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3 years ago

Explain how fusion and fission are similar to and different from each other. Thanks.

2 answers:

5 0

Fusion is the act of putting together ( fusing ) two items so that they become one. Fission is the act of dividing or splitting an item.

Margaret [11]3 years ago

5 0

Nuclear fission and nuclear fusion are similar in that they both involve changes in the nucleus of the atom. Another similarity is that relatively high energies (for the size of the matter involved) are associated with those nuclear changes.

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Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in

sattari [20]

Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

The weight of the bucket of water = mg.

Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

8 0

3 years ago

Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran

alina1380 [7]

Answer:

Datos:

q1 = -50 μC = $50*10^{-6}$

q2 = +30 μC = $30*10^{-6}$

F = 10 N

a) x si la F = 10N

Aplicando la Ley de Coulomb:

x = $\sqrt\frac{k0 * q1 *q2}{F}$ = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10}$ = 1,162m

b) x si la F = 20 N

x= $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}$ = 0,822m

c)x si la F = 50 N

x = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50}$ = 0,520m

3 0

3 years ago

Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of

Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = $\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}$

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = $\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}$

(|r₂₁|)² = 148.25

$F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }$

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0

3 years ago

An atom that has 117 protons in its nucleus has not yet been made. Once this atom is made, to which group will element 117 belon

Pavel [41]

In nomine patris, et filii, et spiritus sancti.

3 0

3 years ago

Someone please help!!!

aivan3 [116]

Answer:

Explanation:

hope this helps!

6 0

2 years ago