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3 years ago

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4kj of energy are supplied to a machine used for lifting a mass.The force required is 800N.If the machine has an efficiency of 5

0%. To what height will it lift the mass?

1 answer:

Bogdan [553]3 years ago

7 0

I do not know shdjfjdjfk

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The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k

sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0

3 years ago

The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated

madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss :Due to friction of pipe surface

2.Minor loss :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

$Losses=K\dfrac{V^2}{2g}$

8 0

3 years ago

A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?

Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

8 0

4 years ago

Let's model this housing price data! Before we can do this, however, we need to split the data into training and test sets. Reme

Lilit [14]

The program reads in a dataset into a pandas dataframe, and uses the train_test_split function in the sklearn library to split the data into <em>training and test sets</em>. The code goes thus :

import pandas as pd

<em>#import</em><em> </em><em>the</em><em> </em><em>pandas</em><em> </em><em>dataframe</em><em> </em><em>and</em><em> </em><em>alias</em><em> </em><em>it</em><em> </em><em>as</em><em> </em><em>pd</em>

from sklearn.model_selection import train_test_split

<em>#import</em><em> </em><em>the</em><em> </em><em>train_test_split</em><em> </em><em>function</em><em> </em>

housing_df = pd.read_csv('housing price.csv')

<em>#read</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>housing</em><em> </em><em>data</em><em> </em>

features_df = df.iloc[:,1:]

<em>#seperate</em><em> </em><em>the</em><em> </em><em>features</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>label</em><em> </em><em>;</em>

target_df = df.iloc[:,0]

<em>#put</em><em> </em><em>the</em><em> </em><em>label</em><em> </em><em>into</em><em> </em><em>a</em><em> </em><em>seperate</em><em> </em><em>dataframe</em><em> </em><em>as</em><em> </em><em>well</em><em>.</em><em> </em>

X_train, X_test, Y_train, Y_test = train_test_split(features_df, target_df, test_size = 0.1, random_state = 1)

<em>#uses</em><em> </em><em>tuple</em><em> </em><em>unpacking</em><em> </em><em>to</em><em> </em><em>randomly</em><em> </em><em>assign</em><em> </em><em>the</em><em> </em><em>data</em><em> </em><em>each</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>4</em><em> </em><em>variables</em><em>.</em><em> </em>

<em>#</em><em>Test</em><em> </em><em>size</em><em> </em><em>is</em><em> </em><em>test</em><em> </em><em>percent</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>entire</em><em> </em><em>dataset</em><em> </em>

Learn more :brainly.com/question/4257657?referrer=searchResults

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3 years ago

A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te

Artyom0805 [142]

Explanation:

a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures

100% (3 ratings)

A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2

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3 years ago