Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale

Physics

1 answer:

gregori [183]4 years ago

8 0

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

Least count of vernier caliper=0.49-0.50=-0.01 mm

Hence, the least count of the instrument=-0.01 mm

Answer: -0.01 mm

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Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,

ikadub [295]

Answer:

$p_t_o_t_a_l=250kg\frac{m}{s}$

Explanation:

The total momentum of a system is defined by:

$(mv)_t_o_t=m_1v_1+m_2v_2+...$

Where,

$(mv)_t_o_t$ is the total momentum or it could be expressed also as $p_t_o_t_a_l$ .

$m_1$ and $m_2$ represents the masses of the objects interacting in the system.

$v_1$ and $v_2$ are the velocities of the objects of the system.

Remember: The momentum is a fundamental physical magnitude of vector type.

We have:

$m_1=725 kg$

$v_1=15\frac{m}{s}\\m_2=625 kg$

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be negative. It goes in a different direction from car A.

$v_2=-17\frac{m}{s}$

Then the total momentum of the system is:

$p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}$