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Lana71 [14]
2 years ago
11

Humid air breaks down (its molecules become ionized) in an electric field of 4.99 x 10^6 N/C. In that field, what is the magnitu

de of the electrostatic force on (a) an electron and (b) an ion with a single electron missing? (a) Number _____ Units _____ (b) Number _____ Units _______
Physics
1 answer:
Zarrin [17]2 years ago
8 0

Explanation:

It is given that,

Electric field, E=4.99\times 10^6\ N/C

The relation between the electric force and the electric field is given by :

F=qE

Where

q is the charge

(a) Charge on the electron, q_e=-1.6\times 10^{-19}\ C

Electrostatic force,

F=-1.6\times 10^{-19}\times 4.99\times 10^6

F=-7.98\times 10^{-13}\ N  

(b) If an electron is missing, the net charge remains the same only sign changes.

So, force becomes, F=7.98\times 10^{-13}\ N  

Hence, this is the required solution.

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A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

7 0
3 years ago
A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a
Zolol [24]

Answer:

\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta} (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L

For small oscillations, the equation can be re-arranged into the following form:

E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L

Where:

\theta = \frac{A}{L^{2}}, measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is 4\theta and the total energy of the pendulum is:

E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L

The factor of change is:

\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}

\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}

3 0
3 years ago
The graphic organizer above shows that the properties of waves are influenced by the energy of waves. Name 2 properties of waves
Stells [14]
Amplitude: the height of the wave<span>, measured in meters
</span><span>Wavelength: the distance between adjacent crests, measured in meters
</span>
3 0
2 years ago
A spring with spring-constant of 100 n/m is compressed 0.10 m. what is its maximum stored elastic potential energy
Ganezh [65]

Elastic potential energy stored in a spring is

(1/2) · (spring constant) · (stretch or compress)² .

PE = (1/2) · (100 N/m) · (0.1 m)²

PE = (50 N/m) · (0.01 m²)

PE = (50 · 0.01) (N · m / m²)

PE = 0.5 N · m

PE = 0.5 Joule

5 0
2 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
vovikov84 [41]

Gravitational force is given by, F= G\frac{mM}{R^{2}}

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, F_{1}= G\frac{m_{1}M}{R^{2}}

Gravitational force of the star on planet 2, F_{2}= G\frac{3m_{1}M}{(3R)^{2}}

Ratio, \frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}

\frac{F_{1}}{F_{2}}=  \frac{3}{1}

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0
2 years ago
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