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1 year ago

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A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

ius is 59 cm. When an object is placed on top of the cylinder, the cylinder compresses by an amount of 5.6 10-7 m. What is the weight of the object?

1 answer:

maksim [4K]1 year ago

4 0

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

Where,

F = Force/Weight

A = Area

$\Delta L$ = Compression

$L_0$ = Original Length

According to the values given we have to

$\Upsilon_{steel} = 200*10^9Pa$

$\Delta L = 5.6*10^{-7}m$

$L_0 = 3.2m$

$r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2$

Replacing this values at our previous equation we have,

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

$200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}$

$F = 38272.5N$

Therefore the Weight of the object is 3.82kN

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A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie

cluponka [151]

Answer:

a) v₃ = 19.54 km, b) 70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

cos 45 = x₁ / V₁

sin 45 = y₁ / V₁

x₁ = v₁ cos 45

y₁ = v₁ sin 45

x₁ = 26 cos 45

y₁ = 26 sin 45

x₁ = 18.38 km

y₁ = 18.38 km

Vector 2 moves 45 km north

y₂ = 45 km

Unknown 3 vector

x3 =?

y3 =?

Vector Resulting 70 km north of the starting point

R_y = 70 km

we make the sum on each axis

X axis

Rₓ = x₁ + x₃

x₃ = Rₓ -x₁

x₃ = 0 - 18.38

x₃ = -18.38 km

Y Axis

R_y = y₁ + y₂ + y₃

y₃ = R_y - y₁ -y₂

y₃ = 70 -18.38 - 45

y₃ = 6.62 km

the vector of the third leg of the journey is

v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

v₃ = √ (18.38² + 6.62²)

v₃ = 19.54 km

to find the angle let's use trigonometry

tan θ = y₃ / x₃

θ = tan⁻¹ (y₃ / x₃)

θ = tan⁻¹ (6.62 / (- 18.38))

θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

θ’= 180 -19.8

θ’= 160.19º

I mean the address is

θ’’ = 90-19.8

θ = 70.2º

70.2º north-west

3 0

2 years ago

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0

1 year ago

A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

$\theta = 234.62 radian$

3 0

1 year ago

In order to get a tree stump out of the ground, chains are connected to two trucks. One truck pulls with a force of 600 N to the

Black_prince [1.1K]

Answer:

The net force on the stump is 1000 N.

Explanation:

Given that,

Force 1 acting on the truck, $F_1=600\ N$ (due north)

Force 2 acting on the truck, $F_2=800\ N$ (due west)

We need to find the net force on the stump. We know that force is a vector quantity. The net force on the stump is given by the the resultant force. It is given by :

$F=\sqrt{F_1^2+F_2^2}$

$F=\sqrt{600^2+800^2}$

F = 1000 N

So, the net force on the stump is 1000 N. Hence, this is the required solution.

3 0

2 years ago

About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot

9966 [12]

Answer:

$2.44156\times 10^{13}\ m^3$

29010.53917 m

Explanation:

$\rho$ = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = $\dfrac{d}{2}=\dfrac{10}{2}=5\ km$

v = Velocity = 11 km/s

$H_v$ = Heat vaporization of water = $2.26\times 10^6\ J/kg$

$\Delta T$ = Change in temperature = 100-20

Mass is given by

$m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg$

The kinetic energy is

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J$

Heat is given by

$Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg$

Mass of water is $2.44156\times 10^{16}\ kg$

Volume is $\dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3$

Amount of water is $2.44156\times 10^{13}\ m^3$

If it were a cube

$h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m$

The height of the water would be 29010.53917 m

4 0

1 year ago