Una llave de tuercas de 0.5kg cae desde una altura de 10m.¿Cual es su cantidad de movimiento inmediato antes de tocar el suelo?
1 answer:
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Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
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● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
Answer:
1.479 A
Explanation:
Expression for de Broglie wave length
λ =
Where m is mass of neutron ,h is plank constant and KE is kinetic energy of
proton.
Kinetic Energy ( KE) = 3 / 2 kT
=1.5 X 1.38 X 10⁻²³ X 290 = 600.3 X 10⁻²³ J
=
44.777 x 10⁻²⁵
λ =
=1.479 A.