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ludmilkaskok [199]

3 years ago

15

A tangential force of 1500 N exerted upon the upper surface of a cube of 20 cm edge. Calulate the shear modulus of the cube mata

rial if the result displacement was 0.1cm

1 answer:

I am Lyosha [343]3 years ago

8 0

<u>Answer:</u>

<em>The shear modulus of the cube material is $7.5 \times 10^6 N/m^2$ . </em>

<u>Explanation:</u>

<em>Given that shearing force applied F = 1500 N </em>

<em>Displacement produced x = 0.1 cm=0.001 m </em>

<em>side of the cube =20 cm = 0.2 m </em>

Since the object is a cube the upper surface is a square and it is on this surface the shearing

force is applied

<em>area of the upper surface $A=a \times a=(20 \times 10^(^-^2^))^2=400 \times 10^(^-^4^) m$ </em>

<em>shear strain = tan⁡ θ = $\frac {x}{h} = \frac {0.001}{0.2} =0.005$ </em>

<em>shearing stress = $\frac {F}{A} = \frac{1500}{0.04} = 37500 N$ </em>

<em>modulus of rigidity η $= \frac{(shearing \ stress)}{(shearing \ strain)}$ </em>

<em> $= \frac{37500}{0.005}=7.5 \times 10^6 N/m^2$ </em>

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A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time

saul85 [17]

Answer:

t = 0.714 s and x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

vₓ = 7.0 m / s

Let's find the time it takes to get to the river

y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{\frac{2y_o}{g} }$

t = ra 2 2.5 / 9.8

t = 0.714 s

the distance traveled is

x = vₓ t

x = 7.0 0.714

x = 5.0 m

3 0

3 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

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3 years ago

An igneous rock becomes buried, is subject to high heat and pressure, and recrystallizes. This rock then is eroded, transported,

nevsk [136]

Answer:

Answered

Explanation:

When an igneous rock becomes buried, is subjected to high heat and pressure, and recrystallizes it is formed into Metamorphic rock. Now this rock is eroded, transported, deposited and subsequently lithified to be converted into Sedimentary rock.

The same igneous rock is first converted into Metamorphic and then into sedimentary by the process of weathering.

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liberstina [14]

K.E=18750J
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Velocity=v

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$\\ \sf\longmapsto 18750=1050v^2$

$\\ \sf\longmapsto v^2=\dfrac{18750}{1050}$

$\\ \sf\longmapsto v^2=17.85m^2$

$\\ \sf\longmapsto v=\sqrt{17.85}$

$\\ \sf\longmapsto v=4.1m/s$

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kap26 [50]

Answer:

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Pluto fills in the first two boxes but it does orbit in the keyperbelt and there are 5 other objects just like it. this is why pluto has been dubbed a dwarf planet.

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3 years ago

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