Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
Answer:
K' = 1777.777 J
Explanation:
Given that
m = 40 kg
v= 15 m/s
K=1000
Given that kinetic energy(K) varies with mass(m) and velocity(v)
K= C(mv²)
Where
C= Constant
m=mass
v=velocity
When
m = 40 kg ,v= 15 m/s ,K=1000
K= C(mv²)
1000 = C( 40 x 15²)
C=0.111111
When m = 40 kg and v= 20 m/s
K' = C(mv²)
K= 0.1111 x (40 x 20²)
K' = 1777.777 J
Answer:
Explanation:
Battery voltage is 6V
A current of 0.361A is draw the voltage reduces to, 5.07V
This shows that the appliances resistance that draws the currents is
Using KVL
The battery has an internal resistance r
V=Vr+Va
Vr is internal resistance voltage
Va is appliance voltage
6=5.07+Va
Va=6-5.07
Va=0.93
Using ohms law to the resistance of the appliance
Va=iR
R=Va/i
R=0.93/0.361
R=2.58ohms
Then if the circuit draws a current of 0.591A
Then the voltage across the load is
V=iR
Va=0.591×2.58
Va=1.52V
Then the voltage drop at the internal resistance is
V=Vr+Va
Vr=V-Va
Vr=6-1.52
Vr=4.48V
Answer:
DE LA SEGUNDA LEY DE NEWTON
Explanation:
<em>m</em> = 16000 N/ = 3200 kg