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3 years ago

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

2 answers:

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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Which does not represent a change in velocity?

alina1380 [7]

Answer:

A balloon is floating up into the air at 1 m/s.

Explanation:

Velocity is the displacement divided by the time;

Velocity = $\frac{displacement}{time}$

It is a vector quantity that has magnitude and direction.

The choice that shows a change in direction is a correct specification of velocity.

A balloon floating up into the air at 1m/s shows no directional change.

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3 years ago

Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizont

vfiekz [6]

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

$F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2$

Now applying Newton's second law to box B. So,

$F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N$

So, 2.75 N is the force that box A exerts on box B.

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3 years ago

Help Please fast like now!!

defon

Answer: d. availability of places to play

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3 years ago

3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed rel

kramer

Answer:

$v_s=\frac{v}{1+\frac{m}{M} }$

Explanation:

Let the shells speed with respect to ground be $v_s$

Let the shells speed with respect to ground be $v_g$

mass of shell, $m$

mass of gun, $M$

The relative velocity of shell with respect to a free unconstrained gun, $v$

According to the Newton's third law of motion the direction of the velocity of gun and shell will be in the opposite direction.

So, the relation between the relative velocity and their individual velocity will be:

$v=v_s+v_g$ ......................(1)

<u>And according to the conservation of momentum (as the condition is very close to the elastic collision):</u>

$M.v_g-m.v_s=0$

substitute the value of $v_g$ from equation (1)

$M\times (v-v_s)=m\times v_s$

$M.v-M.v_s=m.v_s$

$M.v=M.v_s+m.v_s$

$v_s=\frac{M.v}{(M+m)}$

$v_s=\frac{v}{(\frac{(M+m)}{M}) }$

$v_s=\frac{v}{1+\frac{m}{M} }$

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3 years ago

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Sergeu [11.5K]

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Read 2 more answers