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sineoko [7]
3 years ago
14

Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,

and declares he has a field strength of 24.0 N/C. The second friend looks at her meter and realizes she is three times as far away. What field strength does the second friend's meter read?
Physics
1 answer:
yarga [219]3 years ago
5 0

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

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The mole is 6.02 x 10 23 particles. If a person masses out the correct molar mass in grams for a substance then she would have a
Leto [7]

Answer:

1 mole of H2O is 18 grams (2 g H + 16 g Oxygen)

36 / 18 = 2

So 2 moles = 2 * 6.02E23 = 12.04E23 = 1.204E24

7 0
2 years ago
5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.
andreev551 [17]

Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

a = 8/4.2

a = 1.904

6 0
2 years ago
Which of the following results when a crest and trough meet?
kifflom [539]
D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.
6 0
2 years ago
To control whether an object is solid or incorporeal (things can pass through it) you would use the:
Zarrin [17]

Answer:

Gamma radiation or Cathode rays

Explanation:

by striking incident gamma or cathode rays onto the solid when placed on a photographic plate

5 0
2 years ago
A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can
GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

v = speed at which jumper jumps at all time

initial distance jumped is given as

R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}

final distance jumped is given as

R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}

Dividing final distance by initial distance

\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}

\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}

R_{2} =7.38

distance lost is given as

d = R_{1} - R_{2}

d = 7.4 - 7.38

d = 0.02 m

8 0
2 years ago
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