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Elena-2011 [213]

4 years ago

10

How can the same material bend in one situation but break in another

1 answer:

loris [4]4 years ago

5 0

Because it depends on how much pressure you give it

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A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach t

Katarina [22]

Answer:

5.72 seconds

848.27 m/s

97.94 m

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=15-9.81\times t\\\Rightarrow \frac{-15}{-9.81}=t\\\Rightarrow t=1.52 \s$

Time taken to reach maximum height is 0.97 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=15\times 1.52+\frac{1}{2}\times -9.81\times 1.52^2\\\Rightarrow s=11.47\ m$

So, the stone would travel 11.47 m up

So, total height stone would fall is 75+11.47 = 86.47 m

Total distance travelled by the stone would be 75+11.47+11.47 = 97.94 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 86.47=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{86.47\times 2}{9.81}}\\\Rightarrow t=4.2\ s$

Time taken by the stone to travel 86.47 m to the water is is 4.2 seconds

The stone reaches the water after 4.2+1.52 = 5.72 seconds after throwing the stone

$v=u+at\\\Rightarrow v=0+9.81\times 86.47 = 848.27\ m/s$

Speed just before hitting the water is 848.27 m/s

3 0

3 years ago

Tarzan is running with a horizontal velocity, along level ground. While running, he encounters a 2.21 m vine of negligible mass,

Lena [83]

Answer:

a) $a_c = 1.09m/s^2$

b) T = 720.85N

Explanation:

With a balance of energy from the lowest point to its maximum height:

$m*g*L(1-cos\theta)-1/2*m*V_o^2=0$

Solving for $V_o^2$ :

$V_o^2=2*g*L*(1-cos\theta)$

$V_o^2=2.408$

Centripetal acceleration is:

$a_c = V_o^2/L$

$a_c = 2.408/2.21$

$a_c = 1.09m/s^2$

To calculate the tension of the rope, we make a sum of forces:

$T - m*g = m*a_c$

Solving for T:

$T =m*(g+a_c)$

T = 720.85N

3 0

4 years ago

<img src="https://tex.z-dn.net/?f=%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%

Verdich [7]

Answer:

$1.7$ × $10^1^0$

Explanation:

Knowing that, the volume of the sphere is given by, $v=\frac{4}{3}\pi ^3$

Thus, the fractional change in volume is given by,

$=3$ × $\frac{0.02}{100}=\frac{0.06}{100}$

Pressure at the bottom of the sea is,

Δp =pgh = $10^3$ × $10$ × $10^3=10^7$ pa.

Knowing that,

Bulk modulus: $\frac{10^7 * 100}{0.06}=\frac{10^9}{6*10^-^2}=\frac{10^1^1}{6}$

$B =\frac{10}{6}*10^1^0$

$B = 1.7*10^1^0N/M^2$

Answer = $1.7$ × $10^1^0$

[RevyBreeze]

5 0

2 years ago

Read 2 more answers

A spherical mass M is held fixed in deep space, well away from the earth's gravitational pull. A second mass m is released from

Ivahew [28]

Answer:

256 ×a0

Explanation:

I have added an attachment because I am not sure the phrase that negates brainly word why I couldn't submit the answer directly.

View the explanation from attachment 3 followed by 2 and then 1

8 0

4 years ago

A microscope has an eyepiece with a 1.8 cm focal length and a 0.8 cm focal length objective lens. Assuming a relaxed eye, calcul

igomit [66]

Answer:

$0.848\ \text{cm}$

$232.66$

Explanation:

N = Near point of eye = 25 cm

$f_o$ = Focal length of objective = 0.8 cm

$f_e$ = Focal length of eyepiece = 1.8 cm

l = Distance between the lenses = 16 cm

Object distance is given by

$v_o=l-f_e\\\Rightarrow v_o=16-1.8\\\Rightarrow v_o=14.2\ \text{cm}$

$u_o$ = Object distance for objective

From lens equation we have

$\dfrac{1}{f_o}=\dfrac{1}{u_o}+\dfrac{1}{v_o}\\\Rightarrow u_o=\dfrac{f_ov_o}{v_o-f_o}\\\Rightarrow u_o=\dfrac{0.8\times 14.2}{14.2-0.8}\\\Rightarrow u_o=0.848\ \text{cm}$

The position of the object is $0.848\ \text{cm}$ .

Magnification of eyepiece is

$M_e=\dfrac{N}{f_e}\\\Rightarrow M_e=\dfrac{25}{1.8}\\\Rightarrow M_e=13.89$

Magnification of objective is

$M_o=\dfrac{v_o}{u_o}\\\Rightarrow M_o=\dfrac{14.2}{0.848}\\\Rightarrow M_o=16.75$

Total magnification is given by

$m=M_eM_o\\\Rightarrow m=13.89\times 16.75\\\Rightarrow m=232.66$

The total magnification is $232.66$ .

3 0

3 years ago