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Maurinko [17]
3 years ago
12

A horizontal pipe narrows from a radius of 0.250 m to 0.1000 m. If the speed of the water in the pipe is 1.00 m/s in the larger-

radius pipe, what is the speed in the smaller pipe? Given the density of water is 1000 kg/m3 , what is the mass flow rate, kg/s, of the water through the pipe?
Physics
1 answer:
umka21 [38]3 years ago
4 0

Answer:

6.25 m/s

196 kg/s

Explanation:

The areas of the pipe from big to small:

A = \pi R^2 = \pi*0.25^2 = 0.196 m^2

a = \pi r^2 = \pi*0.1^2=0.0314m^2

As the product of speed and cross-section area is constant, the speed in the smaller pipe would be

AV = av

v = \frac{AV}{a} = \frac{0.196 * 1}{0.0314} = 6.25 m/s

The mass flow rate would be:

\dot{m} = \ro AV = 1000 * 0.196 * 1 = 196 kg/s

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A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s.
gtnhenbr [62]

Answer:

(a) 2400N

(b) Yes

Explanation:

(a) Mass (m) = 1.5kg, initial speed (v) = 4m/s, time (t) = 2.5milliseconds = 2.5/1000 = 0.0025seconds

F = mv/t = (1.5×4)/0.0025 = 6/0.0025 = 2400N

(b) The force would be different because the mass of the two hands and forearms is 3kg (2 × 1.5kg = 3kg)

4 0
3 years ago
Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies t
melisa1 [442]

This question is incomplete, the complete question  is;

Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies the initial conditions specified. Units are mks; γ is the damping coefficient, with units of kg/sec

m = 0.2, γ = 1.6 and k = 4

Initial displacement is 1 and initial velocity is -2

x" + _____ x' ____x = 0

x(t) =

Answer:

the solution that satisfies the initial conditions specified is;

x(t) = c_1e^{-4t}cos(2t) + c_2e^{-4t}sin(2t)

Explanation:

Given the data in the question ;

m = 0.2, γ = 1.6, k = 4

x(0) = 1, x'(0) = -2

Now, the differential equation that governs the motions of spring mass system is;

mx" + γx' + kx = 0

so we substitute

0.2x" + 1.6x' + 4x = 0

divide through by 0.2

x" + 8x' + 20x = 0

hence, characteristics equation will be;

m² + 8m + 20 = 0

we find m using; x = [ -b±√(b² - 4ac) ] / 2a

m = [ -8 ± √((8)² - 4(1 × 20 )) ] / 2(1)

m = [ -8 ± √( 64 - 80 ) ] / 2

m = [ -8 ± √-16 ) ] / 2

m = ( -8 ± 4i ) / 2

m = -4 ± 2i

Hence, the general solution of the differential equation is;

x(t) = c_1e^{-4t}cos(2t) + c_2e^{-4t}sin(2t)

From the initial conditions;

c₁ = 1, c₂ = 1

the solution that satisfies the initial conditions specified is;

x(t) = c_1e^{-4t}cos(2t) + c_2e^{-4t}sin(2t)

6 0
3 years ago
Hulk starts at -2 m, jumps to -15 m, and settles at 6 m. What is Hulk's displacement?
Lorico [155]

Answer:

Sorry don't know the answer

6 0
3 years ago
Two objects (38.0 and 17.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
fgiga [73]

Answer:

a) 3.7 m/s^2

b) 231.8 N

Explanation:

Let m1 be mass of the first object (m1 = 38.0 kg) and let m2 be the mass of the second object (m2 = 17.0 kg ). Let a be the acceleration of the two objects. Let F1 be the force of gravity exerted on m1 and F2 be the force of gravity exerted on m2. Let M = m1 +m2

a)

F1 = m1g and F2 = m2g

So Fnet = F1 + F2

Since the pulleys will move in different directions when accelerating...

Fnet = F1 - F2

M×a = m1g - mg2

M×a = g×(m1 -m2)

a = g×(m1 - m2)/M

a = 9.8×(38 - 17)/(38 + 17)

a = 3.7 m/s^2

b)

Looking at the part for m2

Fnet  = T - m2g

-m2×a = T - m2g

T = m2(g - a)

T = 231.8 N

7 0
3 years ago
The temperature of a heat engine is 500 K. Some of the heat generated by the engine flows to the surroundings, which are
Naddik [55]

Answer:

The efficiency of the engine is 75%  under given conditions.

<u>Explanation: </u>

<em>Efficiency is defined as the ratio of work done by the engine to the total heat supplied to a heat engine</em>. We know work done by the heat engine be difference between total heat supplied \left(Q_{h}\right) by total heat utilized \left(Q_{c}\right). We can derive formula for efficiency by considering the above factors, \text { efficiency }=\frac{\text { work done }}{\text { heat suplied }}=\frac{Q_{h}-Q_{c}}{Q_{h}}=1-\frac{Q_{c}}{Q_{h}}=1-\frac{T_{1}}{T_{2}} \therefore \text { efficiency }=1-\frac{T_{1}}{T_{2}}  

Now consider a heat engine is at a temperature of 500K and the heat produced by the engine released to the surroundings at a temperature of 125K. The efficiency is calculated by using the above formula, \text { efficiency }=1-\frac{T_{1}}{T_{2}}=1-\frac{125}{500}=1-0.25=0.75

∴ Efficiency of the engine is 75%

8 0
3 years ago
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