Distance of lake a is 200 km at 20 degree north of east
distance between lake a and b is 230 km at 30 degree west of north
now the distance between base and lake b is given as
given that
now the total distance is
now the magnitude of the distance is given as
also the direction is given as
<em>so it is 277.4 km at 74.7 degree North of East</em>
The law of conservation of momentum tells us that momentum
is conserved, therefore total initial momentum should be equal to total final
momentum. In this case, we can expressed this mathematically as:
mA vA + mB vB = m v
where, m is the mass in kg, v is the velocity in m/s
since m is the total mass, m = mA + mB, we can write the
equation as:
mA vA + mB vB = (mA + mB) v
furthermore, car B was at a stop signal therefore vB = 0,
hence
mA vA + 0 = (mA + mB) v
1800 (vA) = (1800 + 1500) (7.1 m/s)
<span>vA = 13.02 m/s</span>
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
Because it is if you know you know and it is also helping the sentwcnde and air and confiscation
Answer:
A. To find the mass flow rate.
We use= 220 x 0.355/ 60
= 1.3kg/s
B. Volume flowrate is = mass flowrate / density
But density is 1000kg/m³
= 1.3kg/s/ 1000kg/m³
= 0.0013m³/s
C. Flow speead at 1
= 0.0013m³/s / (2 x 10-2m)²
= 6.5m/s
D.flow speed at 2
0.0013m³/s / (8x 10-2m)²
=1.63m/s
E. Gauge pressure at point 1
= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)
= 119kpa
