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adelina 88 [10]
3 years ago
6

4. There is a car with a large mass and a car with a small mass. If you push both cars with the

Physics
1 answer:
svlad2 [7]3 years ago
3 0
I think the small mass Sorry if I’m incorrect
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A proton is released such that its initical velocity is from right to left across this page. THe proton's path, however, is defl
eduard

Answer:

the magnetic field is leaving the sheet

Explanation:

The magnetic force is given by the expression

        F = q v x B

where bold letters indicate vectors, the modulus of this expression is

       F = q v B sin θ

the direction of the force is given by the right hand rule, for a positive charge

the thumb indicates the direction of the speed, in this case from right to left

the palm the direction of the force, in our case upwards

the fingers extended the direction of the magnetic field, this case after fixing the other two components it points out of the blade

In short the magnetic field is leaving the sheet

7 0
2 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
What is the value of the activation energy of the uncatalyzed reaction in reverse?
alexgriva [62]
It would be: Activation Energy = 300 KJ

Hope this helps!
5 0
3 years ago
If you go twice as fast, your kinetic energy becomes.
xenn [34]

Answer:

2 times bigger

Explanation:

hope this helps if so, please vote and mark as Brainlist! thanks and good luck!

5 0
3 years ago
Read 2 more answers
PLZ HELP!!!!
zavuch27 [327]

Answer:

AAAAAA!!!!!!!

Explanation:

Hope that helped lol!

8 0
2 years ago
Read 2 more answers
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