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2 years ago

5

A car starts from rest and accelerates at 5 m/s/s.

1 answer:

goldenfox [79]2 years ago

3 0

Please find attached photograph for your answer. Hope it helps. Please do comment

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A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average

Maksim231197 [3]

Answer: $72.66 AU=1.089(10)^{10} km$

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period $T$ of a planet is proportional to the cube of the semi-major axis $a$ of its orbit”:

$T^{2}\propto a^{3}$ (1)

Now, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes:

$T^{2}=a^{3}$ (2)

So, knowing $T=619.36 years$ and isolating $a$ from (2) we have:

$a=\sqrt[3]{T^{2}}$ (3)

$a=\sqrt[3]{(619.36 years)^{2}}$ (4)

Finally:

$a=72.66 AU$ T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

$a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km$

4 0

3 years ago

Aunt Matilda goes to a well and throws a penny straight down the well at 3.0 m/s. She hears a splash after 0.5 seconds. How deep

nevsk [136]

Answer : The correct option is (d) 2.73 m

Explanation :

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$

where,

s = distance or height = ?

u = initial velocity = 3.0 m/s

t = time = 0.5 s

a = acceleration due to gravity = $9.8m/s^2$

Now put all the given values in the above equation, we get:

$s=(3.0m/s)\times (0.5s)+\frac{1}{2}\times (9.8m/s^2)\times (0.5s)^2$

$s=2.73m$

Therefore, the correct option is (d) 2.73 m

8 0

3 years ago

A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti

Korvikt [17]

Answer:

The strength of the electric field is $1.35\times10^{4}\ N/C$ .

Explanation:

Given that,

Speed $v= 5.05\times10^{5}\ m/s$

Time $t= 3.90\times10^{-7}\ s$

Angle = 45°

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$5.05\times10^{5}=0+a\times3.90\times10^{-7}$

$a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}$

$a=1.29\times10^{12}\ m/s^2$

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

$F= ma=qE$

$ma = qE$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}$

$E=1.35\times10^{4}\ N/C$

Hence, The strength of the electric field is $1.35\times10^{4}\ N/C$ .

3 0

3 years ago

Any Help??? Dued before 2:30pm and it's 9:30am by me... so please take yuh time and ans.... jus ans 1 if you want... This is for

Kryger [21]

Answer:

The methodology employed by Galileo contributed to the development of Physics by find moons of Jupiter. (I think)

sorry if it's wrong

5 0

3 years ago

enyata [817]

The sun <u><em>appears</em></u> brighter than any other star.

(It isn't really, but it looks that way because it's much much much much much much closer to us than any other star.)

7 0

3 years ago