Answer:
yes u can flag football has everything that pad football has so you can enlist on being offensive position but you have to play like you want that position
Explanation:
Answer:
E = 307667 N/C
Explanation:
Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.
This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:
0.0098 N = T * cos(37)
then T = 0.0098/cos(37) N = 0.01227 N
Knowing the tension's magnitude, we can find its horizontal component:
T * sin(37) = 0.007384 N
and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:
0.007384 N = E * 2.4 * 10^(-8) C
Then E = 0.007384/2.4 * 10^(-8) N/C
E = 307667 N/C
If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.
<h3>
What are planets?</h3>
Planets are the large spherical shaped objects that rotate about the Sun in the elliptical orbits.
Planets are shaped from Planetary cloud. The dust storm and gases gathers under its own weight. The dense matter beginnings pivoting at high paces and accumulates more mass. The center structures, the star and rest of it ultimately levels into a curved plate from which planet is formed.
Thus, if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.
Learn more about planets.
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A theory can help create a model
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge



The charge on the inner surface is q.


We need to calculate the electric field outside the shell
Using formula of electric field

Put the value into the formula



Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 