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Amiraneli [1.4K]
3 years ago
11

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

e 0.8 and 0.4, respectively. A horizontal string is attached to the block and a constant tension is maintained in the string. What is the force of friction acting on the block if the tension is 24 N? Answer in units of N.
Physics
1 answer:
Ne4ueva [31]3 years ago
6 0

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

f=\mu mg

Put the value into the formula

f=0.8\times37

f=29.6\ N

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

f = \mu N

Put the value in to the formula

f=0.4\times37

f=14.8\ N

Hence, The frictional force acting on the block is 14.8 N.

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3 years ago
7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How
Lera25 [3.4K]

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

Y = \frac{FL}{Ae}

<u></u>

<u>Part A</u>

cross sectional area and applied force are the same as the original but the length is doubled

\frac{FL_1}{A_1e_1} =  \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} =  \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o

The second wire will be stretched by 2 mm

<u>Part B</u>

a third wire with the same length but twice the diameter of the first

\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o

e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm

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4 years ago
PLEASE HELP!!!!!!! MY HOMEWORK IS DUE TODAY !!!!!!!!
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Answer:

Distance = 25000000 miles

Time = 50 hours

Explanation:

Venus is the closest planet to Earth. It is about 25 million miles away from Earth. Its precise distance depends on where both Venus and Earth are in their respective orbits

Given that

Speed V = 500000 mph

Distance d = 25 000,000 miles

Speed = distance/ time

Time = distance/speed

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3 years ago
A 26.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m th
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Answer:

18 m/s

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Force is given by mass × acceleration

<em>F = ma</em>

<em>a = F/m</em>

For the first 11.0 m, there is no friction. Hence, the constant force is applied fully on the crate. The acceleration is

<em>a = </em>225/26.0<em> </em>m/s²

By the equation of motion, <em>v² = u² + 2as</em>

where <em>v</em> is the final velocity, <em>u</em> is the initial velocity, <em>a</em> is the acceleration and <em>s</em> is the distance travelled.

Using the values for the first part of the motion,

<em>v² = </em>0² + 2 × 8.65 × 11.0

<em>v = </em>13.8 m/s

This is the initial velocity for the second part of the motion. This part has friction of coefficient 0.20.

The frictional force = 0.20 × weight = 0.20 × 26.0 × 9.8 = 50.96 N

The effective force moving the crate in the second motion = 225 - 50.96 N = 174.04 N

This gives an acceleration of 174.04/26.0 = 6.69 m/s².

Using parameters for the equation of motion, <em>v² = u² + 2as</em>

<em>v² = </em>13.8² + 2 × 6.69 × 10 = 324.24

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The mechanical energy of a 2kg body is 35J and its potential energy is 10J calculate speed energy​
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Answer:

20

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hope i hv answered ur question

5 0
3 years ago
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