The bone in direct contact with the main metatarsal is that the carpal bone. The major metatarsal is that the bone behind the enormous toe of the foot. Of every the metatarsal bones, the main metatarsal bone is that the shortest, thickest and strongest. The main metatarsal is break-up into head, body and also base.

You are watching: Which bone is in direct contact with the first metatarsal

**Explanation:**

princetoneclub.org:

6.67 ohm

Explanation:

From the question provided above, the adhering to data to be 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 identical 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

Invert

R = 20/3

R = 6.67 ohm

Therefore, the indistinguishable resistance is 6.67 ohm.

Two horses begin at rest. After a couple of seconds, equine A is traveling v a velocity that 10m/s west, while equine B is traveling wi

How much net force is forced to advice a 0.5 kg toy car, initially at remainder to a velocity the 2.4 m/s in 6s?

A rubber ball is dropped from remainder from a elevation h. The round bounces off the floor and also reaches a height of 2h/3. How have the right to we usage

**princetoneclub.org:**

**Option e.**

In this situation the occupational is done on the ball by nonconservative pressures that brought about the ball having less complete mechanical power after the bounce.

**Explanation:**

**nonelastic collision**when a moving ball access time the ground.Although the

**conservation the mechanical power possessed**by the round which is the sum of

**P.E and K.E.,**but

**kinetic power is no conserved.**

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The no conservative force did the work-related on the sphere that after ~ bouncing

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**lost some the the mechanical energy**of the ball. The kinetic energy in the

**beginning is converted**in some other power like

**friction and air resistance**in this case.

A car enters a tunnel in ~ 24 m/s and increases steadily at 2.0 m/s2. In ~ what rate does it leaving the tunnel, 8.0 seconds later?

Given: V1 = early velocity = 24 m/sa = acceleration = 2.0 m/s^2s = time = 8 sV2 = final velocity = ?For linear-motion difficulties with those provided terms, the adhering to formula is used:V2 = V1 + asSubstituting the given values:V2 = 24 + 2(8)V2 = 24 + 16V2 = 40 m/sTherefore the automobile will have actually a rate of 40 m/s together it leaves the tunnel.