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JulsSmile [24]
2 years ago
14

19. a) Name the type of Joint present in i) Neck ii) Shoulder b) Which of the skull bones are movable ?​

Physics
1 answer:
Mila [183]2 years ago
4 0
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In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s
atroni [7]

Answer:

    K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

         Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

        Em = K + U₂

       K = Em –U₂

We substitute

     K = m g A - m gA2

    K = m g (A - A2)

3 0
3 years ago
Read 2 more answers
How can a moving coil galvanometer can be made into a dc ammeter?
Dennis_Churaev [7]
I am absolutely sure that the way how can a moving coil galvanometer can be made into a dc ammeter is of course by connecting a. low resistance across the meter. You should remember that you must connect <span>a shunt resistor straight across the galvanometer. Do hope this answer will help you! Regards.</span>
7 0
3 years ago
The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single
Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W =  mg = (689/1000)9.8 = 6.75 N,  D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

7 0
3 years ago
Please I really need help on these
blagie [28]

Answer:

  3, 4, 2

Explanation:

I can't see the final ones.

4 0
3 years ago
Point masses m1 m2 are placed at opposite ends
tiny-mole [99]

(a) x = \frac{m_2L}{m_1+m_2}

<u>Explanation:</u>

Given:

Moment of Inertia of m₁ about the axis, I₁ = m₁x²

Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²

Kinetic energy is rotational.

Total kinetic energy is E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)

Work done is change in kinetic energy.

To minimize E, differentiate wrt x and equate to zero.

m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at \frac{m_2L}{m_1 + m_2}

7 0
3 years ago
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