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valina [46]
2 years ago
6

Question 7 Points 2

Physics
1 answer:
uranmaximum [27]2 years ago
8 0

Answer:

The dynamo has a wheel that touches the back tyre. As the bicycle moves, the wheel turns a magnet inside a coil. This induces enough electricity to run the bicycle's lights. The faster the bicycle moves, the greater the induced voltage - and the brighter the lights.

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5 Muscular and strength

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Stars are called blackbody radiators because they
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The answer is D) absorb most of the light that falls on them

A blackbody is an object that absorbs most (theoretically all) of the light that falls on it. Stars are called black bodies because they demonstrate something close to this property. Why do they appear white and not black? As black body radiators heat up, they slowly turn from black to grey to white, according to the human eye. Since stars are incredibly hot, they appear white to us. 
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3 years ago
Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o
Ainat [17]

Answer:

  • <em>Abbie’s acceleration is (1/2) Zak’s acceleration.</em>

Explanation

1. <u>Data</u>:

a) ω = constant

b) Abbie: r₁ = 1 m

c) Zak: r₂ = 2 m

d) Ac₁ = ? Ac₂

2. <u>Formulae</u>

  • Ac = ω² r

3. <u>Solution</u>:

a) Abbie:

  • Ac₁ = ω² r₁  =  ω² (1m)

b) Zack:

  • Ac₂ = ω² r₂  = ω² (2m)

c) Divide Ac₁ / Ac₂

  • Ac₁ / Ac₂ =  ω² (1m) / [ω² (2m) ] = 1/2

⇒      Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂

5 0
3 years ago
Question 12 (1 point) Question 12 Unsaved
Bezzdna [24]

Slope is your answer

7 0
3 years ago
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!!!

4 0
3 years ago
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