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2 years ago

Ang kongreso ang gumagawa ng pambansang Batas. Ito ay binubuo ng _____ kapulungan.

Physics

1 answer:

yan [13]2 years ago

8 0

Answer:

The answer is A.Dalawang

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For a proton (mass = 1.673 x 10–27 kg) moving with a velocity of 2.83 x 104 m/s, what is the de Broglie wavelength (in pm)?

Oduvanchick [21]

Answer:

The value of de Broglie wavelength is 14.0 pm

Explanation:

Given;

mass of proton, m = 1.673 x 10⁻²⁷ kg

velocity of the proton, v = 2.83 x 10⁴ m/s

De Broglie wavelength is given as;

$\lambda = \frac{h}{mv}$

where;

h is planck's constant = 6.626 x 10⁻³⁴ kgm²/s

m is mass of the proton

v is the velocity of the proton

$\lambda = \frac{6.626*10^{-34}}{(1.673*10^{-27})(2.83*10^4})} \\\\\lambda = 1.40 *10^{-11} \ m\\\\\lambda = 14.0 \ pm$

Therefore, the value of de Broglie wavelength is 14.0 pm

6 0

3 years ago

Mechanical advantage of a simple machine is 4,what does it mean?

adelina 88 [10]

Answer:

I hope this answer is correct

Explanation:

The mechanical advantage of a machine is 4. Mechanical advantage MA is the ratio of output (generated by the machine) force to input (applied to the machine) force. So MA = 4 means that for example if you apply 100 N then your machine will multiply that force and generate 400 N.

7 0

3 years ago

Which of the following properties of materials does NOT affect resistance?

Shalnov [3]

Answer:

B thickness

Explanation:

thickness

4 0

3 years ago

A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin

Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

$Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa$

The shearing strain is defined as the tangent of the displacement that the block over its length:

$SS=tan\theta =\frac{Displacement}{L} =\frac{10}{40} =0.25$

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

$Sm=\frac{10208.3333}{0.25} =40833.3332Pa$

6 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

Ang kongreso ang gumagawa ng pambansang Batas. Ito ay binubuo ng _____ kapulungan​.

Ang kongreso ang gumagawa ng pambansang Batas. Ito ay binubuo ng _____ kapulungan.