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vovangra [49]
3 years ago
13

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

between the boxes and the horizontal surface.
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
5 0
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
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A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0
3 years ago
Identifying Characteristics Common to All Protists Use the drop-down menus to complete the statements. All protists live in envi
Annette [7]

Answer:

Use the drop-down menus to complete the statements.

All protists live in environments that are

✔ wet

.

All protists must

✔ reproduce

.

All protists have

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All protists make or consume

✔ food

Explanation:

I got it right hope its the right one

3 0
3 years ago
Read 2 more answers
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon'
kramer
Chameleon's tongue is more fast than thought. Its long sticky tongue moves at an amazing ballistic speed which lashes out unsuspecting insects and bugs. Now let us see how fast it is.

GIven:

acceleration of the chameleon's tongue- 260 m/s
2 for 20 ms
constant speed 30 ms
50 ms total time
1/20 of a second

solution:

<u>260</u> = <u>  n</u><u>  </u>
20       50
<u>20 n </u>= <u>13, 000</u>
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Setler79 [48]

Answer:

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Explanation:

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The resultant speed of the plane  is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}

ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}

Hence, we have that the resultant speed of the plane  is (3) 226 m/s

Have a nice day!

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