Answer:
Lifting a bag of groceries
Answer:
T = 0.0088 m²/s
Explanation:
given,
initial piezometric elevation = 12.5 m
thickness of aquifer = 14 m
discharge = 28.24 L/s = 0.02824 m³/s
we know
k = 0.629 mm/sec
Transmissibilty
T = k × H
T = 0.629 × 14 × 10⁻³
T = 0.0088 m²/s
Answer:
(a) 1320 W
(b) 480 W
(c) E':E ≈ 11:2
Explanation:
(a) Applying,
P' = VI'................. Equation 1
Where P' = Power of the blow-dryer, V = Voltage, I = current rating of the blow-dryer.
From the question,
Given: V = 120 V, I' = 11 A
Substitute these values into equation 1
P = (120×11)
P = 1320 W
(b) Similarly,
P = VI................... Equation 2
Where P = Power of the vacuum cleaner. I = current rating of the vacuum cleaner.
Also Given: I = 4 A,
Therefore
P = 4(120)
P = 480 W
(c)
E' = P'/t'............. Equation 3
E = P/t................ Equation 4
Where E' = Energy of the blow-dryer, t' = time of use of the blow-dryer, E = Energy of the vacuum cleaner, t = time of use of the vacuum cleaner
From the question,
Given: t' = 15 minutes = (15×60) = 900 seconds, t = 30 minutes = (30×60) = 1800 seconds
Substitute these values into equation 3 and 4
E' = 1320/900
E' = 1.47 J,
E = 480/1800
E = 0.267
Therefore,
E':E = 1.47:0.267
E':E ≈ 11:2
Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm
Solution :
Let 
kg
m/s
Let 
and 
are the speeds of the disk 
and 
after the collision.
So applying conservation of momentum in the y-direction,
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
m/s
So, 
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
