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ValentinkaMS [17]
3 years ago
12

Can somebody pls pls help

Physics
1 answer:
Elina [12.6K]3 years ago
4 0

Answer:

i cant see it

Explanation:

huh i cant see it

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Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What
vesna_86 [32]

Answer:

1.02\cdot 10^{-8} N, repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have  same  sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

q_1=q_2=1.6\cdot 10^{-19}C is the magnitude of the two electrons

r=1.5\cdot 10^{-10} m is their separation

Substituting into the formula, we find the electric force between them:

F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N

And the force is repulsive, since the two electrons have same sign charge.

4 0
3 years ago
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and ampl
scoundrel [369]

Answer:

Explanation:wave=wavelength×frequency,

3 0
3 years ago
Nuclei of u-238 atoms are
Elis [28]
 <span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
5 0
3 years ago
20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?
Tresset [83]
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance 
so Work done = 0.085 x 20g x 28
 = 466.956 J

7 0
4 years ago
What would the final temperature be if 8.94 X 10 3 joules of heat were added to 454 grams of copper at 23.0 o C?
ozzi

Answer: 74.1^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=8.94\times 10^3 Joules

m= mass of copper = 454 g

c = specific heat capacity = 0.385J/g^0C

Initial temperature of the copper = T_i = 23.0°C

Final temperature of the water = T_f  = ?

Change in temperature ,\Delta T=T_f-T_i

Putting in the values, we get:

8.94\times 10^3=454\times 0.385\times (T_f-23.0)^0C

T_f=74.1^0C

The final temperature of copper will be 74.1^0C

8 0
3 years ago
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