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anzhelika [568]
3 years ago
12

Using the principle of the conservation of mechanical energy, show that the acceleration a of a freely falling body has the valu

e g.​
Physics
1 answer:
andrew-mc [135]3 years ago
8 0

Answer:

ytuiugfugygoxjguohfukfjhkcjhkxjghkxhgifju

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How does the number of neutrinos passing through your body at night compare with the number passing through your body during the
lapo4ka [179]

Answer:

c)While the amount of neutrinos passing through the Earth does not change much, there is something that changes to you at night. You stop moving and sleep for several hours.  

Think of Neutrinos like rain. During the day you can avoid rain, and run from one dry spot to another(like from a car to a house). At night, however, your asleep, so the rain will hit you for eight hours(if your sleeping outside with no tent for some reason).

5 0
3 years ago
A 4,000 kg boat floats with one-third of its volume submerged. if two more people get into the boat, each of whom weighs 690 n,
AveGali [126]

 

The weightiness of the added water displaced is equivalent to the joined weight of the two extra people who come to be into the boat:


<span>m water g                   = 2 x 690 N</span>

<span>                                   = 1,380 N</span>

<span>
</span>

The mass of the water displace is then


<span>m water g                   = 1,380 N</span>

<span>                                   = 1,380 N / 9.8 m/s^2</span>

<span>                                   = 141 kg</span>

<span>
</span>

Compute the calculation for density for the volume of water displace and practice this outcome for the mass of the water displace to get the answer:


<span>p water                      = mass of water / volume of water</span>

<span>
</span>

<span>volume of water        = mass of water / p water</span>

<span>                                  = 141 kg / 1000 kg /m^3 eliminate kilogram</span>

<span>                                  = 0.14 m^3 the additional volume of water that is displaced</span>

4 0
4 years ago
If the radius of an object in circular motion is doubled, what change will occur in the centripetal force?
cestrela7 [59]

Answer:

The centripetal force will be 1/2 as big as it was. (option c)

Explanation:

Recall that centripetal force (F_c) is defined as: F_c=m\,* \frac{v^2}{r} where "v" is the tangential velocity of the object in circular motion, "r" is the radius of rotation and "m" is the object's mass.

So if we start with such formula with a given mass, radius, and tangential velocity, and then we move to a situation where everything stays the same except for the radius which doubles, then the new centripetal force (F'_c) will be given by: F'_c=m\,* \frac{v^2}{2r}

and this is half (1/2) of the original force:

F'_c=m\,* \frac{v^2}{2r}\\F'_c=m\,* \frac{v^2}{r}*\frac{1}{2} \\F'_c=F_c\,*\,\frac{1}{2}

which is expressed by option "c" of the provided list.

8 0
3 years ago
Read 2 more answers
A square sheet of rubber has sided that are 20 cm long. If you stretch one side so that it is twice as long. What is the new are
Anika [276]

Answer:ignore this it was wrong

Explanation:

3 0
4 years ago
A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra
sergey [27]

Answer:

\frac{di}{dt}  = 7.31 \  A/s

Explanation:

From the question we are told that  

     The  number of turns is  N =  450 \  turns

      The  radius is  r =  1.17 \ cm =  0.0117 \ m

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  e  =  8.20 *10^{-6} \  V/m

Generally according to Gauss law

        \int\limits { e } \, dl  =  \mu_o *  N  *  \frac{di}{dt }  *  A

=>    e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  A

Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                A =  \pi r ^2

=>      e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  \pi r^2

=>       \frac{di}{dt}  =  \frac{2e * x  }{\mu_o * N  * r^2}ggl;

Here  \mu_o is the permeability of free space with value

          \mu_o  =  4\pi * 10^{-7} \  N/A^2

=>     \frac{di}{dt}  =  \frac{2 *  8.20*10^{-6} *  0.0345  }{ 4\pi * 10^{-7} * 450  * (0.0117)^2}

=>      \frac{di}{dt}  = 7.31 \  A/s

6 0
4 years ago
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