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Elis [28]
3 years ago
7

What is the best way to ensure that the results of your experiment are accurate?

Physics
1 answer:
Ainat [17]3 years ago
6 0
Re do the experiment 3 times over again and if you get the same results then it should be correct
You might be interested in
Which kind of light is used to carry information through optical fibers?
ZanzabumX [31]
Red light would be commonly used due to the longer wavelength
3 0
3 years ago
How much force is required to move a electron through an electric field with strength of 1.375 x 10^19 N/C?
Luda [366]

Answer:

The magnitude of the force required to move the electron through the given field is 2.203 N

Explanation:

Given;

The field strength of the electron, E = 1.375  x 10¹⁹ N/C

charge of electron, q = 1.602 x 10⁻¹⁹ C

The magnitude of the force required to move the electron through the given field is calculated as follows;

F = Eq

F = (1.375  x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)

F = 2.203 N

Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N

4 0
3 years ago
What is the atmospheric pressure and temperature at sea level in a standard<br> atmosphere?
Lady bird [3.3K]

Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

Hope this helped, Have a wonderful day!!

4 0
2 years ago
A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/
Artyom0805 [142]

Answer:

I=182.97\ kg-m^2

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, \tau=860\ N

Angular acceleration of the shell, \alpha =4.7\ m/s^2

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

\tau=I\alpha

I is the rotational inertia of the shell

I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2

So, the rotational inertia of the shell is 182.97\ kg-m^2.

7 0
3 years ago
A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.
kodGreya [7K]

Answer:

The  z-component of the force is  \= F_z  =  0.00141 \ N    

Explanation:

From the question we are told that

          The charge on the particle is q =  7.76 *0^{-8} \  C    

           The magnitude of the magnetic field is  B =  0.700\r i \ T

            The  velocity of the particle toward the x-direction is  v_x  =  -1.68*10^{4}\r  i  \ m/s

           The  velocity of the particle toward the y-direction is

v_y  =  -2.61*10^{4}\ \r j  \ m/s

           The  velocity of the particle toward the z-direction is

v_y  =  -5.85*10^{4}\ \r k  \ m/s

Generally the force on this particle is mathematically represented as

          \= F  =  q (\= v   X  \= B )

So  we have    

          \= F  =  q ( v_x \r  i + v_y \r  j  +  v_z \r k  )  \ \ X \ (  \= B i)

         \= F  = q (v_y B(-\r  k) + v_z B\r j)      

  substituting values

       \= F  = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r  z) + [(5.58*10^{4}) (0.700)]\r y)    

      \= F=  0.00303\ \r j +0.00141\ \r k                  

So the z-component of the force is  \= F_z  =  0.00141 \ N    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).

7 0
3 years ago
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