Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Answer:
Explanation:
Electric field E = 4 x 10⁷ V / m
Dielectric constant k = 24
capacitance of capacitor
C = kε₀ A / d
d = plate separation
A = plate area
C = .89 x 10⁻⁶
V / d = electric field
for minimum d , electric field will be maximum
V / d = 4 x 10⁷
1930 / d = 4 x 10⁷
d = 1930 / 4 x 10⁷
d = 482.5 x 10⁻⁷ m
= 48.25 x 10⁻⁶ m
C = kε₀ A / d
.89 x 10⁻⁶ = 24 ε₀ A / d
A = .89 x 10⁻⁶ X d / 24 ε₀
A = .89 x 10⁻⁶ X 48.25 x 10⁻⁶ / 24 x 8.85 x 10⁻¹²
= 42.9 / 212.4
= .2019 m²
Substract two consecutive terms of the sequence to see if there is a common difference:

As we can see, there is a common difference of -6.
Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).
Notice that the first term of the sequence is 3.
Then, the rule for the sequence is to start with 3 and add -6 repeatedly.
Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.
Answer:
v₀ = 280.6 m / s
Explanation:
we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,
We write the mechanical energy when the shock has passed the bodies
Em₀ = K = ½ (m + M) v²
We write the mechanical energy when the spring is in maximum compression

½ (m + M) v² = ½ k x²
Let's calculate the system speed
v = √ [k x² / (m + M)]
v = √[152 ×0.78² / (0.012 +0.109) ]
v = 27.65 m / s
This is the speed of the bullet + Block system
Now let's use the moment to solve the shock
Before the crash
p₀ = m v₀
After the crash

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

m v₀ = (m + M) v
v₀ = v (m + M) / m
let's calculate
v₀ = 27.83 (0.012 +0.109) /0.012
v₀ = 280.6 m / s
Answer with Explanation:
We are given that
Diameter of pipe,


Volume flow rate of the petroleum along the pipe=

By equation of continuity







1 m=100 cm