I have 6 questions from newtons first law, Second law and third law topic.

An object, with mass 72 kg and speed 28 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the oth

IRINA_888 [86]

Answer:

14112 J

Explanation:

When the 72 Kg mass explodes into two, one mass is twice the other so 72/3=24 Kg

M1= 24 kg, M2= 72-24=48 kg

From law of conservation of linear momentum, the sum of initial and final momentum are equal. p=mv where p is momentum, m is mass and v is velocity. Fir this case, since the less massive piece stops, its final velocity is zero.

72*28=48v2

V2=72*28/48=42 m/s

Difference between initial and final kinetic energy will be

$\triangle KE= 0.5(Mv^{2}-m2v2^{2})\\\triangle KE= 0.5(72*28^{2}-48*42^{2})=-14112 J$

Therefore, from observers reference, kinetic energy of 14112 J is added

5 0

4 years ago

Using the force table, components of a vector can be found experimentally by suspending masses from 2 orthogonal strings which o

SCORPION-xisa [38]

Answer: 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).

Explanation:

1) Since the mass of 155 g is suspended at 210 degrees, you need to find the components of its weight on the orthogonal coordinate system (0° and 90°).


2) You do that using the trignometric ratios sine and cosine.


Weight is mass × g.

Weight of the object = 155g × g

Angle, α = 210°


Horizontal component (0°)

cosα = horizontal / hypotenuse ⇒ horizontal = hypotenuse × cosα

⇒ horizontal = 155g × g × cos(210°) = - 134.23g × g

Vertical component
sinα = vertical / hypotenuse ⇒ vertical = hypotenuse × sinα
⇒ vertical = 155g × g × sin(210°) = -77.5g × g

3) Conclusion:

Therefore, the masses that must be suspended to balance the forces of the 155g mass are 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).

8 0

3 years ago

Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add

jok3333 [9.3K]

Answer:

When x = 2.8 cm, $B_{x1} = 0.0265 T$