Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t
Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C
Case II
----------
When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C
Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C
Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C
Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
U=10 m/s
v=30 m/s
t=6 sec
therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2
now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:
v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m
True, they had a hole in their hip socket that allowed them to run faster than other reptiles of their size at the time. As well as most reptiles besides reptiles had legs to the side, rather than under them like dinosaurs did.
Hope this helps!
Answer:
distance = 21.56 m
Explanation:
given data
mass = 50 kg
initial velocity = 18.2 m/s
force = -200 N ( here force applied to opposite direction )
final velocity = 12.6 m/s
solution
we know here acceleration will be as
acceleration a = force ÷ mass
a =
= -4 m/s²
we get here now required time that is
required time =
...............1
put here value
required time =
so distance will be
distance =
........2
distance =
distance = 21.56 m
Answer:
Pressure, 
Explanation:
It is given that,
Combined mass of the man and the chair, m = 95 kg
Radius of the leg of chair, r = 0.5 cm = 0.005 m
A large man sits on a four-legged chair with his feet off the floor. The force acting per unit area is called the pressure exerted.


Area of 4 legs, A = 4 A



So, the pressure each leg exert on the floor is
. Hence, this is the required solution.