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dimaraw [331]
1 year ago
13

Does gravitational potential energy increase when getting closer to planet universal gravitation law.

Physics
1 answer:
LuckyWell [14K]1 year ago
4 0

Answer:

Yes

Explanation:

When the planet moves farther away, the speed and kinetic energy decrease, and the gravitational potential energy increases.

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A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
What is the definition of work when net force is parallel to the distance?
goldfiish [28.3K]
Work in general is given by W=F·d where F is the force vector and d is the displacement vector.  The dot symbol is the dot product which is a measure of how parallel two vectors are.  It can be replaced by the cosine of the angle between the two vectors and the vectors replaced by their magnitudes.  If F and d are parallel then the angle is zero and the cosine is unity.  So in this case work can be defined as the product of the magnitudes of the force and distance:
W=Fd
6 0
2 years ago
An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car
Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)=3.2 m/s^2

Speed of car after 4.5 s

using equation of motion

v=u+at

v=4.92+3.2\times 4.5=4.92+14.4

v=19.32 m/s

Displacement of the car after 4.5 s

v^2-u^2=2as

19.32^2-4.92^2=2\times 3.2\times s

349.05=2\times 3.2\times s

s=54.54 m

4 0
2 years ago
Forces that act in equal and opposite directions on an object
Akimi4 [234]
These are known as balanced forces because they will not change the motion of the object, and it will remain at rest unless forces become unbalanced- meaning they would be unequal and not opposing. 
5 0
2 years ago
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
Zinaida [17]

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = 12.0v+4.00v^2

Equating both, we have

    784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

 So terminal velocity of object = 12.58 m/s    

3 0
3 years ago
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