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coldgirl [10]
3 years ago
13

According to the Law of the Conservation of Matter, if you dissolve 25 grams of sugar into 150 grams of water, the mixture shoul

d weigh how much?
Physics
1 answer:
levacccp [35]3 years ago
7 0
19grams becauce if u
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nirvana33 [79]

Kepler's first law - sometimes referred to as the law of ellipses - explains that planets are orbiting the sun in a path described as an ellipse. ... The resulting shape will be an ellipse. An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points is a constan

4 0
2 years ago
1. What is the kinetic energy of a frog that has mass of 13kg and can hop<br> 9m/s? *
emmasim [6.3K]

Answer:526.6J

Explanation:I did a test

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3 years ago
What is the orbital velocity in km/s and period in hours of a ring particle at the outer edge of Saturn's A ring
mote1985 [20]

Answer:

The orbital velocity v  =  16.4 \ km/s

The period is  T =  14.8 \ hours

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the planet , this is mathematically represented as

       \frac{GM_s  *  m }{r^2 }  = m w^2 r

=>    w = \sqrt{ \frac{GM}{r^3} }

Here G is the gravitational constant with value  G = 6.67*10^{-11}

        M_s  is the mass of with value  M_s  =5.683*10^{26} \  kg

        r is the is distance from the center of the  to the  outer edge of the  A ring

i.e r = R  + D  

Here R  is the radius of the planet   with value  R  = 60300 \ km

         D  is the distance from the  equator to the outer edge of the  A ring with value  D = 80000 \  kg

So  

       r =80000 + 60300

=>    r =140300 \ km  = 1.4*10^{8} \  m

So

    =>    w = \sqrt{ \frac{ 6.67*10^{-11}*  5.683*10^{26}}{[1.4*10^{8}]^3} }

    =>    w =  1.175*10^{-4} \ rad/s

Generally the orbital velocity is mathematically represented as

       v  = w * r

=>     v  = 1.175*10^{-4}   * 1.4*10^{8}

=>     v  = 1.64*10^{4} \  m /s =  16.4 \ km/s

Generally the period is mathematically represented as

     T   =  \frac{2 \pi }{w }

=> T   =  \frac{2 *  3.142  }{ 1.175 *10^{-4} }

=> T   = 53473 \ second = 14.8 \ hours

Answer:

The orbital velocity v  =  16.4 \ km/s

The period is  T =  14.8 \ hours

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the , this is mathematically represented as

       \frac{GM_s  *  m }{r^2 }  = m w^2 r

3 0
3 years ago
The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,
omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

g = \frac{GM}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius

The values given are based on the constants of the earth, so they can be expressed as

M_p = \frac{1}{100} M_e

r_p = \frac{1}{4} r_e

The relationship of gravity would then be given:

g_e = \frac{GM_e}{r_e^2}

The relationship with the new planet, from the gravity of the earth would be given

g_p = \frac{GM_p}{r_p^2}

g_p = \frac{G(1/100)M_e}{(1/4 r_e)^2}

g_p = \frac{GM_e 16}{100 r_e^2}

g_p = 0.16 \frac{GM_e}{r_e^2}

g_p = 0.16g_e

The relationship with the weight of the earth would be given as:

W_e = m*g_e = 600N

W_p = m*g_p = m(0.16g_p)

W_p = (m*g_p)(0.16)

W_p = 600*0.16

W_p = 96N

Therefore the weigh on this planet would be 96N

3 0
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Describe the velocity of three objects you encounter during the day. If you know the numerical value of the velocity, write it d
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We use these objects during day will be actually hands nose eyes
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