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MArishka [77]
3 years ago
10

All of the following are functions of the skeletal system EXCEPT:

Physics
2 answers:
sasho [114]3 years ago
8 0
I'm pretty sure the answer would be D
VashaNatasha [74]3 years ago
6 0

Answer:

D. Producing ligaments and tendons.

just to the quiz on edge

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A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)
Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0
3 years ago
reasons. 5. Why is the unit of temperature called a fundamental unit? Give reasons. ring derived unit.​
Furkat [3]

Explanation:

It doesn't depends upon other.

It have it's own identity.

It's a lot easier to measure temperature than to measure the motion of component particles.

8 0
3 years ago
Newtons laws of motion
Viktor [21]

Answer:

Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.

Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.

Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.

7 0
3 years ago
Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el
kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

E =\frac{Kq}{r^2}

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9} = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

8 0
3 years ago
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
yaroslaw [1]

Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

3 0
3 years ago
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