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bearhunter [10]
3 years ago
14

How are contact forces different from non-contact forces?

Physics
2 answers:
madam [21]3 years ago
8 0
A non-contact force is a force which acts on an object without coming physically in contact with it. In contrast a contact force is a force which acts on an object coming physically in contact with it. ...
Sedaia [141]3 years ago
8 0
Contact force needs to be in contact to make “contact force”. Its the opposite for non-contact force, it has to have NO contact to make “non-contact force” happen, i hope this helps
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When you must add three vectors together, what is not true this process? You must only give a magnitude of the resultant vector
vovangra [49]

The addition of any numbers of vector provide the magnitude as well as the direction of the resultant vector, hence the mentioned first option is not true.

The addition of vector required to connect the head of the one vector with the tail of the other vector and any vector can be moved in the plane parallet to the previous location, so, the mentioned second and third options are true.

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What will happen to force?
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Answer:

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8 0
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A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest
34kurt

Displacement = (straight-line distance between the start point and end point) .

Since the road east is perpendicular to the road north,
the car drove two legs of a right triangle, and the magnitude
of its final displacement is the hypotenuse of the triangle.

    Length of the hypotenuse = √ (215² + 45²)

                                              =  √ (46,225 + 2,025)

                                              =   √ 48,250

                                              =       219.7 miles .

8 0
3 years ago
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What two measuring units can i use to measure the distance
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8 0
3 years ago
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

7 0
3 years ago
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