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Natasha2012 [34]
2 years ago
13

What reactions do catalytic converters catalyze

Physics
1 answer:
Yuliya22 [10]2 years ago
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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Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a sc
Sladkaya [172]

Answer:

0.3376 mm

Explanation:

The computation of the spacing in mm between the slits is shown below:

As we know that

d = \frac{m\lambda L}{\Delta y}

where,

\lambda = wavelength

L = distance from the scrren

\Delta y = spanning distance

As there are 11 bright fingers seen so m would be

= 11 - 1

= 10

Now placing these values to the above formula

So, the spacing is

= \frac{(10)(633 \times 10^{-9})(3.2m)}{60 \times 10^{-3}}

= 0.3376 mm

We simply applied the above formula.

6 0
3 years ago
Read 2 more answers
When light hits the water droplets in a cloud, the cloud appears white. The light waves are being:
Tamiku [17]

Answer:

scattered

Explanation:

7 0
3 years ago
A particle covers 10m distance in 2sec. if final velocity is 8m/s, find initial velocity​
klasskru [66]

Explanation:

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion

Case I

s=ut+

2

1

at

2

10=5u+

2

1

a(5)

2

20=10u+25a

4=2u+5a..............(1)

Case 2

In next 3s the particle covers more 10m distance. So

20=8u+

2

1

a(8)

2

5=2u+8a.........(2)

On solving equation (1) and (2)

4=2u+5a

5=2u+8a

a=

3

1

m/s

2

Put the value of a in equation (1)

u=

6

7

m/s

Now to find distance in next 10 s. total time will be 10s

s=

6

7

×10+

2

1

×

3

1

×(10)

2

s=28.33m

Distance travelled in next 2 sec

s=28.33−20=8.33m

4 0
3 years ago
Read 2 more answers
A 50 kg child runs off a dock at 2.0 ms (horizontally) and lands in a waiting rowboat of mass 150 kg. At what speed does the row
Alex73 [517]

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

Hi there!

Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

Then:

50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

5 0
3 years ago
Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
10 months ago
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