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Pachacha [2.7K]

2 years ago

12

Why isnt this technology used in coal plants

1 answer:

n200080 [17]2 years ago

5 0

Can cause atmospheric problems.

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¿A qué velocidad debe circular un auto de carreras para recorrer 87 km en 20 min? (pasar a metros por segundo m/s)

Alexxx [7]

Answer:

Velocidad en m / s = 72,25 m / s

Explanation:

Dado

Distancia a recorrer por el coche de carreras = 87 Km

1 km = 1000 m

Por lo tanto, 87 km = 87000 m

Tiempo necesario para viajar 87 km / 87000 metros = 20 minutos o 20 * 60 = 1200 segundos

Velocidad en m / s = 87000/1200

Velocidad en m / s = 72,25 m / s

7 0

2 years ago

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

DedPeter [7]

Answer:

Explanation:

3.4 m/s due North, -1.1 m/s due East

7 0

3 years ago

Which interaction of nature depends on the distance through which it acts and is involved in beta decay? a. strong c. electromag

USPshnik [31]

The answer is weak. The interaction of nature that will depend on the distance through the way it acts and involved in beta decay is the weak interaction or the weak force. This interaction is the responsible for radioactive decay which also plays a significant role in nuclear fission.

5 0

2 years ago

Which is NOT something that all green plants have in common?

Anvisha [2.4K]

Answer:

Number 4

Explanation:

7 0

2 years ago

Read 2 more answers

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

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

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

2 years ago