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2 years ago

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A 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time t=0, th

e resulting mass-spring system is disturbed from its rest state by the force F(t)=140cos(8t). The force F(t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds.

1 answer:

Nitella [24]2 years ago

5 0

Answer:

K= 1000 N-m

Explanation:

It is assumed that we asked to find the spring constant k of the spring

We know that under equilibrium condition

weight of the object = force applied by the spring

given m =10 Kg

x= extension in the spring = 9.8 cm

mg=kx

$10\times9.8=k\times9.8\times10^(-2)$

K= 1000 N-m

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The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t

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Answer:

$g'=\frac{g__R}{4}$

Explanation:

Given:

gravitational field strength of moon at a distance R from its center, $g__R$

Distance of the satellite from the center of the moon, $h=2R$

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

$g'=g__{R}} \times \frac{R^2}{(2R)^2}$

$g'=\frac{g__R}{4}$

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

6 0

2 years ago

100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

$A=\frac{\pi d^2}{4}$

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

8 0

3 years ago

A catcher stops a 0.15-kg ball traveling at 40 m/s in a distance of 20 cm. what is the magnitude of the average force that the b

iren2701 [21]

The magnitude of the average force that the ball exerts against his glove is 600 N

$\texttt{ }$

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

mass of ball = m = 0.15 kg

initial speed of ball = u = 40 m/s

final speed of ball = v = 0 m/s

distance = d = 20 cm = 0.2 m

<u>Asked:</u>

average force = F = ?

<u>Solution:</u>

<em>We will use </em><em>Newton's Law of Motion</em><em> to solve this problem as follows:</em>

$F = m a$

$F = m (\frac { u^2 - v^2 } { 2d } )$

$F = 0.15 \times \frac { 40^2 - 0^2 } { 2 \times 0.2 }$

$F = 0.15 \times \frac { 1600 } { 0.4 }$

$F = 0.15 \times 4000$

$\boxed {F = 600 \texttt{ N}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0

2 years ago

Read 2 more answers

The total energy of a 0.050 kg object travelling at 0.70 c is:

finlep [7]

What is the unit c denotes here

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2 years ago

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Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed t

poizon [28]

Answer:

(E) The two objects reach the bottom of the incline at the same time.

Explanation:

Given;

first object with mass, m

second object with mass, 5m

The acceleration of gravity for both object is the same = 9.8 m/s²

Since both objects have the same acceleration of gravity, and no external force due friction (frictionless inclined plane), they will reach bottom of the inclined at the time.

Thus, the acceleration due to gravity is constant for all objects regardless of their masses.

Therefore, the correct option is E;

(E) The two objects reach the bottom of the incline at the same time.

5 0

2 years ago