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3 years ago

If two lines in a system of linear equations have the same slope and same y-intercept, how many solutions will the system have?

Physics

1 answer:

arsen [322]3 years ago

8 0

Answer:

We will have infinite solutions to the system of linear equations.

Explanation:

Well, when we have two lines with the same slopes and the same y-interception, both of them are overlapped, so we will have infinite solutions to the system of linear equations.

This kind of system is called dependent system.

I hope it helps you!

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A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe

Levart [38]

Answer:

Coefficient of static friction will be equal to 0.642

Explanation:

We have given acceleration $a=6.3m/sec^2$

Acceleration due to gravity $g=9.8m/sec^2$

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to $a=\mu g$ , here $\mu$ is coefficient of static friction and g is acceleration due to gravity

So $\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642$

So coefficient of static friction will be equal to 0.642

3 0

2 years ago

Why were the rings of Uranus not observed directly from telescopes on the ground on Earth? How were they discovered?

leonid [27]

Answer:Explained below.

Explanation:

Uranus rings is made up of jet black, coal-like particles in small bands, making them difficult to perceive from Earth.This indicates that they are probably composed of a mixture of the ice and a dark material. The nature of material is dismal, but it might be some organic compounds greatly darkened by the charged particle irradiation from the Uranian magnetosphere. Rings were discovered by using a infrared telescope throughout the occultation of a star as Uranus passed in front of it. The light from the star dimmed many times before it was obstructed by the disk of Uranus and subsequently, showing the presence of various distinct rings.

6 0

3 years ago

KIIS FM broadcasts at 102.7 MHz, what is the wavelength of that radio wave?

marysya [2.9K]

Answer:

2.92 m

Explanation:

As we know, frequency × Wavelength = Speed of light

so here frequency of 102.7 MHz can be written as 102.7× 10⁶ Hz..

So Lambda (wavelength) = 3×10⁸/ 102.7 × 10⁶ which gives 2.92 metres or 2.92 × 10¹⁰ Å

3 0

2 years ago

A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p

DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

$t_{total}=t_{up}+t_{down}$

Time upriver is calculated with the net speed of student and 0.500 km:

$t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s$

(Becareful with units 0.5 km= 500m) Similarly of downriver:

$t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s$

So the sum is:

$t_{total}=1111.0s$

b) Still water does not affect student speed, so total time would be simply:

$t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s$

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.

6 0

2 years ago

Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c

Nata [24]

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0

2 years ago