Answer:
See the answers below
Explanation:
We can solve both problems using vector sum.
a)
Let's assume the forces that help the diver dive as positive downward, and the forces that oppose upward, as negative
![F_{resultant}=100+30-85+900\\F_{resultant}=845[N]](https://tex.z-dn.net/?f=F_%7Bresultant%7D%3D100%2B30-85%2B900%5C%5CF_%7Bresultant%7D%3D845%5BN%5D)
The drag force is horizontal d this way in the horizontal direction we will only have the drag force that produces the water stream.
![F_{drag}=50[N]](https://tex.z-dn.net/?f=F_%7Bdrag%7D%3D50%5BN%5D)
b)
Let's assume the forces that propel the rocket upwards as positive and forces like the weight of the rocket and other elements as negative forces.
![F_{resultant}=960+7080-7700\\F_{resultant}=340 [kN]](https://tex.z-dn.net/?f=F_%7Bresultant%7D%3D960%2B7080-7700%5C%5CF_%7Bresultant%7D%3D340%20%5BkN%5D)
when a field is changing in time
Answer:
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Explanation:
What do you mean??
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Answer:
An electric generator creates a time-dependent induced voltage.
An electric generator converts mechanical energy into electrical energy.
Explanation:
An electric generator works on the principle of electromagnetic induction. This phenomenon hinges on the fact that moving a conducting wire through a magnetic field produces electricity. Electricity is generated as long as there is relative motion between the conductor and the magnetic field. The emf generated is known as induced emf or induced voltage.
The electric generator converts mechanical energy to electrical energy.
Answer:
E = 1,873 10³ N / C
Explanation:
For this exercise we can use Gauss's law
Ф = E. dA =
/ ε₀
Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.
The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.
The surface of a sphere is
A = 4π r²
E 4π r² = q_{int} /ε₀
The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is
q_{int} = q₁ + q₂
q_{int} = (530 - 200) 10⁻⁹
q_{int} = 330 10⁻⁹ C
The electric field is
E = 1 / 4πε₀ q_{int} / r²
k = 1 / 4πε₀
E = k q_{int}/ r²
Let's calculate
E = 8.99 10⁹ 330 10⁻⁹/ 1.32²
E = 1,873 10³ N / C