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3 years ago

The equation for distance is d = st. If a car has a speed of 77.0 m/s and travels for 45.0 seconds, how far does it go?

2 answers:

8 0

The equation is d = s t , just like it says right at the beginning there.

-- Take the equation.

-- In the place where the equation says "s", erase 's' and write "(77 m/s)".

-- In the place where the equation says "t", erase 't' and write "(45 sec)".

-- NOW the equation says d = (77 m/s) (45 sec) .

-- When there are two numbers right next to each other like that,
it means you're supposed to multiply them.

-- So multiply them.

-- Their product is equal to "d", and that's the answer.

elena-14-01-66 [18.8K]3 years ago

3 0

Since you are given the formula all you need to do is "plug and chug"!

d = 77(45) = 3465 m = 3.465 km.

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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

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

Please help.. urgent Which statement is equivalent to Newton's first law? a. 15,300 N b. 1.20*10^3 N c. 2,030 N d. 1,560 N

taurus [48]

According to Newton laws of motion,
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values,
F = 1,560 * 1.30
F = 2028 N ~ 2030 N [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!

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3 years ago

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3 years ago

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professor190 [17]

Answer:

The correct option is D

Explanation:

This question is incomplete because of the absence of the setup which as been attached below. The setup shows/determines/tests the friction of wood (which is a block material), since Jerry wants to test the friction between different types of materials, he will have to replace the wooden block with another type of block material of choice so as to determine the friction of that also.

In order to have a comprehensive experiment, Jerry can use 4-5 different types of block material in the course of the experiment.

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3 years ago

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coldgirl [10]

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