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MAVERICK [17]
3 years ago
15

Which of the following codes is used to report the removal of 37 skin tags by electrosurgical destruction?

Physics
1 answer:
Tema [17]3 years ago
7 0
Are there any answer choices??
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A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls without slipping down an inclined plane of height 5.3 m. what is t
inn [45]
<span>Answer: Total kinetic energy at the bottom = 0.5(1+0.4) mv^2 = mgh V^2 = 7*9.8/0.7 V = 9.9m/s ω = V/r = 9.9/1.7 = 5.8rad/s Answer c. 5.8 rad/s</span>
7 0
3 years ago
Average velocity is different than average speed because calculating average velocity involves ?
VashaNatasha [74]

direction ................................................................

6 0
3 years ago
Read 2 more answers
An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming
Vlad [161]

Answer:

Explanation:

Given

mass of archer m=0.3\ kg

Average force F_{avg}=201\ N

extension in arrow x=1.3\ m

Work done to stretch the bow with arrow

W=F\cdot x

W=201\times 1.3=261.3\ m

This work done is converted into kinetic Energy of arrow

W=\frac{1}{2}mv^2

where v= velocity of arrow

261.3=\frac{1}{2}\times 0.3\times v^2

v=\sqrt{1742}

v=41.73\ m/s

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

W=mgh

261.3=0.3\times 9.8\times h

h=\frac{261.3}{0.3\times 9.8}

h=88.87\ m

   

4 0
3 years ago
What happens in a tug of war if the net forces are balanced and why?
FinnZ [79.3K]

Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

6 0
3 years ago
1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame
IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0
2 years ago
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