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2 years ago

What force is required to accelerated a body with a mass of 15 kilograms at a rate of 8 m/s2

Physics

1 answer:

rjkz [21]2 years ago

7 0

Force = (mass) · (acceleration)

= (15 kg) · (8 m/s²)

= 120 kg-m/s² = 120 newtons

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

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2 years ago

A light-year is the distance light travels in one year (at speed = 2.998 × 108 m/s). (a) how many meters are there in 11.0 light

larisa [96]

<span>The answers are as follows:

(a) how many meters are there in 11.0 light-years?

11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m

(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?

1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au

(c) what is the speed of light in au/h? au/h

</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h

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2 years ago

The old rubber boot has two leaks. The top of the boot is 0.3 m higher than the leaks. What is the velocity of the water coming

Otrada [13]

Answer:

Leak 1 = 3.43 m/s

Leak 2 = 2.42 m/s

Explanation:

Given that the top of the boot is 0.3 m higher than the leaks.

Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2

From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.

Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2

Find the attached files for the solution and figure

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