All categories

3 years ago

What are pollutants of greatest concern and who is at risk?

1 answer:

6 0

Oil leakage in aquatic areas. (fish)
Burning fuel. (Air/global warming)
Toxic waste burried in dirt (living Creatures)
Lights/Electricity (destroyes Birds migration)
Noise (migration on dolphines and other sea mammals)

You might be interested in

Action-reaction forces : Select one:

ladessa [460]

i think its B)... im not sure so dont take it seriously

4 0

3 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with

Tanya [424]

Answer:

(a) $K_{small}=4839.3J$

(b) $K_{larger}=17406.4J$

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

$K=\frac{1}{2}Iw^2$

Where I is rotational of inertia of solid cylinder about its central axis.

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2$

Substitute the given values

$K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J$

For Part (b)

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2$

Substitute the given values

$K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J$

8 0

3 years ago

Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha

egoroff_w [7]

The centripetal acceleration of an object is given by the relation,

$Ac =V^2/R$

where Ac = centripetal acceleration = $98 m/s^2$

R = radius of rotation = 15 m

V = speed of astronaut

Hence, $\frac{V^2}{15} =98$

solving this we get, V = 38.34 m/s

3 0

3 years ago

Read 2 more answers

Fossils found in the La Brea tar pits indicate a California climate that was A) similar to today's climate. B) similar to the pr

NikAS [45]

Answer:

i believe it is D but not 100% sure

Explanation:

3 0

3 years ago

Read 2 more answers

A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to

dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So $1.41=2\times 3.14\sqrt{\frac{0.683}{K}}$

$0.2245=\sqrt{\frac{0.683}{K}}$

Squaring both side

$0.0504=\frac{0.4664}{K}$

$K=9.255N/m$

So spring constant of the spring will be equal to 9.255 N /m

7 0

3 years ago