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3 years ago

There are two unitless vectors:

Physics

1 answer:

Hunter-Best [27]3 years ago

4 0

Answer:

$200.38^0$

Explanation:

Given the forces:

F1 = 8.92 i + 17.37 j

F2 = 8.31 i - 10.97 j

If a third vector is added to them such that they add up to to the null vector as F1 + F2 + F3 = 0

then to get F3:

F3 = -F2-F1

F3 = -(8.31 i - 10.97 j)-(8.92 i + 17.37 j )

F3 = -8.31 i + 10.97 j-8.92 i - 17.37 j

F3 = -8.31i-8.92i+10.97j-17.37j

F3 = -17.23i-6.4j

from the vector:

x = -17.23 and y = -6.4

angle of the third vector with respect to the +x-axis is expressed as:

$\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-6.4}{-17.23}\\\theta = tan^{-1}0.3715\\\theta = 20.38^0$

Hence the angle the vector makes with the x axis will be $\theta = 180+20.38 = 200.38^0$

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Vinil7 [7]

Answer:

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3 years ago

Read 2 more answers

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

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7 0

3 years ago

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

3 years ago

. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required

Delicious77 [7]

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

<h3>Magnitude of electric field </h3>

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

E is the electric field
m is mass of the particle
g is acceleration due to gravity
q is charge of the particle

<h3>For an electron</h3>

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

<h3>For proton</h3>

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

Learn more about electric field here: brainly.com/question/14372859

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4 0

2 years ago

When hydrogen is burned in the presence of oxygen it will form water as per the equation: 2h2

Natasha_Volkova [10]

Answer:

B. In the first case, a combination reaction takes place and in the second case, a decomposition reaction takes place.

Explanation:

When hydrogen burns in oxygen, water is formed and when water is electrolysed then hydrogen and oxygen are produced. What type of a reaction takes place?

(i) In the first case,

(ii) In the second case?

A. In both the cases, i.e. (i) and (ii), a combination reaction takes place.

B. In the first case, a combination reaction takes place and in the second case, a decomposition reaction takes place.

C. In the first case, a combination reaction takes place and in the second case, a displacement reaction takes place.

D. In the first case, a displacement reaction takes place and in the second case, a decomposition reaction takes place.

Answer :

Hint: When hydrogen burns in air, H2H2 reacts with O2O2 to form water molecules. Here two reactants combine to form a single product. And, when water is electrolysed into hydrogen and oxygen, here a single compound decomposes into more than one product.

Complete step by step answer:

In the First case:

When hydrogen burns in oxygen, water is formed.

Hydrogen + Oxygen = Water

2H2+O2→2H2O2H2+O2→2H2O

When two reactants elements combine to form a single component, then this reaction is called the combination reaction. So, in this case, two reactants i.e. Hydrogen and oxygen combine to form a single product i.e. water molecules and heat is also released.

In the Second case:

When water is electrolysed then hydrogen and oxygen are produced.

2H2O→2H2+O22H2O→2H2+O2

When a single reactant is decomposed or broken down into two or more products, it is known as decomposition reaction. Here, a single reactant i.e. water, breaks or decomposes up to form two or more simple products i.e. hydrogen and oxygen. This decomposition reaction takes place by the action of electricity. So, we can see that in the first case, a combination reaction takes place and in the second case, a decomposition reaction takes place.

Therefore, the correct answer is option (B).

Note: Hydrogen is combustible and it burns with a popping sound in oxygen with an almost colourless flame to form water molecules and release heat. This reaction is highly exothermic. Hydrogen must be handled very carefully as the mixture of hydrogen and oxygen can be explosive when two are present in a particular ratio.

7 0

2 years ago