R R it matters on the arrows pay attention to the directions take the arrows away then put them back to see what u get 1 at a time its newtons third law
Explanation:
<em>Ten games</em> will be played.
The first team can be any one of 5 . For each of those, the opponent can be any one of the other 4 .
Number of ways to pair them up = (5 x 4) = 20 ways .
<em>BUT</em> ... Whether the Reds play the Blues, or the Blues play the Reds, it's the SAME GAME. Each possible game shows up twice in the list of 20 ways. So there are really only 10 different pairs ==> <em>10 games</em> .
Answer:
As per Coulomb's law we know that force between two charges is given as
F = \frac{kq_1q_2}{r^2}F=
r
2
kq
1
q
2
here we know that
q_1 = 2.5 \times 10^{-6} Cq
1
=2.5×10
−6
C
q_2 = -5.0 \times 10^{-6} Cq
2
=−5.0×10
−6
C
r = 0.0050 mr=0.0050m
now from above formula we will have
F = \frac{(9 \times 10^9)(2.5 \times 10^{-6})(5 \times 10^{-6})}{(0.0050)^2}F=
(0.0050)
2
(9×10
9
)(2.5×10
−6
)(5×10
−6
)
F = 4500 NF=4500N
so they will attract towards each other as they are opposite in nature with force F = 4500 N
Answer:
14.625 years
Explanation:
Because acceleration/deceleration are negligent we can simplify this problem into 2 basic parts:
First: how long before the astronaut arrives? Since the speed of the craft is 0.8 times the speed of light and the distance to travel is 6.5 light-years, the equation to answer this part is:
0.8t=6.5
solve for t by dividing by 0.8
t=6.5/0.8 or 8.125
so, she will arrive in 8.125 years.
Then, once she sends her message, it will travel towards earth at 1 times the speed of light or:
1.0t=6.5 which is just t=6.5
so, her message will arrive back to Earth (8.125+6.5) 14.625 years after takeoff.
