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3 years ago

10

If you drop a gum rapper on the floor is that pollution

2 answers:

sdas [7]3 years ago

8 0

Not exactly, it is littering. littering makes the place look unpleasing and unclean, but it wont necessarily kill us. now if a factory is dumping their sewage and whatnot in the water, that is considered as industrial pollution because it is killing almost all the wildlife that live in the water, and it can contaminate the water so if you take a swim in it and accidentally swallow some of the water, it can make you sick and can possibly kill you.

Dmitry [639]3 years ago

5 0

No pollution is in the air, dripping a gum wrapper on the floor is littering

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Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at

Soloha48 [4]

Answer:

v = 11.0 m/s at 198.6° (18.6° south of west)

ΔKE = -145 kJ

Explanation:

I assume you want to find the final velocity and the change in kinetic energy.

Take east to be +x and north to be +y.

Momentum is conserved in the x direction:

(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ

vₓ = -10.4 m/s

Momentum is conserved in the y direction:

(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ

vᵧ = -3.50 m/s

The magnitude of the final velocity is:

v² = (-10.4 m/s)² + (-3.50 m/s)²

v = 11.0 m/s

The direction of the final velocity is:

θ = atan(-3.50 m/s / -10.4 m/s)

θ = 198.6°

The initial kinetic energy is:

KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²

KE₀ = 253,275 J

The final kinetic energy is:

KE = ½ (1800 kg) (11.0 m/s)²

KE = 108,682 J

The change in kinetic energy is:

ΔKE = 108,682 J − 253,275 J

ΔKE ≈ -145,000 J

7 0

3 years ago

two equal and unlike parallel forces of magnitude 34N act on a rigid body,such that the moment of couple is 8.50 Nm. calculate t

goldfiish [28.3K]

The moment of a couple is Force × perpendicular distance from the arm of the line of action

so the arm of the couple= moment of couple/force=8.5/34=0.25m

the arm is 0.25m

5 0

3 years ago

A car (m = 1302 kg) traveling along a road begins accelerating with a constant acceleration of 1.50 m/s2 in the direction of mot

antiseptic1488 [7]

First we will use the concepts of motion kinetics for which the final speed is defined as shown below,

$v_f^2=v_i^2+2as$

Here,

$v_f$ = Final velocity

$v_i$ = Initial velocity

a = Acceleration

s = Distance

Replacing,

$(35)^2 = v_i^2+2(1.5)(392)$

$v_i = 7m/s$

Using the conservation of energy for kinetic energy we have,

$KE = \frac{1}{2}mv_i^2$

$KE = \frac{1}{2}(1302)(7)^2$

$KE = 31900J$

Therefore the kinetic energy of the car is 31900J

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If the Coast Guard warns of a giant wave of water approaching the shore as a result of a major earthquake, they are warning of

shepuryov [24]

The answer is A Tsunami

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3 years ago

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Help please (20 pts) I think it's C

gregori [183]

Answer:

I'm not sure it is c I'm sure it is d

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