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balandron [24]
3 years ago
11

In 3-5 sentences, explain the difference between mass and weight and summarize how mass and weight are related to gravitational

force.
Use the "RAP" method to answer this short-answer question:

Restate the question.
Answer the question.
Prove your answer citing textual evidence from the course.
Don't forget to use complete sentences and proofread your answer.
Physics
2 answers:
earnstyle [38]3 years ago
5 0
Bruh..........................
Damm [24]3 years ago
5 0
I’m so sorry.... BUE
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A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the
Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁  = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

    m₁v₁ + m₂v₂  = v(m₁ + m₂)

Insert parameters and solve for v;

   5 x 1  + 15 x 0 = v (5 + 15 )

          5  = 20v

          v = \frac{5}{20}   = 0.25m/s

5 0
3 years ago
You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.
Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

F_{a} =\frac{F_{f}+F_{i}  }{2}  Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

W = F_{a} (x_{f} -x_{i} )   :Formula (3)

x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

:

4 0
3 years ago
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