The bone in direct contact with the primary metatarsal is that the carpal bone. the primary metatarsal is that the bone behind the massive toe of the foot. Of all the metatarsal bones, the primary metatarsal bone is that the shortest, thickest and strongest. the primary metatarsal is split into head, body and base.

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6.67 ohm


From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20


R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

Two horses begin at rest. After a few seconds, horse A is traveling with a velocity of 10m/s west, while horse B is traveling wi

How much net force is required to accelerate a 0.5 kg toy car, initially at rest to a velocity of 2.4 m/s in 6s?

A rubber ball is dropped from rest from a height h. The ball bounces off the floor and reaches a height of 2h/3. How can we use

Option e.

In this case the work is done on the ball by nonconservative forces that resulted in the ball having less total mechanical energy after the bounce.


This is the type of nonelastic collision when a moving ball hits the ground.Although the conservation of mechanical energy possessed by the ball which is the sum of P.E and K.E., but kinetic energy is not conserved.

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The non conservative force did the work on the ball that after bouncing lost some of the mechanical energy of that ball. The kinetic energy in the beginning is converted in some other energy like friction and air resistance in this case.
A car enters a tunnel at 24 m/s and accelerates steadily at 2.0 m/s2. At what speed does it leave the tunnel, 8.0 seconds later?
Given: V1 = initial velocity = 24 m/sa = acceleration = 2.0 m/s^2s = time = 8 sV2 = final velocity = ?For linear-motion problems with those given terms, the following formula is used:V2 = V1 + asSubstituting the given values:V2 = 24 + 2(8)V2 = 24 + 16V2 = 40 m/sTherefore the car will have a speed of 40 m/s as it leaves the tunnel.