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3 years ago

What reactions do catalytic converters catalyze

1 answer:

3 0

A catalytic converter is an exhaust emission control device that reduces toxic gases and pollutants in exhaust gas from an internal combustion engine into less-toxic pollutants by catalyzing a redox reaction (an oxidation and a reduction reaction).

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Which of the following forces does NOT act on an object sliding down an incline plane?

Oduvanchick [21]

A.Force of tensión is the answer

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3 years ago

Which of the following statements best summarizes the main points of the passage?

Deffense [45]

Answer:the summery is ...i really don't know the picture is blurry and I cant see can you make it clear?

Explanation:

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2 years ago

Calculate the gravitational potential energy of the interacting pair of the Earth and a 4 kg block sitting on the surface of the

STALIN [3.7K]

Answer:

- 2.425 x 10^5 J

Explanation:

The gravitational potential energy between earth and the bock is given by

$U=-G\frac{Mm}{r}$

Where, G is the universal gravitational constant = 6.67 x 10^-11 Nm^2/kg^2

M is the mass of earth = 5.8 x 10^24 kg

m is the mass of block = 4 kg

r be the radius of earth = 6380 km = 6380 x 10^3 m

by substituting the values in the above expression, we get

$U=-6.67\times10^{-11}\frac{5.8\times 10^{24}\times 4}{6380\times 10^{3}}$

U = - 2.425 x 10^5 J

5 0

3 years ago

Read 2 more answers

A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.

Kay [80]

Answer:

a.6.5025 J

b.6.5025 J

Explanation:

We are given that

Mass of pellet,m=0.27 g= $0.27\times 10^{-3} kg$

1 kg=1000 g

Spring constant,k=1800 N/m

x=8.5 cm= $8.5\times 10^{-2} m$

1m=100 cm

a.Potential energy stored in the compressed spring is given by

P.E= $\frac{1}{2}kx^2$

$P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2$

$P.E=6.5025 J$

b.By using law of conservation of energy

P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J

6 0

4 years ago

At some instant, a particle traveling in a horizontal circular path of radius 7.90 m has a total acceleration with a magnitude o

DochEvi [55]

Answer:

a) Speed of the particle at this instant

v = 8.43 m/s

b) Speed of the particle at (1/8) revolution later

v = 14.83 m/s

Explanation:

We apply the equations of circular motion uniformly accelerated :

$(a_{T}) ^{2} = (a_{n} )^{2} +(a_{t} )^{2}$ Formula (1)

$a_{n} = \frac{v^{2} }{r}$ Formula (2)

$a_{t} = \alpha *r$ Formula (3)

v= ω*r Formula (4)

ω² = ω₀² + 2*α*θ Formula (5)

Where:

$a_{T}$ : total acceleration, (m/s²)

$a_{n}$ : normal acceleration, (m/s²)

$a_{t}$ : tangential acceleration, (m/s²)

$\alpha$ : angular acceleration (rad/s²)

r : radius of the circular path (m)

v : tangential velocity (m/s)

ω : angular speed ( rad/s)

ω₀: initial angular speed ( rad/s)

θ : angle that the particle travels (rad)

Data:

$a_{T}$ = 15 m/s²

$a_{t}$ = 12 m/s²

r=7.90 m :radius of the circular path

Problem development

In the formula (1) :

$a_{n} = \sqrt{(a_{T})^{2} -(a_{t})^{2} }$

We replace the data

$a_{n} = \sqrt{(15)^{2} -(12)^{2}}$

$a_{n} = 9 \frac{m}{s^{2} }$

We use formula (2) to calculate v:

$9 = \frac{v^{2} }{7.9}$ Equation (1)

a)Speed of the particle at this instant

in the equation (1):

$v=\sqrt{9*7.9} = 8.43 \frac{m}{s}$

b)Speed of the particle at (1/8) revolution later

We know the following data:

θ =(1/8) revolution=( 1/8) *2π= π/4

$a_{t}$ = 12 m/s²

v₀= 8.43 m/s

r=7.9 m

We use formula (3) to calculate α

$12 = \alpha *7.90$

$\alpha =\frac{12}{7.9} = 1.52 \frac{rad}{s^{2} }$

We use formula (4) to calculate ω₀

v₀= ω₀ *r

8.43 = ω₀*7.9

ω₀ = 8.43/7.9 = 1.067 rad/s

We use formula (5) to calculate ω

ω² = ω₀² + 2*α*θ

ω²= (1.067)² + 2*1.52*π/4

ω² =3.526

ω = 1.87 rad/s

We use formula (4) to calculate v

v= 1.87 rad/s * 7.9m

v = 14.83 m/s : speed of the particle at (1/8) revolution later

5 0

3 years ago