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mamaluj [8]
3 years ago
5

Discuss the difference between short-term and long-term fitness goals. Provide an example of each

Physics
1 answer:
photoshop1234 [79]3 years ago
5 0

Answer:

. A long-term goal is a goal set for you to achieve with 3-5 years, while short-term goals take a few days, weeks, or maybe months. An example of a short-term is eating like less carbs. An example of a long-term running like a 5k marathon.

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A cubic meter (m³) is ______ a cubic centimeter (cm³).
valentina_108 [34]

Answer:

C. equal to

Explanation:

1 Cubic meter (m³) is equal to 1000000 cubic centimeters (cm³). To convert cubic meters to cubic centimeters, multiply the cubic meter value by 1000000.

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3 years ago
Phys-1A Horizontal Practice 2
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Defination wheel and axle​
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3 years ago
An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car
saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

F = ma

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

-10000 =2000\times a

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

a = -\dfrac{10000}{2000}

a=-5\ m/s^2

For the distance,

Using third equation of motion

v^2-u^2=2as

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

0-30^2=2\times(-5)\times s

s = 90\ m

Hence, The car traveled the distance before stopping is 90 m.

6 0
3 years ago
Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o
Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, M=1.99\times 10^{30}\ kg

Radius of Mercury's orbit, r=5.79\times 10^{10}\ m

Radius of discovered planet, R=\dfrac{2}{3}r

R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

T^2\propto R^3

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

T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}

T=\sqrt{1.71\times 10^{13}}

T = 4135214.625 s

or

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0
3 years ago
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