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Darina [25.2K]
3 years ago
10

Complete this equation that represents the process of nuclear fusion. Superscript 226 Subscript 88 Baseline R a yields Superscri

pt A Subscript B Baseline R n + Superscript 4 Subscript 2 Baseline H e A: B:
Physics
2 answers:
Talja [164]3 years ago
7 0

Answer:

I know that the b portion for the bottom is 86 because thats the atomic number of Rn

Explanation:

Tom [10]3 years ago
6 0

Answer:

A: 222

B: 86

Explanation:

Correct on E2020

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Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!
krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

L = 10 Log\frac{I}{I_0}

20 = 10 Log{I}{10^{-12}}

I_1 = 10^{-10}W/m^2

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

L = 10 Log\frac{I}{I_0}

50 = 10Log\frac{I}{10^{-12}}

I_2 = 10^{-7}W/m^2

now we know that

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}

r_2 = 47 cm

so now the distance from friend must be 47 cm

8 0
3 years ago
I need this right ASAP!!! plz plz plz
UkoKoshka [18]

Answer:

The answer is D

Nucleus is neutral

8 0
3 years ago
Read 2 more answers
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

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4 years ago
A 25 kh inductor carries a current of 60 ma. how much energy is stored in this system
IgorLugansk [536]
Energy stored= Li^2/2
=45 W
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3 years ago
How do the most common forms of energy work?
DiKsa [7]

Answer:

activity 2

Explanation:

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