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bearhunter [10]
3 years ago
11

an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2

Physics
2 answers:
Degger [83]3 years ago
7 0
60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N
Ainat [17]3 years ago
6 0

Answer:

The weight on planet x is 60 N

Explanation:

It is given that,

Weight of an object on earth, W = 98 N

Value of acceleration due to gravity, g=9.8\ m/s^2

Weight, W=mg

m=\dfrac{W}{g}=\dfrac{98\ N}{9.8\ m/s^2}=10\ kg

Since, mass of an object remains same everywhere.

If the value of acceleration due to gravity on planet x is, a=6\ m/s^2

So, weight on the planet x is,

W=m\times a

W=10\ kg\times 6\ m/s^2

W = 60 N

So, the weight on planet x is 60 N. Hence, this is the required solution.

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the earth has a radius of 6.38×10^16 meter and turns around once on its axis in 24 hour.what is the radial acceleration of perso
Scrat [10]

337493603.8m/s²

Explanation:

Radius of the earth = 6.38 x 10¹⁶m

time = 24hr (86400s)

Unknown:

Centripetal acceleration = ?

Solution:

The centripetal acceleration is directed inward to keep the body from falling off the surface of the earth.

     centripetal acceleration  = \frac{v^{2} }{r}

  where v is the velocity and r is the radius

   also;

            v  = wr

  where w is the angular velocity

substituting in the equation for centripetal acceleration gives;

                 

          a = w²r

 also w = \frac{2 x pi}{T}

      therefore;

                 a = \frac{4 \pi  ^{2} r }{T^{2} }

 a = \frac{ 4 x 3.142^{2}  x 6.38 x 10^{16} }{86400^{2} }

 a = 337493603.8m/s²

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

5 0
3 years ago
A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin
blsea [12.9K]

Answer:

The stress is calculated as 1.003\times 10^{9}\ Pa

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = 0.560\times 10^{- 3}\ m

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, \Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m

Now,

The stress in the wire is given by:

Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = 25.2\times 9.8 = 246.96\ N

Cross  sectional Area, A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}

Using these values in eqn (1):

\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa  

8 0
2 years ago
Explain why the elements of the same group exhibit the same chemical behavior​
ElenaW [278]

Answer:

All the elements of a period have similar chemical properties because they have the same number of valence electrons in their outermost shell. Their atoms have the same number of electrons in the highest occupied energy level

4 0
3 years ago
US
erastovalidia [21]

Answer:11.5m

Explanation:

6 0
2 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
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