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Ulleksa [173]
4 years ago
7

Which type of wire would have the least resistance?

Physics
1 answer:
telo118 [61]4 years ago
4 0
The answer is E. Short and thick while cold.
You might be interested in
17 points
const2013 [10]

The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>
  • MV = mv................. Equation 1

<h3>Where:</h3>
  • M = mass of the cannon
  • m = mass of the cannonball
  • V = velocity of the cannon
  • v = velocity of the cannonball

Make v the subject of the equation.

  • v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>
  • M = 500 kg
  • V = 3 m/s
  • m = 10 kg.

Substitute these values into equation 2.

  • v = (500×3)/10
  • v = 150 m/s.

 

Hence, The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

3 0
2 years ago
In a displacement/time graph the slope of the line is equal to the ________.
mamaluj [8]
In a displacement/time graph, the slope of the line is equal to the velocity
3 0
3 years ago
Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli
Masja [62]

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0
3 years ago
In a convex lense f=20cm, m=1,then what is u and v?​
katen-ka-za [31]

Answer:

I'm not sure if I know whatever the answer is

3 0
3 years ago
An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len
worty [1.4K]

Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

7 0
3 years ago
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