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Afina-wow [57]
2 years ago
14

Review. A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250-kg block on the left and a 0

.500-kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is(b) 0.100
Physics
1 answer:
photoshop1234 [79]2 years ago
7 0

The answer is 0.245N.

<h3>What is kinetic energy?</h3>
  • A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
  • Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
  • The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
  • Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

(b) 0.100

For the block on the left, f_{k} =u_{k} n= 0.100(2.45N)=0.245N.

∑F_{x}=ma_{x}

–0.308N+0.245N=(0.250kg)a

a=−0.252m/s^{2} if the force of static friction is not too large.

For the block on the right, f_{k} =u_{k} n=0.490N. The maximum force of static friction would be larger, so no motion would begin, and the acceleration is zero

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

#SPJ4

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The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi
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Answer:

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dielectric constant, K = 7

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The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

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Take the ratio of the two forces;

\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N

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5 0
3 years ago
On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

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4 years ago
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