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stira [4]
3 years ago
7

a student compared two samples of matter he recorded the results in this chart which property do the two samples have in common

Physics
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

私は知らない

Explanation:

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Pls help I’m being timed!!
Firdavs [7]

Answer:

B

Explanation:

Heat increase molecular motion

6 0
3 years ago
Read 2 more answers
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

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)

■■■■■■■■■■■■■■■■■■■■■■■■■■

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

■■■■■■■■■■■■■■■■■■■■■■■■■■

● 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

■■■■■■■■■■■■■■■■■■■■■■■■■■

● 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

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
Read 2 more answers
What is the kinetic energy of a 9.0 kg steelhead if its speed is 16 m/s?
Vesnalui [34]

<u>We are given:</u>

Mass of the Steelhead(m) = 9 kg

Velocity of the Steelhead(v) = 16 m/s

<u>Calculating the Kinetic Energy:</u>

KE = 1/2mv²

replacing the variables

KE = 1/2 * 9 * (16)²

KE = 1152 Joules

8 0
2 years ago
A ball accelerates uniformly at +4.0 m/s2
kirill115 [55]

Answer:

483883838939384847848484

8 0
2 years ago
In one experiment the electric field is measured for points at distances r from a uniform line of charge that has charge per uni
Yuki888 [10]

Answer:

(a) A. Uniform line of charge and B. Uniformly charged sphere

(b) To three digits of precision:

λ = 1.50 * 10^-10 C/m

p = 2.81 * 10^-4 C/m^3

Explanation:

8 0
3 years ago
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