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1 year ago

A ball dropped from a window strikes the ground 2.71 seconds later. How high is the window above the ground?

Physics

1 answer:

koban [17]1 year ago

4 0

$s=36.52m$

Given ,

Time =2.71 second

We apply the formula

$s=ut+12at^2$

$u=0,t=2.73 \\a=g \\So, the \\height \\ s=36.52m$

When a body is lying on a horizontal surface, the object is subjected to a downward force due to the action of gravity. These gravitational forces always aim to attract the item, confirming the force's nature as an attractive force. Consider an object of mass "M" resting on a horizontal surface; the force exerted to the object by gravity is equal to the weight of the object "W." The acceleration owing to gravity, indicated as "g," is the acceleration by which an object is drawn towards the ground. As a result, Newton's second law states that when the values in the above connection are substituted.

To learn more about Motion due to gravity from the given link:

brainly.com/question/14047086

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where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

2 years ago

Based on Newton’s 3rd Law of motion, if an archer shoots an arrow at the target, the action force is the bowstring against the a

maw [93]

Answer:

maybe target, not sure.

6 0

2 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

3 years ago

Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group.

serg [7]

Well formation of metallic bond depends on free electrons.smaal sized atoms hold their electrons more firmly as compared to large size atoms ,this z due to distance of outer shell electrons by nucleus..in this way no of free electrons affect strength of metallic bond..smaal sized atoms release less free electrons..

8 0

3 years ago

Read 2 more answers

A striped billiard ball moves toward the right with speed 3 m/s. A solid billiard ball with the same mass moves toward the left

Helen [10]

Answer:

Final speed of striped ball is 3 m/s in left direction .

Explanation:

Given :

Two billiard ball with the same mass moves toward the left at the same speed 3 m/s .

Let , us assume right hand side direction to be positive and left hand side direction to be negative .

Also , let speed of ball after collision is (striped ball ) u and (solid ball) v .

It is also given that the collision is elastic .

Therefore , kinetic energy is conserved .

$\dfrac{m(3)^2}{2}+\dfrac{m(3)^2}{2}=\dfrac{mu^2}{2}+\dfrac{mv^2}{2}\\\\u^2+v^2=18$ ...... ( 1 )

Also , by conserving linear momentum .

We get :

$3m-3m=mu+mv\\u=-v$ ...... ( 2 )

Putting value of u from equation 2 to equation 1 .

We get :

$2v^2=18\\v=3\ m/s$

And , u = -3 m/s .

Therefore , final speed of striped ball is 3 m/s in left direction .

Hence , this is the required solution .

4 0

3 years ago