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3 years ago

Chivalry is the

Physics

2 answers:

grin007 [14]3 years ago

5 0

The answer is A.) ..

s344n2d4d5 [400]3 years ago

4 0

Answers is A. European code of honorable behaviors for Knights

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ΡΟ. . When preparing for any of the golf variation games (golf foot golf, disc golf), a person should consider the following bef

attashe74 [19]

Answer:

All of the above

Explanation:

To ensure the ball go to where we expected it go

5 0

3 years ago

A solid metal sphere with radius 0.430 m carries a net charge of 0.270 nC . Part A Find the magnitude of the electric field at a

rodikova [14]

Answer:

8.46 N/C

Explanation:

Using Gauss law

$E=\frac {kQ}{r^{2}}$

Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge

Here, K is Coulomb's constant whose value is $9\times 10^{9} Nm^{2}/C^{2}$

r = 0.43 + 0.106 = 0.536 m

$E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C$

8 0

4 years ago

A 17.3 eV electron has a 0.295 nm wavelength. If such electrons are passed through a double slit and have their first maximum at

4vir4ik [10]

Answer:

0.541 nm

Explanation:

The condition for maxima is,

$dsin\theta=m\lambda$

Here, m=0,1,2,.....

And d is the slit separation, m is the order of maxima, $\lambda$ is the wavelength.

Given that, the 17.3 eV electron posses a wavelength of

$\lambda=0.295 nm\\\\\lambda=0.295\times 10^{-9}m$

And the order of maxima is $m=1$ .

And the angle at which first order maxima occur is, $\theta=33^{\circ}$ .

Put these values in maxima condition while solving for d.

$d=\frac{1\times 0.295\times 10^{-9}m}{sin33^{\circ}} \\d=\frac{0.295\times 10^{-9}m}{0.545} \\d=0.541\times 10^{-9}m}\\d=0.541 nm$

Therefore, the slit separation is 0.541 nm.

7 0

3 years ago

If the mass is 22 g and the volume is 11 ml, what is the density of the object

andriy [413]

2 g/mL there’s your answer

8 0

3 years ago

Read 2 more answers

What energy is needed to raise the temperature of 2 kg of water from 20 ºC to 100 ºC. The heat capacity of water is 4190 J /kg/

aev [14]

Answer:

670400 J

Explanation:

m = mass of water = 2 kg

T₀ = initial temperature of water = 20 ºC

T = initial temperature of water = 100 ºC

c = heat capacity of water = 4190 J /kg ºC

Q = energy needed to raise the temperature of water

Energy needed to raise the temperature of water is given as

Q = m c (T - T₀)

Inserting the values

Q = (2) (4190) (100 - 20)

Q = 670400 J

5 0

4 years ago