In a clock, the average speed is maximum for the tip of

Physics

2 answers:

seraphim [82]3 years ago

7 0

The answer is A) seconds hand

aleksandrvk [35]3 years ago

7 0

A) Seconds hand

There are 60 minutes in an hour, 1 hour in an hour, and 3,600 seconds in an hour. Since there are more seconds in an hour than minutes, the seconds hand moves at a much quicker rate than the other two hands.

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A thin block of soft wood with a mass of 0.078 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

natali 33 [55]

Answer:

Final speed of the bullet is 228.3 m/s

Explanation:

As we know that there is no external force on the system of wooden block and the bullet

so we can say momentum of the system is conserved here

so here we can say

$P_i = P_f$

$m_1v_1 = m_1v_{1f} + m_2v_{2f}$

$4.67\times 10^{-3}(613) = 4.67 \times 10^{-3}v_{1f} + (0.078)(23)$

so we will have

$2.86 = 4.67\times 10^{-3} v_{1f} + 1.794$

$v_{1f} = 228.3 m/s$

7 0

3 years ago

The terminal speed of a sky diver is 130 km/h in the spread-eagle position and 326 km/h in the nosedive position. Assuming that

SIZIF [17.4K]

Answer:

$\frac{A_{slow} }{A_{fast} }=6.2885$

Explanation:

Given data

Terminal velocity for spread eagle position vt=130 km/h

Terminal velocity for nosedive position vt=326 km/h

The terminal speed of the diver is given by

$v_{t}=\sqrt{\frac{2mg}{CpA} }$

Therefore the area is given by

$A=\frac{2mg}{Cp(v_{t})^{2} }\\$

Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is

$\frac{A_{slow} }{A_{fast} }=\frac{326^{2} }{130^{2} }\\\frac{A_{slow} }{A_{fast} }=6.2885$