Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where =Resistance of wire at Temperature T
= Resistivity at temperature T
Where
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows
T = 81.52 Deg C
Mass doesn't depend on where it is, and doesn't change.
Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right
Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s
Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right
Answer:
t = 0.354 hours
Explanation:
given,
coefficient of rolling friction μr=0.002
mass of locomotive = 180,000 Kg
rolling speed = 25 m/s
The force of friction = μ mg
= (.002) x (180000) x (9.8)
= 3528 N
F = m a
now,
m a = 3528 N
180000 x a = 3528
a = 0.0196 m/s²
Then apply
v = u + at
0 = 25 - 0.0196 x t
t = 1275.51 sec
t = 1275.61/3600 hours
t = 0.354 hours
time taken by the locomotive to stop = t = 0.354 hours
Answer:
W = 2.3 10²
Explanation:
The force of the weight is
W = m g
let's use the concept of density
ρ= m / v
the volume of a sphere is
V = π r³
V = π (1.0 10⁻³)³
V = 4.1887 10⁻⁹ m³
the density of water ρ = 1000 kg / m³
m = ρ V
m = 1000 4.1887 10⁻⁹
m = 4.1887 10⁻⁶ kg
therefore the out of gravity is
W = 4.1887 10⁻⁶ 9.8
W = 41.05 10⁻⁶ N
now let's look for the electric force
F_e = q E
F_e = 12 10⁻¹² 15000
F_e = 1.8 10⁻⁷ N
the relationship between these two quantities is
= 41.05 10⁻⁶ / 1.8 10⁻⁷
\frac{W}{F_e} = 2,281 10²
W = 2.3 10²
therefore the weight of the drop is much greater than the electric force