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Ira Lisetskai [31]
3 years ago
13

The Clean Air Act emphasizes that one way to prevent and reduce air pollution is to involve public participation

Physics
1 answer:
Marina CMI [18]3 years ago
7 0
I think the answer would be true that’s my opinion
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Sorry that i ask for too much but I'm horrible at this ⭐-_-
Evgesh-ka [11]

Answer:

1. A

2. C

3. D

4.A

Explanation:

6 0
2 years ago
Read 2 more answers
0.01
katen-ka-za [31]

Answer:

A

Explanation:

60 + 30 = 90

90 divided by 3 = 30

30 divided by 60 = 0.5

so your answer is 0.5 m/s

4 0
3 years ago
If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet.
crimeas [40]

Answer:

8.1 m/s^2

Explanation:

The strength of the gravitational field at the surface of a planet is given by

g=\frac{GM}{R^2} (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2

For the unknown planet,

M_X = 0.231 M_E\\R_X = 0.528 R_E

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E

And substituting g = 9.8 m/s^2,

g_X = 0.829(9.8)=8.1 m/s^2

3 0
3 years ago
Speed is usually what
kolbaska11 [484]

Answer:

measured in GHz?

Explanation:

im not sure what the context is it depends on what your lesson is on

3 0
3 years ago
A block is thrown with an initial velocity of 30.0 m/s at an angle of 25.0o above the horizontal. What is the highest elevation
Serga [27]

The highest elevation reached by the ball in its trajectory is 16.4 m.

To find the answer, we need to know about the maximum height reached in a projectile.

What's the mathematical expression of the maximum height reached in a projectile motion?

  • The maximum height= U²× sin²(θ)/g
  • U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity

What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?

  • Here, U = 30.0 m/s and θ= 25°
  • Maximum height= 30²× sin²(25)/9.8

= 16.4m

Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ4

3 0
2 years ago
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