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3 years ago

A body moving with a velocity of 20m/s begins to accelerate at 3 m/s 2 . How far does the body move in 3 seconds

1 answer:

5 0

Given v_in = 20 m/s and a = 3 m/s2, assuming that the body moves at constant acceleration, the motion is modeled by the equation:

s(t) = (v_in)t + (1/2)a(t^2)

where s(t) is the distance traveled

substituting the given,

s(t) = 20t + (3/2)(t^2)

at t = 3

s(t) = 20(3) + (3/2)(3)^2

= 73.5 m

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Scientists think the is a solid layer of iron with a liquid layer of iron around it

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A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

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Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$

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3 years ago

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, aft

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Since the object is dropped from some height so its initial speed must be zero

acceleration of the object is due to gravity

so we can use kinematics to find the time it will take to drop by x = 22 m

$\delta x = v_i * t + \frac{1}{2}at^2$

$22 = 0 + \frac{1}{2}*9.8*t^2$

$t = 2.12 s$

Now the speed after 2.12 s will be given as

$v_f = v_i + at$

$v_f = 0 + 9.8 * 2.12$

$v_f = 20.8 m/s$

so above is the speed and time

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3 years ago

What happens to electron flow with a conductor of the voltage source is removed?

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The electrons stop flowing

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3 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

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2 years ago