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Evgen [1.6K]
3 years ago
8

You happen to know that the coefficient of static friction between your patio table and the ground is 0.42. You decide you want

to move it (45kg) by pushing on it. Are you able to move the table?
Physics
1 answer:
Deffense [45]3 years ago
4 0

Answer:

If the force applied is larger than 185.2 N, yes.

Explanation:

In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:

F_f = \mu mg

where

\mu=0.42 is the coefficient of static friction

m=45 kg is the mass of the table

g=9.8 m/s^2 is the gravitational acceleration

Substituting,

F_f=(0.42)(45 kg)(9.8 m/s^2)=185.2 N

So, we are able to move the table if we push with a force larger than 185.2 N.

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A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

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Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

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m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

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3 0
3 years ago
An 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55
Sindrei [870]

Answer:

a)Yes will deform plastically

b) Will NOT experience necking

Explanation:

Given:

- Applied Force F = 850 lb

- Diameter of wire D = 0.15 in

- Yield Strength Y=45,000 psi

- Ultimate Tensile strength U = 55,000 psi

Find:

a) Whether there will be plastic deformation

b) Whether there will be necking.

Solution:

Assuming a constant Force F, the stress in the wire will be:

                       stress = F / Area

                       Area = pi*D^2 / 4

                       Area = pi*0.15^2 / 4 = 0.0176715 in^2

                       stress = 850 / 0.0176715

                       stress = 48,100.16 psi

      Yield Strength < Applied stress > Ultimate Tensile strength

                        45,000 < 48,100 < 55,000

Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.

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Answer:

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Explanation:

F = G . m(goku) . m(planet) / d²

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Answer:

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5 0
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