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3 years ago

Learning Task 1: Determine what kind of carbon allotropes are the given pictures

Physics

2 answers:

const2013 [10]3 years ago

5 0

Answer:

1.buckball c70

2. buckballs c60

3. ionsdaleite

4.graphite

5.diamond

6.amorphous carbon

Explanation

kotegsom [21]3 years ago

4 0

Answer:

1.C70

2.Fullerene

3.Lonsdaleite

4.Graphite

5.Diamond

6.Amorphous carbon

I hope this helps. thank you

Explanation:

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Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a const

vesna_86 [32]

Answer:

84π m²/s or 263.76 m²/s

Explanation:

see the attached file

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3 years ago

A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and

balandron [24]

The upward force exerted on the board by the support is mathematically given as

Fu= 764.8 N

<h3>What is the upward force exerted on the board by the support?</h3>

Generally, the equation for is mathematically given as

Considering that the Net Force on the system is null

The weight of the children plus the weight of the board equals the upward force imposed on the support.

The upward force

Fu= 440 + 272 + 52.8 N

Fu= 764.8 N

In conclusion, he upward force exerted on the board by the support

Fu= 764.8 N

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8 0

2 years ago

Heat engines take input energy in the form of heat, use some of that energy to do work, and exhaust the remainder. Similarly, a

Elanso [62]

Answer:

a). $Q_{h}=1.5x10^5J$

b). $Q_{c}=13.2x10^3J$

Explanation:

a).

Using the law of thermodynamic and using to isovolumetric

ΔU=Qh

$Q_{h}=\frac{W_{net}}{Efficiency}$

So to determine the internal energy knowing the work it do

$Q_{h}=\frac{1.5x10^4J}{0.10}=1.5x10^5J$

b).

Efficiency is the work done in relation of the work apply in this case to produce heat so:

$E=\frac{Q_{h}-Q_{c}}{Q_{h}}$

Solve to Qc

$Q_{c}=Q_{h}*[1-E]$

$Q_{c}=1.5x10^4J*(1-0.10)$

$Q_{c}=13.2x10^3J$

3 0

3 years ago

A motorboat accelerates uniformly from a ve- locity of 7.0 m/s to the west to a velocity of 2.4 m/s to the west. If its accelera

Lena [83]

Answer:

Distance, d = 8 meters

Explanation:

Given that,

Initial velocity of the motorboat, u = 7 m/s (west)

Final velocity of the motorboat, v = 2.4 m/s (west)

Acceleration of the motorboat, $a=-2.7\ m/s^2$ (east)

We need to find the distance covered by it. It can be calculated as :

$v^2-u^2=2as$

$(2.4)^2-(7)^2=2\times (-2.7)s$

s = 8 meters

So, the distance covered by the motor boat is 8 meters. Hence, this is the required solution.

8 0

3 years ago

A 600-g mass traveling at 8.0 m/s undergoes a head-on elastic collision with a 200-g mass traveling toward it also at 8.0 m/s. W

Pepsi [2]

Answer:

vf₂ = 16 m/s

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

P=m*v

where

P:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ : Initial linear momentum quantity

Pf : final linear momentum quantity

Data

m₁ = 600 g

m₂= 200 g

v₀₁ = 8 m/s :initial velocity of m₁

v₀₂ = - 8 m/s :initial velocity of m₂ ; (-) indicates that m₂ is moving in the opposite direction to m₁

Problem development

We appy the formula (1):

We assume that at the end of the collide both masses move in the initial direction of m₁ , then vf₁ and vf₁ have a sign (+)

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

(600) *(8)+ (200)*( -8) = (600) *vf₁ + (200)*vf₂

4800 - 1600 = (600) *vf₁ + (200)*vf₂

3200 = (600) *vf₁ + (200)*vf₂ : We divide both sides of the equation by 100

32= 6*vf₁ +2*vf₂ Equation (1)

For elastic collision, the elastic restitution coefficient (e) is equal to 1 :

$e= \frac{ v_{f2}- v_{f1} }{ v_{o1}- v_{o2} }$

$1= \frac{ v_{f2}- v_{f1} }{ 8- (-8) }$

16 = vf₂ - vf₁

vf₁ = vf₂ - 16 Equation (2)

We replace Equation (2) in the Equation (1)

32= 6*vf₁ +2*vf₂

32= 6*( vf₂ - 16) +2*vf₂

32= 6*vf₂ - 96 + 2*vf₂

32 + 96= 6*vf₂ + 2*vf₂

128 = 8 *vf₂

vf₂ = 128 / 8

vf₂ = 16 m/s

6 0

4 years ago