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tia_tia [17]
2 years ago
15

What is an object's mass if it accelerates at 5 m/s2 when a force of 0.5 N is applied?

Physics
1 answer:
LenaWriter [7]2 years ago
6 0

Answer:

0.1kg

Explanation:

mass = force / acceleration

0.5 / 5

0.1kg

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to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on
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Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

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If F1 = 104 and F2 = 104, what will be the net force?
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Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total di
S_A_V [24]

Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.

Now the spring is in series combination

F\alphae

F=ke

k=f/e.........*

where k is the force constant or the constant of proportionality

k=f/e

f_{eff} =f_{1} +f_{2}............................1

also for effective force constant

divide all through by extension

1) Total force is

Ft=F1+F2

Ft=k1e1+k2e2

F = k(e1+e2) 2)

Since force on the 2 springs is the same, so

k1e1=k2e2

e1=F/k1 and e2=F/k2,

and e1+e2=F/keq

Substituting e1 and e2, you get

1/keq=1/k1+1/k2

Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.

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