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3 years ago

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Alanna spots a bird in her back yard. The bird is sitting on a tree. Explain how the outer coverings of the bird and tree are di

fferent and have different functions.

1 answer:

eduard3 years ago

6 0

Answer:

the outer covering of a bird is usually for camoflauge (idk how to spell) or to attract mates and the outer part of a tree is used for protection i think, correct me if im wrong ;-;

Explanation:

You might be interested in

A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that h

RoseWind [281]

Answer:

a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

F -W = m a

Force is electrical force

F = k q₁ q₂ / r²

k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

q₁ = q₂ = q

a = (k q² / r² - m g) / m

a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

a = -7.29 m / s²

The negative sign indicates that the direction of this acceleration is downward

6 0

2 years ago

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m

anzhelika [568]

Answer:

Momentum after collision will be 6000 kgm/sec

Explanation:

We have given mass of the whale = 1000

Initial velocity v = 6 m/sec

It collides with other mass of 200 kg which is at stationary

Initial momentum of the whale = 1000×6 = 6000 kgm/sec

We have to find the momentum after collision

From conservation of momentum

Initial momentum = final momentum

So final momentum = 6000 kgm/sec

5 0

3 years ago

A bus that was moving at a high speed stopped suddenly. One of

fiasKO [112]

Answer:

this doesn't make sinces

7 0

2 years ago

A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t

jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is $9.8 \mathrm{m} / \mathrm{s}^{2}$

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

$\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})$

$\text { Work }=-15.03 \mathrm{J}$

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is $\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

$\mathrm{F}=13 \times 9.8$

F = 127.4N

The force exerted on the object is 127.4N.

4 0

2 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

n200080 [17]

a) $8.93\cdot 10^{-30} N$ , downward

b) $1.76\cdot 10^{-17} N$ , upward

c) $1.20\cdot 10^{-17} N$ , downward

Explanation:

a)

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

$F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N$

And the direction is downward, towards the Earth's centre.

b)

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$E=110 N/C$ is the strength of the electric field

Substituting,

$F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N$

And since the electron has negative charge, the direction of the force is opposite to that of the electric field, so upward.

c)

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

$F=qvB$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$v=1.50\cdot 10^6 m/s$ is the velocity

$B=50.0\mu T = 50.0 \cdot 10^{-6} T$ is the strength of the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N$

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

5 0

3 years ago