Using the circuit diagram <br>find the emf

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

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$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

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Definition of Newton’s first law of motion

kipiarov [429]

Newton's first law of motion - sometimes referred to as the law of inertia. Newton's first law of motion is often stated as. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

3 0

3 years ago

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A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

nlexa [21]

Answer:

$a_c=1.44\ m/s^2$

Explanation:

<u>Centripetal Acceleration</u>

It's the acceleration that an object has when traveling on a circular path to take into consideration the constant change of velocity it must have in order to keep going in the circular path.

Being v the tangent speed, and r the radius of curvature of the circle, then the centripetal acceleration is given by

$\displaystyle a_c=\frac{v^2}{r}$

We can compute the value of v by using the distance and the time taken to travel:

$\displaystyle v=\frac{x}{t}=\frac{200\ m}{26.4\ s}$

$v=7.58\ m/s$

Now we calculate the centripetal acceleration

$\displaystyle a_c=\frac{7.58^2}{40}=1.44\ m/s^2$

$\boxed{a_c=1.44\ m/s^2}$

4 0

3 years ago

True or false? Internal motions within the object are used in creating a particle model

nikitadnepr [17]

The answer is true because it is true

5 0

3 years ago

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A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

aleksandrvk [35]

Answer:

a) in order to be maximum the length of the square wire should be 10m and no cut for making the triangle

b) in order to be minimum the length of the square wire should be 3.84 m and the triangle wire should be 6.16 m

Explanation:

for a square with a side length of a and a equilateral triangle of length b

area square = a²

area triangle = base* height/2

for an equilateral triangle height = base * sin 60 = (√2 / 2 )* base

therefore

area triangle = base* height/2 = base² (√2 /4) =(√2 /4) b²

the total length of the wire is = square length + triangle length = 4*a + 3*b

therefore

A=a²+(√2 /4) b²

4*a + 3*b=L→ b = (L-4*a)/3

A= a²+(√2 /4) (L-4*a)²/ 9

the maximum and minimum amount can be found taking the derivative of the area respect with a:

dA/da= 2*a + (√2 /36) 2*(L-4*a)*(-4) = 0

a -(√2 /9) (L-4*a) = 0

a - √2 /9 * L + √2 /9*4*a =0

(√2 /9*4+1)*a = √2 /9 * L

a= √2 /9 * L / (√2 /9*4+1) = √2 /9 * L / (√2 /9*4+1) = √2 /9 * 10 m / (√2 /9*4+1)

a= 0.96 m

therefore

b = (L-4*a)/3 = (10 m - 4*0.96m)/3= 2.053 m

A=a²+(√2 /4) b² = (0.96 m)²+ (√2 /4) (2.053m)² = 2.411 m²

in the extreme cases

a= 10 m/4=2.5 m and b=0

thus A= (2.5 m)² = 6.25 m²

b=10 m/3= 3.33 m and a=0

thus A= (√2 /4) (3.33 m)² = 3.92 m²

therefore minimum area A=3.92 m² with Length 1=4*0.96 m=3.84 m , Length 2 =2.053 m*3 = 6.16 m

the maximum area is A=6.25 m² with Length 1=10 m and Length 2=0 m

7 0

3 years ago

Using the circuit diagram find the emf ​

Using the circuit diagram find the emf