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Neporo4naja [7]
3 years ago
12

if 1.5kg block is pulled across a horizontal surface that has a coefficient of kintic friction of 0.60. if a block is pulled by

a 12n force, what is the net force acting on the block
Physics
1 answer:
inysia [295]3 years ago
4 0

Answer:

3.2N

Explanation:

Given parameters:

Mass of block  = 1.5kg

Coefficient of kinetic friction  = 0.6

Force of pull on block  = 12N

Unknown:

Net force on the block  = ?

Solution:

Frictional force is a force that opposes motion:

  Net force  = Force of pull  - Frictional force

 Frictional force  = umg

   u is coefficient of kinetic friction

   m is the mass

    g is the acceleration due to gravity

 Frictional force  = 0.6 x 1.5 x 9.8  = 8.8N

 Net force  = 12N  - 8.8N  = 3.2N

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Ideally, the resistance of an ammeter should be:
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Ideally the resistance should be ZERO
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According to Newton's 2nd Law of Motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is
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According to Newton's second Law of motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is 1m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a moving body can be calculated by dividing the force of the body by its mass.

According to this question, the mass of an object is 10 kg and the force is 10 newtons, then the acceleration can be calculated as follows:

acceleration = 10N ÷ 10kg

acceleration = 1m/s²

Therefore, according to Newton's second Law of motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is 1m/s².

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8 0
2 years ago
A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch
Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

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3 years ago
Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne
VLD [36.1K]

Answer:

The the maximum emf is 2.52\times10^{-11}\ V

Explanation:

Given that,

Magnetic field B = 1.40\times10^{-3}\ T

Frequency = 60 Hz

Diameter = 7.8 μm

We need to calculate the maximum emf

Using formula of emf

\epsilon=NBA\omega

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B= magnetic field

A = area

Put the value in to the formula

\epsilon=1\times1.40\times10^{-3}\times\pi\times(\dfrac{7.8\times10^{-6}}{2})^2\times2\times\pi\times60

\epsilon=2.52\times10^{-11}\ V

Hence, The the maximum emf is 2.52\times10^{-11}\ V

5 0
3 years ago
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