<span>At the top of the waterfall, the water has potential energy. Once it goes over</span>
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
Magnitude of normal force acting on the block is 7 N
Explanation:
10N = 1.02kg
Mass of the block = m = 1.02 kg
Angle of incline Θ
= 30°
Normal force acting on the block = N
From the free body diagram,
N = mgCos Θ
N = (1.02)(9.81)Cos(30)
N = 8.66 N
Rounding off to nearest whole number,
N = 7 N
Magnitude of normal force acting on the block = 7 N
Answer:
a)3312 x 10⁴ J
b)I = 57.5 A
c)9200 W
Explanation:
Given that
P =4600 W
Time t= 2 h = 2 x 3600 s= 7200 s
We know that
1 W = 1 J/s
a)
Energy stored in the battery = P .t
=4600 x 7200 J
=3312 x 10⁴ J
b)
We know that power P given as
P = V .I
V=Voltage ,I =Current
4600 = 80 x I
I = 57.5 A
c)
The energy supplied = 4600 x 2 = 9200 W
