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3 years ago

When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the

Physics

1 answer:

Aleksandr [31]3 years ago

7 0

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

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Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to report the mass truthfully). (a) W

bija089 [108]

Answer:

1411.8 N/m

Explanation:

From Hooke's law;

F= Ke

Where

F= force on the spring

K= force constant

e = extension

But e= 8.50 × 10^-2m

F= weight = 12.0 kg × 10 = 120 N

K = F/e = 120/8.50 × 10^-2

K= 1411.8 N/m

7 0

3 years ago

How are land-use laws in the United States different from the land-use laws inherited from England?

exis [7]

private property owners in the United States can restrict public access to their land

5 0

3 years ago

Three 3-ohm resistors are connected in parallel. The total resistance for the resistors is

Anna11 [10]

Explanation:

I think the total resistance of the resistors is 1 ohm . it is also known as equivalent resistance.

hope it helps.

3 0

3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

Which of the following statements is true?

likoan [24]

Answer:

d.Energy as heat transferred into an object is determined by the amount of work done on the object.

Explanation:

4 0

3 years ago

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