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Sunny_sXe [5.5K]
3 years ago
7

Moving a distal part of arms or legs to follow a circular path is:

Physics
1 answer:
azamat3 years ago
8 0
Moving a distal part of arms or legs to follow a circular path is called circumduction. This medical term is sometimes described as circular movement because the movements of hands and arms are tracing circles in the air or space. Circumduction is also described as conical movement.<span> </span>
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A screw is a type of simple machine. If we look closely at a screw, we see that it is really a _________ wrapped around a centra
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Is is really an inclined plane 
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4 years ago
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A rocket engine provides 28,913 Newtons of thrust. The rocket has a mass of 2,350 kilograms. Calculate its acceleration if it mo
Dafna11 [192]
1. The unknown is acceleration
2. The givens are the force(28,913N) and the mass(2350kg)
3. The equation is a=f/m where a is acceleration,f is force and m is mass
4.a=28913/2350 a=12.30340426
5. Therefore its acceleration is 12.30m/s²(rounded to nearest hundredth)

6 0
3 years ago
If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0
3 years ago
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Calculate the heat flux (in W/m2) through a sheet of a metal 11-mm thick if the temperatures at the two faces are 350 and 110 ˚C
Masja [62]

q = 1156363.6W/m².

To calculate the heat flux per unit area (W/m²) of a sheet made of metal:

q = -k(ΔT/Δx)

q = -k[(T₂ - T₁)/Δx]

Where, k is the thermal conductivity of the metal, ΔT is the temperature difference and Δx is the thick.

With Δx = 11 mm = 11x10⁻³m, T₂ = 350°C and T₁ = 110°C, and k = 53.0 W/m-K:

q = -53.0W/m-K[(110°C - 350°C)/11x10⁻³m

q = 1156363.6W/m²

3 0
3 years ago
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so
lubasha [3.4K]

Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

4 0
4 years ago
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