Ask question

Ask question

All categories

3 years ago

12

Which of the following materials is brittle? wood metal ceramic plastic

1 answer:

Ira Lisetskai [31]3 years ago

5 0

Ceramic is the most brittle of the materials you have specified.

You might be interested in

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

An object moves in a circle with a period of 0.025 hours. What is its frequency in Hz?

Sedaia [141]

Answer:

40 Hz

Explanation:

f = 1/T = 1 / 0.025 = 40 Hz

7 0

3 years ago

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0

3 years ago

you suspect a network cable has a break in it somewhere. which tool would be best to use to determine the location of the break

34kurt

Answer:

Optical Time Domain Reflector

Explanation:

Localized the break, supplies a graphical trace of where the break occurs in order to detect high loss splice point as far as 25 miles.

3 0

3 years ago

Read 2 more answers

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

garri49 [273]

Answer:

$w_{0}=14$

$t=\frac{\pi }{14}$

Explanation:

<u>Data</u>

<u>mass m= 100g</u>

<u>Length L= 5cm</u>

<u>we can use:</u>

<u>gm-kL= 0</u>

<u>divide both side by m</u>

<u>g - </u> $\frac{kL}{m}$ <u>=0</u>

<u>where</u>

$\frac{k}{m}$ = $\frac{g}{L}$

$\frac{k}{m}=w_{0}$ ^{2}

so now

$w_{0}^{2}$ = $\frac{9.8*100}{5}$

$w_{0}^{2}=\frac{980}{5}$

$w_{0}^{2}=196$

square both side

$w_{0}=\sqrt{196}$

$w_{0}=14$

We can apply:

u(t)=Acoswt +Bsinwt

u(t)=Acos14t +Bsin14t

u(0)=0 where A=0

therefore

u(0) = Bsin14t

$u^{'}$ (0) = 10 ⇒ 10=14B ⇒ B= $\frac{14}{10}$ B= $\frac{5}{7}$

so now u(t)= $\frac{5}{7}$ sin14t

so t will be:

t= $\frac{\pi }{14}$

t= $\frac{3.14}{14}$

t=0.22 seconds

8 0

3 years ago