The magnitude of their resultant vector is 4.6 meters/seconds
Since we are to add the velocity vectors in order to find the magnitude of their resultant vector.
Hence:
Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)
Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds
Resultant vector magnitude 4.6 meters/seconds
Inconclusion The magnitude of their resultant vector is 4.6 meters/seconds
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The velocity of the body is zero; option A
<h3>What is the motion of an oscillating body?</h3>
The motion of an oscillating body is known as simple harmonic motion.
Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.
For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.
In conclusion, oscillating bodies undergo simple harmonic motion.
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Your answer would be A. You divide 96 by 16 to find the answer
The velocity of the submarine immediately after firing the missile is 0.0104 m/s
Explanation:
Mass of the submarine M=50 tonne=
Mass of the missile m=40 kg
velocity of the missile v= 13m/s
we have to calculate the velocity of the submarine after firing
This is the recoil velocity and its expression is derived from the law of conservation of momentum
recoil velocity of the submarine

Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.