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2 years ago

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5 What is the kinetic energy of a 28 kg car moving at a velocity of 7 m/s O 1963 98 686 J O 350 nocter diagram, at which point w

ould the kinetic energy of the

1 answer:

Shtirlitz [24]2 years ago

4 0

Answer:

98 joules

Explanation:

1/2 of 28 = 14

14 multiply by 7= 98 joules

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The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscop

Blababa [14]

Answer:

The focal length of eye piece is 6.52 cm.

Explanation:

Given that,

Angular Magnification of the microscope M = -46

the distance between the lens in microscope L= 16 cm

The focal length of objective f₀ = 1.5 cm

Normal near point N = 25 cm

Have to find focal length of eye piece f ₙ =?

The angular magnification is given by

M ≈ - (L-fₙ)N/f₀fₙ

Rearranging for fₙ

fₙ =L(1 - Mf₀/N)⁺¹

=18/2.76

fₙ = 6.52 cm

The focal length of eye piece is 6.52 cm.

6 0

3 years ago

Which of the following accurately describes the way in which a muscle moves?

Vaselesa [24]

<h3><u>Answer</u>;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3><u>Explanation;</u></h3>

<em><u>The events of muscle fiber shortening occurs with in the sacromeres in the fibers. </u></em>
<em><u>Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.</u></em>
<em><u>Filament movement starts at the region or zone where thin and thick filaments overlap. </u></em>
<em><u>Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.</u></em>

7 0

2 years ago

Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci

EleoNora [17]

We subtract the velocity of car 1 from the velocity of car 2:
$v=(30\ mph\ North)-(20\ mph\ South)$
$=(30\ mph\ North)+(20\ mph\ North)$
$=50\ mph\ North$

4 0

2 years ago

Read 2 more answers

A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

$T^2=\frac{4\pi ^2}{GM} a^2$

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

$M = 3.5M_{sun} = 6.96*10^{30} kg$

$a=4.2AU = 6.28*10^{11} meters$

$G = 6.67*10^{-11}m^3kg^{-1}s^{-2}$

We put this into Kepler's law and get:

$T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.$

This when converted to years is 4.6 years.

4 1

2 years ago

Read 4 more answers

A stationary boat in the ocean is experiencing waves from a storm. the waves move at 56 km/h and have a wavelength of 160 m, bot

Lera25 [3.4K]

The wavelength $\lambda$ of the wave is 160 m, and this is the distance between two consecutive crests. The boat is located at a crest of the wave, this means that the first trough is located 80 meters from the boat (because the distance between a crest and a trough is half the wavelength).

The speed of the wave is
$v=56 km/h = 15.6 m/s$
so the time the boat takes to reach the first trough is
$t= \frac{S}{v} = \frac{80 m}{15.6 m/s}=5.1 s$

5 0

3 years ago