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DiKsa [7]
2 years ago
14

Large and small numbers are often best entered in exponential notation. For example, the mass of the earth is 5.98x1024 kg and t

he mass of an electron is 9.11x10-31 kg. Canvas does not have the capability to handle scientific notation. We will include the power of 10 with the units needed in the answer. The charge of an electron is -1.60x10-19 C. Determine the charge to mass ratio of the electron (in 1011 C/kg). (Hint: divide the electron's charge by its mass; the answer is a negative number with a positive exponent)
Physics
2 answers:
stich3 [128]2 years ago
5 0

Answer:

The charge to mass ratio is

m/e= -0.17*10^12kg/c

Explanation:

Step one :

Given data

mass of an electron m= 9.11x10-31kg

charge of electron q= -1.60x10-19C

Applying the equation

mv²/r=Bev

Where m =mass of charge

v= velocity of charge

r= distance

B=magnetic field

e= charge

Rearranging to get the charge-mass ratio we have

mv²/r=Bev

m/e=Br/v

(-1.60x10-19)/(9.11x10-31kg)=Br/v

-0.17*10^12

m/e= -0.17*10^12kg/c

Doss [256]2 years ago
4 0

Answer:

The charge to mass ratio is -1.76\times10^{11}

Explanation:

Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\

We need to find how much charge is contained  in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}

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A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft>s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
      = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
2 years ago
How would reversing the wheel’s initial direction of rotation affect the result??
d1i1m1o1n [39]
It would change the sign on the vector quantities and have no change to the scalar quantities
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Randy wants to know whether a soil's porosity affects how easily seedlings grow in it.
telo118 [61]
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he plants seedlings in soils with different levels of porosity and equal levels of permeability. 

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4 0
2 years ago
Read 2 more answers
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
2 years ago
What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J
FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0
2 years ago
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