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3 years ago

A car moves in a straight line at a speed of 55.4 km/h.

Physics

1 answer:

Misha Larkins [42]3 years ago

6 0

55.4 km => 1hr (60 min) (60×60=120 sec)

1 km => 120 ÷ 55.4

1.21 km => (120 ÷ 55.4) × 1.21

= 2.6209 sec ( 5sf )

= 2.62 sec ( 3sf )

i hope you are able to understand my solution :))

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g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is plac

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Answer:

Explanation:

We shall find electric field at origin due to two given charges sitting on the either side of origin .

Total field will add up due to their same direction .

Field due to a charge Q

= 9 x 10⁹ x Q / R² ; R is distance of point , Q is charge

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= 112.5 x 10¹⁰ N/C

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= 202.5 x 10¹⁰ N/C

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4 years ago

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3 years ago

What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is

QveST [7]

Missing detail in the text:
"A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
 $F=qE$
Considering the charge on the bead as a single point charge, the electric field generated by it is
 $E=k_e \frac{Q}{r^2}$
with $k_e = 8.99\cdot 10^9 Nm^2/C^2$ , $Q=+25 nC=25 \cdot 10^{-9}C$ is the charge on the bead. We want to calculate the field at $r=4.0 cm=0.04 m$ :
$E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m$
The proton has a charge of $q=1.6\cdot 10^{-19}C$ , therefore the force exerted on it is
$F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N$

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, $m=1.67\cdot 10^{-27} kg$ , we have
$F=ma$
$a= \frac{F}{m}= \frac{2.25\cdot 10^{-14} N}{1.67\cdot 10^{-27} kg}=1.35 \cdot 10^{13} m/s^2$

The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
$a=-1.35 \cdot 10^{13} m/s^2$

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3 years ago