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dedylja [7]
3 years ago
11

1. It s defficult to lift a heavier stone than the lighter one. why​

Physics
1 answer:
stiks02 [169]3 years ago
4 0

Answer:

lThe effect of the attraction of the earth on a bigger stone can be observed more than the effect of attraction of the earth on a smaller one. hence it is difficult to lift a large stone than the smaller one on the earth surface.

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stiks02 [169]
Neap tide is your answer
7 0
3 years ago
Some people say you can tell if someone has HIV by looking at him or her. True or false? Why?
Alex17521 [72]

Answer:False.

Explanation: HIV is a viruse that only shows symptoms after a few weeks. in Fact HIV does not show any physical symptoms. So, no you can not tell if someone has or is been infected by HIV by directly looking at them.

Have a Nice Day!!! =)

5 0
3 years ago
Read 2 more answers
An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect
Dvinal [7]

Answer:

The distance is d =1.66*10^{-9}m

Explanation:

From the question we are told that

         The initial speed of the  electron is v_i  = 3.2 *10^5 m/s

         The mass of electron is m = 9.11*10^{-31}kg

         Let d be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

         Let KE_i be the initial kinetic energy of the electron \

          Let KE_d be the kinetic energy of the electron at the distance d from the proton

  Considering that energy is conserved,

  The energy at the initial position of the electron = The energy at the final position of the electron

      i.e

             KE_i +U_1 = KE_d + U_2

U_1 \ and \ U_2 are the potential energy at the initial  position of the electron and at distance d of the electron to the proton

                Here U_1 = 0

So the equation becomes

                   \frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2  + \frac{kq_1 q_2}{d}

Here q_1 \ and  \ q_2 are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

 k is electrostatic constant with value 8.99*10^9 N \cdot m^2 /C^2

i.e q = 1.602 *10^{-19}C

           v_d is the velocity at distance d from the proton = 2v_i

  So the equation becomes

             \frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}

            \frac{1}{2} mv_i^2  = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}

           3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}

Making d the subject of the formula

           d = \frac{2k(q)^2}{3mv_i^2}

              = \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}

              =1.66*10^{-9}m

             

           

         

                 

   

6 0
3 years ago
How does velocity describe an objects motion?
rjkz [21]

Answer:

Velocity is how fast and in what direction it moves.

Explanation:

5 0
3 years ago
Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5​
juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

T=2\pi \sqrt{\dfrac{l}{g}}

or

T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}

Where

l is length of the pendulum

l\propto T^2

or

\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2 ....(1)

ATQ,

\dfrac{T_1}{T_2}=1.5

Put in equation (1)

\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}

So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0
3 years ago
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