Answer:
Va = 5000 m / 3600 s = 1.39 m/s
(Va - Vb) 60 = 10
Vb = Va - .167 = 1.22 m/s
(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap
.17 T = 4200 Π
T = 24700 Π time for A to again catch B
N = 1.39 * 24700 Answer:
Va = 5000 m / 3600 s = 1.39 m/s
(Va - Vb) 60 = 10
Vb = Va - .167 = 1.22 m/s
(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap
.17 T = 4200 Π
T = 24700 Π time for A to again catch B
N = 1.39 * 24700 Π / (4200 Π) = 8.2 laps
A will make 8 but not 9 rounds before catching B
Answer:
given -
initial velocity = 4.09 m/s
acceleration = 1.01 m/s²
distance = 23.4 m
time = ?
using second formula of motion,
s = ut + 1/2 at².
where, s = distance
u = initial velocity
t = time
a = acceleration
23.4 = 4.09(t) + 1/2(1.01)(t) ²
23.4 = 4.09t + 2.02t²
2.02t² + 4.09t - 23.4 = 0
solving the equation by using quadratic formula
Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :
a = 2.02
b = 4.09
c = -23.4
we get t=2.539 or t= -4.563
time cannot be negative so
t= 2.539 sec = 2.6 Sec is the answer
Component ' W ' is the standard schematic symbol for a switch.
The short line between the two circles can be shown straight,
like it is now, when the switch is 'closed' and current can flow.
Or, it can also be drawn slightly tilted, so it's not touching one
of the circles. In that case, the switch is 'open', no current can
flow through it, and the whole circuit is shut down.
t = time of contact of hammer with the nail = 0.4 sec
I = impulse exerted by the hammer on the nail = 32 N s
F = force applied by the hammer on the nail
we know that , impulse is given as
I = F t where I = impulse , F = force , t = time
hence the impulse of hammer on nail is given as
I = F t
inserting the values
32 Ns = F (0.4 sec)
F = 32/0.4
F = 80 N
hence the force exerted by the hammer comes out to be 80 N
GPE = mgh Where m is the mass of the object (kg), g is acceleration due to gravity (which I will assume to be 9.81ms-2), and h is the height of the object above ground level. This is used in cases where the object is close to the earth, since any change in gravitational force is negligible.
Substituting in the numbers: