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Mariana [72]
3 years ago
13

String 1 is pulled on with a force of 50 N to the right. String 1 is attached to a mass of 20 kg on a level surface. String 2 co

nnects the 20 kg mass to the second mass of 40 kg. Friction can be ignored. Find the tension force in string 2.
​
Physics
1 answer:
vladimir1956 [14]3 years ago
7 0

The masses are pulled in only one dimension, so we can ignore the forces acting in the vertical direction. We take the positive direction to be to the right.

For the 20 kg mass, the net force is

∑ <em>F</em>₁ = 50 N - <em>T</em>

where <em>T</em> is the tension in string 2.

For the 40 kg mass, the net force is

∑ <em>F</em>₂ = <em>T</em>

For both masses combined, the net force is

∑ <em>F</em> = 50 N

Use Newton's second law to compute the acceleration of the system:

50 N = (20 kg + 40 kg) <em>a</em>

<em>a</em> = (50 N) / (60 kg) ≈ 0.83 m/s²

Then use the net force equation for the 40 kg mass to compute the tension:

<em>T</em> = (40 kg) <em>a</em> ≈ 33 N

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Answer:

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Explanation:

given,

equilateral triangle of base = 25 cm

electric field strength = 260 N/C

Area of triangle = \dfrac{\sqrt{3}}{4}a^2

                          =  \dfrac{\sqrt{3}}{4} 0.25^2

                          = 0.0271 m³

electric flux = E. A

                   = 260 × 0.0271

                   = 7.046 N m²/C

since, tetrahedron does not enclose any charge so, net flux through tetrahedron is zero.

electric flux through the three side = (electric flux through base)/3

                               = \dfrac{7.046}{3}

electric flux through the three side = 2.35 N m²/C

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maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
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When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
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