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Dennis_Churaev [7]

2 years ago

14

A 144-g baseball is dropped from a tree 11.0m above the ground.A.With what speed would it hit the ground if air resistance could

be ignored?Express your answer to three significant figures and include the appropriate units.B.If it actually hits the ground with a speed of 7.50m/s , what is the magnitude of the average force of air resistance exerted on it?Express your answer to three significant figures and include the appropriate units.

1 answer:

nikitadnepr [17]2 years ago

5 0

Answer:

a. 14.68m/s

b. 1.043N

Explanation:

a.

Given

Mass = m = 144g = 0.144kg

Height = h = 11m

Using energy conservation,

½mv² = mgh where g = 9.8m/s² ----- divide both sides by m

½v² = gh ---- divide both sides by ½

v² = 2gh

v = √2gh

v = √(2 * 9.8 * 11)

v = 14.68332387438212m/s

v = 14.68m/s ----- Approximated

b.

v = 7.5m/s

By definition, “the average force of air resistance,”

When work done against air resistance is solved by multiplying height by the average force of resistance ------ by definition

i.e. W = -Fh

Minus is used because F and h points to different direction (when viewed as vectors).

This is proved by:

W = Fhcos(180) because h points toward the direction of motion (downward) while F points opposite.

W = Fh * -1

W = Fh

Using energy conservation,

mgh - Fh = ½mv² ---- Make Fh, the subject of formula

Fh = mgh - ½mv² ---- Divide through by h

F = mg - mv²/2h

F = 0.144 * 9.8 - 0.144 * 7.5²/(2 * 11)

F = 1.043N

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Answer:

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How much will the kinetic energy of a school bus increase if its velocity is tripled?

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Answer:

Explanation:

Given a school bus.

Let say initially the school bus is traveling with speed "v"

Let assume mass of school bus is "m"

Then, the initial kinetic energy is

K.E_initial = ½mv²

Now, if the initial velocity is tripled,

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v_new = 3v.

Note: the mass of the school does not change it is constant

Then, new kinetic energy is

K.E_new = ½m(v_new)²

v_new = 3v

Then,

K.E_new = ½m(3v)²

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Since K.E = ½mv²

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4 0

2 years ago

A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv

olchik [2.2K]

Answer:

magnification is - 159

objective distance is 3.85 cm

Explanation:

Given data

focal length f1 = 1.40 cm

focal length f2 = 2.20 cm

separated d = 19.6 cm

to find out

angular magnification and How far from the objective

solution

we know magnification formula that is

magnification = ( - L / f1 ) (D/f2)

here D = 25 cm put all value

magnification = ( - 19.6 / 1.40 ) (25/2.20)

magnification = - 159

and

now we apply lens formula

i/f = 1/q + 1/p

p = f2 = 2.20

so

q = f2 p / p -f2

q = 1.4(2.20) / ( 2.2 - 1.4 )

q = 3.85 cm

so objective distance is 3.85 cm

3 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

2 years ago