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3 years ago

What part of an atom exists in areas of probability outside the nucleus?

Physics

2 answers:

spin [16.1K]3 years ago

4 0

Electrons is the answer

dusya [7]3 years ago

3 0

Electrons exist in areas of probability around the nucleus.
Neutrons and protons are in the atom's nucleus.
Photons are little packets of electromagnetic energy.

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Which of the following is a contact force?

erica [24]

Option C Magnetic force

5 0

2 years ago

Read 2 more answers

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

3 years ago

Which of the following expressions for power or dimensionally correct?

Slav-nsk [51]

Power=F.V

dimension: ML^2T^-2

8 0

3 years ago

To move a heavy object, like a refrigerator what could be used to help decrease the frictional force?

Olenka [21]

Answer:

A. Remove everything in the refrigerator to lighten the load.

B. Put a lubricant between the surface of the object and the floor

C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily

Explanation:

Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.

F = μN where N is the normal force which depends on the mass.

Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.

8 0

3 years ago

A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the part

marta [7]

Answer:

Explanation:

Given that,

Charge q=-5.10nC

Magnetic field B= -1.2T k

And the magnetic force

F =−( 3.30×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F= q(v×B)

−( 3.30×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( 1.2xj - 1.2y i )

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2y i + 1.2x j)

So comparing comparing coefficients

let compare x axis component

-( 3.30×10−7N )i=-1.2qy i

−3.30×10−7N = -1.2qy

y= -3.3×10^-7/-1.2q

y= -3.3×10^-7/-1.2×-5.10×10^-9)

y=-53.92m/s

Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

x=-124.18m/s

a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-124.18i-53.92j)•−(3.30×10−7N)i+(7.60×10−7 N )j =

4.1×10^-5 - 4.1×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cos

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

6 0

2 years ago