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4 years ago

Which is an example of a covalent bond?

1 answer:

7 0

Hello there.

Question: <span>Which is an example of a covalent bond?

Answer: Water and Diamonds are good examples of covalent bonds.

Hope This Helps You!
Good Luck Studying ^-^</span>

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How can I protect my computer from being shot by an MBT

Pie

Answer: While Technology is Evolving so is Hackers that being said "Norton Vpn" would be beneficial for computer protection.

Explanation:

4 0

2 years ago

Determine the average normal stress developed in rod AB if the load has a mass of 55 kg . The diameter of rod AB is 8 mm.

Fudgin [204]

Answer:

The normal stress is 10.7[MPa]

Explanation:

The normal stress can be calculated with the following equation:

$S_{norm} =\frac{F}{A} \\where:\\F= force [Newtons]\\A=area [m^2]\\S_{norm} = Normal stress [\frac{N}{m^{2} }] or [Pa]$

The area of the rod can be calculated using the equation:

$A=\frac{\pi }{4}*d^{2} \\d=8[mm]=0.008[m]\\A=\frac{\pi }{4}*(0.008)^{2} \\A=5.02*10^{-5} [m^{2} ]$

The force is the result of the mass multiplied by the gravity.

$F=55[kg]*9.81[m/s^{2} ] = 539.6[N]\\\\S_{norm} = 539.6/5.02*10^{-5} \\S_{norm} = 10.7*10^{6}[Pa] = 10.7[MPa]$

3 0

4 years ago

A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exe

RideAnS [48]

Answer:

The pressure exerted by the brick on the table is 18,933.3 N/m².

Explanation:

Given;

height of the brick, h = 0.1 m

density of the brick, ρ = 19,300 kg/m³

acceleration due to gravity, g = 9.81 m/s²

The pressure exerted by the brick on the table is calculated as;

P = ρgh

P = (19,300)(9.81)(0.1)

P = 18,933.3 N/m²

Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².

4 0

4 years ago

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0

4 years ago

A person uses a constant force to push a 14.0 kg crate 1.80 m up a frictionless 10⁰ incline and to also increase its speed from

zubka84 [21]

a) Making h as the shorter cathetus, we have,

$h=1.8*sin10=0.312$

We proceed to calculate the work done by gravity

$W_g=-mgh$

$W_g=-140*0.312=-43.76J$

b) It's necessary to calculate the Kinetic Energy, we use the equation of KE,

$\Delta KE = \frac{1}{2}m(v_2^2-v_1^2)$

$\Delta KE = \frac{1}{2}14(1.5^2-0.5^2)=14J$

c) By work energy theorem

$W_{net}=\Delta KE \rightarrow W_net = 14J$

d) we calculate net work through the law of conservation

$W_{net}=W_f+W_g=14$

$\rightarrow W_f = 14-W_g=14-(-43.76)$

$W_f=57.76J$

6 0

4 years ago