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kati45 [8]
3 years ago
15

Describe four consequences of having a different type of joint in the thumb.​

Physics
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

Essentially, your thumb is the main piece of your body that has saddle joints. The bones in your seats joint are in charge of moving forward and backward, side to side.

When all is said in done, the piece of the thumb joint that is subjected to extreme anxiety is that known as CMC joint or carpometacarpal joint. This joint is fundamentally shaped by the metacarpal bone and it explains with the trapezium bone of the wrist.

Explanation:

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If there is an object in top of another object, why does the upper object exert a downward normal force?​
rusak2 [61]

Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.  

Explanation:

This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.

4 0
3 years ago
What do radio waves and microwaves have in common?
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I Will say the Answer is A

Explanation:

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Un carro de montaña rusa parte del reposo desde una primera cumbre, desciende una distancia vertical de 45 metros y luego sube u
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3 years ago
50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
attashe74 [19]

Answer:

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8 0
2 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
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