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3 years ago

Help plzzz!!! I only have 10 minutes to turn in

Physics

1 answer:

Lostsunrise [7]3 years ago

4 0

The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car. When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

Kinetic energy = (1/2) (mass) (speed²)

And there we have it

The car's mass is 3,600 kg.
Its speed is 'v' m/s .
(1/2) (mass) (v²) = 5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

v² = (5406/1800) (Newton-meter/kg)

v² = (5406/1800) (kg-m²/s² / kg)

v² = (5406/1800) (m²/s²)

= = = = =

v = √(5406/1800) m/s

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A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

$\omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}$

$\omega = 1.864\times 10^{-7}\ rad/s$

b) tangential speed

$v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}$

v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

$a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}$

a = 3.62 x 10⁻³ m/s²

8 0

3 years ago

A 5.00 kg mass is placed on top of a vertical spring, which compresses a distance of 3.13 cm. Calculate the force constant (in N

77julia77 [94]

Answer:

Explanation:

Step one:

given data

mass m=5kg

compression x= 3.13cm to m= 0.0313m

<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>

F=ke

where

k= spring constant in N/m

e= extension/compression in

Step two:

assume g= 9.81m/s^2

F=mg

F=5*9.81

F=49.05N

substitute in the expression F=ke

49.05=k*0.0313

k=49.05/0.0313

k=1567.09 N/m

<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>

8 0

3 years ago

The stars in the model below are arrange by.... *

brilliants [131]

Answer:

where is the picture of star

5 0

3 years ago

PLEASE HELP WILL GIVE EXTRA POINTS

wolverine [178]

Sorry but you tell me not to ask for help or answers but you want them yourself

6 0

3 years ago

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

nirvana33 [79]

(a) $7.25\cdot 10^{-8}m$

Wien's displacement law is summarized by the equation

$\lambda = \frac{b}{T}$

where

$\lambda$ is the peak wavelength

$b=2.898\cdot 10^{-3} m \cdot K$ is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

$\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m$

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays

X-rays 1 nm - 1 pm

ultraviolet 380 nm - 1 nm

visible light 750 nm - 380 nm

infrared $25 \mu m - 750 nm$

microwaves 1 mm - $25 \mu m$

radio waves > 1 mm

The peak wavelength of this star is

$\lambda=7.25\cdot 10^{-8}m=72.5 nm$

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

7 0

3 years ago