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Georgia [21]
2 years ago
7

Two boys want to balance a seesaw perfectly. One boy weighs 120 pounds and is sitting four feet from the fulcrum. The other boy

weighs 80 pounds. Where should the lighter boy sit to balance the seesaw?
Physics
1 answer:
OLEGan [10]2 years ago
6 0

Basically a distance multiplied by a weight that is equal to the distance that is going to be multiplied by the weight. (for the equation we will use X for the distance).

equation: 4 x1 20 = ? x 80

Now step one: 4(120) = X(80)  

Or another way is 480 = 80X  

480/80 = X  

48/8= X  

X = 6

I hope this could help! Sorry if it didn't make much sense otherwise!


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Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23
Sloan [31]

Answer:

The ratio is   \frac{T_1}{T_2}  = 3.965

Explanation:

From the question we are told that

   The  radius of Phobos orbit is  R_2 =  9380 km

    The radius  of Deimos orbit is  R_1  =  23500 \  km

Generally from Kepler's third law

    T^2 =  \frac{ 4 *  \pi^2 *  R^3}{G * M  }

Here M is the mass of Mars which is constant

        G is the gravitational  constant

So we see that \frac{ 4 *  \pi^2  }{G * M  } =  constant

   

    T^2 = R^3   *  constant      

=>  [\frac{T_1}{T_2} ]^2 =  [\frac{R_1}{R_2} ]^3

Here T_1 is the period of Deimos

and  T_1 is the period of  Phobos

So

      [\frac{T_1}{T_2} ] =  [\frac{R_1}{R_2} ]^{\frac{3}{2}}

=>    \frac{T_1}{T_2}  =  [\frac{23500 }{9380} ]^{\frac{3}{2}}]

=>    \frac{T_1}{T_2}  = 3.965

   

8 0
2 years ago
You apply the brakes of your car abruptly and your book starts sliding off the front seat. Three observers sitting in the car ex
krek1111 [17]

Answer:

All the observers are correct.

Explanation:

This is simply a problem of reference frames from which the motion of the book is being viewed by the various observers.

From their various reference frames, they are all correct.

Observer A must be in the inertial reference frame.

<em>Observers who can explain the behavior of the book  and the car by using the relationship between the sum of  the forces and changing velocity are said to be observers in inertial reference frames.</em>

This is clearly shown by what observer A noticed. There was a relative motion between the book and the car as she pointed out, making her to be in an inertial reference frame.

<em>Similarly, observers in inertial reference frames can also explain the changes in velocity of objects  by considering the forces exerted on them by other objects.</em>

This is shown by observer B as he is able to notice how the force of the car affects the velocity of the book.

Observer C is actually in a non-inertial reference frame, as newtons law of force motion relationship are no longer observed. This occurs in the non inertial reference frame.

7 0
2 years ago
Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocit
ser-zykov [4K]
The reaction time of Boris is t(r), so before that, Boris will not have jumped. Thus, H(b)(t) = 0
The vertical displacement will simply be
D(t) = H(a)(t)
7 0
2 years ago
a pencil is pushed horizontally off a desk with a speed of 1.2m/s if it takes 0.4 seconds for the pencil to reach the ground how
kari74 [83]
The pencil has a horizontal velocity of 1.2 m/s.  To see how far it travels in the horizontal direction over the 0.4 seconds it is falling, use the relation x=vt, where x is the distance, v is the x-directed velocity, and t is time.
x=1.2m/s*0.4s = 0.48m 
3 0
2 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

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

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

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

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
2 years ago
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