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3 years ago

What are the consequences of sediment pollution?

1 answer:

6 0

Loss of habitats for fish, birds, and other wildlife. Sediment pollution is one of the leading causes of the loss of the wetlands, but it’s not just the wetlands. Changes in the nutrients in your water. The same problem that affects the fish in your area may also affect you. Other drinking water contamination.

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023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0

3 years ago

- The school zone in front of your school has a posted speed limit of 25 mi/h, which is about 11 m/s. Let's

miv72 [106K]

Answer:

s = 38.7 m

Explanation:

First we calculate the distance covered during uniform motion of reaction time.

s₁ = vt

where,

s₁ = distance covered during uniform motion = ?

v = uniform speed = 11 m/s

t = time = 2.3 s

Therefore,

s₁ = (11 m/s)(2.3 s)

s₁ = 25.3 m

Now, we calculate the distance covered during decelerated motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = -4.5 m/s²

s₂ = distance covered during decelerated motion = ?

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 11 m/s

Therefore,

2(-4.5 m/s²)s₂ = (0 m/s)² - (11 m/s)²

s₂ = (-121 m²/s²)/(-9 m/s²)

s₂ = 13.4 m

the total distance will be:

s = s₁ + s₂

s = 25.3 m + 13.4 m

s = 38.7 m

5 0

3 years ago

How do you think overpumping groudwater is related to the formation of sinkholes?

vladimir2022 [97]

Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps

3 0

4 years ago

What is the definition of binary star system

lakkis [162]

Is a system of two objects in space which are so close that their gravitational interaction causes them to orbit about a common center of mass.

8 0

3 years ago

Read 2 more answers

An elevator is accelerating upward at a rate of 3.6 m/s2. a block of mass 24 kg hangs by a low-mass rope from the ceiling, and a

kozerog [31]

The answer:

When the elevator accelerates upward at a rate of 3.6 m/s², the value of the acceleration becomes

A=g+3.6=13.4 m/s²
and by using the newton's law, F=mass x A, we have
T1= (24 + 90 )x 13.4= 1527.6 N, where T1 is the Tension in upper rope
 and
T2= ( 90 )x 13.4= 1206N, where T2 is the Tension in lower rope

When the elevator accelerates downward at a rate of 3.6 m/s², the value of the acceleration becomes
A=9.8 - 3.6 = 6.2 m/s²

T1= (24 + 90 )x 6.2= 706.8 N, where T1 is the Tension in upper rope
and
T2= ( 90 )x 6.2= 558N, where T2 is the Tension in lower rope

5 0

3 years ago