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Free_Kalibri [48]
3 years ago
7

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate

the charge on the ion.
Physics
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer: 6.408(10)^{-19} C

Explanation:

This problem can be solved by the following equation:

\Delta K=q V

Where:

\Delta K=7.37(10)^{-17} J is the change in kinetic energy

V=115 V is the electric potential difference

q is the electric charge

Finding q:

q=\frac{\Delta K}{V}

q=\frac{7.37(10)^{-17} J}{115 V}

Finally:

q=6.408(10)^{-19} C

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The basketball coach tells his team to run sprints back and forth across the court, which is 30 m long. They start at the left e
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Distance is the total length of an object's path. Displacement is the overall change in position, ie. how far an object is from its initial position.


The court is 30 m long, so a path going back and forth once is 60 m long. Going along this path 6 times totals 360 m.


The end point is the same as the starting point, so the displacement is 0 m.

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3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
3 years ago
A 66​-foot-tall woman walks at 55 ​ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th
I am Lyosha [343]

Answer:

a. \frac{dx}{dt}=20ft/s

b. \frac{d(x+L)}{dt}==25ft/s

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}

Solve to find the rate relation

x=\frac{24}{6}*L

x=4*L

Now the rate of the change rate

\frac{dx}{dt}=4*\frac{dL}{dt}

\frac{dx}{dt}=4*5ft/s=20ft/s

Finally the rate of her shadow moving

\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}

\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s

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