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Pani-rosa [81]
3 years ago
12

What is the type of friction force if the object appears just about to start motion? Why?

Physics
1 answer:
Y_Kistochka [10]3 years ago
7 0

Answer:

Sliding friction

Explanation:

friction that acts on objects when they are sliding over a surface. Sliding friction is weaker than static friction

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Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri
irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

d=\frac{2g(m1-m2)}{k}

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

E_{i}=E_{f}

K_{i}+U_{i}=K_{f}+U_{f} (1)

With K_{i} the initial kinetic energy, U_{i} the initial potential energy, K_{f} the final kinetic energy and U_{f} the final potential energy. Note that initialy the masses are at rest so K_{i} = 0, when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so K_{f} =0. So, equation (1) becomes:

U_{i}=U_{f} (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so U_{i}=U_{gi} and at final moment we have potential gravitational energy and potential elastic energy so U_{f}=U_{gf}+U_{ef}, using this on (2)

U_{gi}=U_{gf}+U_{ef} (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), U_{gi}=0 and we have by (3) :

0= U_{gf}+U_{ef} (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

0=-m1gd+m2gd+\frac{kd^{2}}{2}

dividing both sides by d

0=-m1g+m2g+\frac{kd}{2}

g(m1-m2)= \frac{kd}{2}

d=\frac{2g(m1-m2)}{k}, with k the constant of the spring and g the gravitational acceleration.

7 0
4 years ago
How are crystals prepared?
Contact [7]

Answer:

The process of crystal forming is called crystallization. Crystals often form in nature when liquids cool and start to harden. Certain molecules in the liquid gather together as they attempt to become stable. In nature, crystals can form when liquid rock, called magma, cools.

<h2><u><em>Hope it will help you !</em></u></h2>

4 0
3 years ago
Read 2 more answers
Eddie and Val observed the picture of an athlete running in a race.
dedylja [7]

Answer:

Your answer would be C <u><em>Hope this helps</em></u>

8 0
3 years ago
Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

Where

f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

n = f_2-f_1

f_2 = n+f_1

Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

f_2 = 206Hz

The frequency of the tightened is 205Hz

7 0
3 years ago
A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor
Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force F_{1}=6\ N

Second force F_{2}=10\ N

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

\tau=Force\times lever\ arm

So, Here,

\tau_{2}=5 \tau_{1}

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

F_{2}\times d_{2}=5\times F_{1}\times r_{1}

r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}

Where, F_{1}=First force

F_{1}=First force

F_{2}=Second force

r_{1}= distance

Put the value into the formula

r_{2}=\dfrac{5\times6\times0.2}{10}

r_{2}=0.6\ m

Hence, The moment arm is 0.6 m

6 0
4 years ago
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