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3 years ago

The most common type of plant on earth is

Physics

1 answer:

xxMikexx [17]3 years ago

6 0

Mosses, algae and flowering plants.

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How do you calculate the electric force by object a exerted on object a (itself)?

max2010maxim [7]

I don't think an object can exert a force on itself.

Try it: Get up on a skateboard, and see if you can do anything to yourself that makes you start moving ... without touching anything else.

It'll be easy to tell if you succeed. If you actually do exert an unbalanced force on yourself, then you'll begin to accelerate.

5 0

3 years ago

Oxygen enters the blood through the.... I do not know the answer. Can someone please help me?!?!

frez [133]

Oxygen enters the blood Through your lungs

8 0

3 years ago

Read 2 more answers

. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th

mixer [17]

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

$F \Delta t = m (v-u)$

where

F is the average force

$\Delta t$ is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

$\Delta t = 60.0 ms = 0.06 s$

Solving for F,

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N$

And since we are interested in the magnitude only,

F = 106.7 N

5 0

3 years ago

Read 2 more answers

Heat energy is:

Archy [21]

I claim that it's d moving electrons.

6 0

3 years ago

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0

3 years ago