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Dennis_Churaev [7]
2 years ago
6

A length of copper wire has a resistance 29 Ω. The wire is cut into three pieces of equal length, which are then connected as pa

rallel lengths between points A and B. What resistance will this new "wire" of length L0 3 have between points A and B? Answer in units of Ω.
Physics
1 answer:
erik [133]2 years ago
6 0

Answer:

3.222 ohms

Explanation:

If the total wire had a resistance of 29 ohms, when cut in three, each piece will have a resistance of 9.666 ohms.

As these three pieces (R1, R2 and R3) are now connected in parallel, the equivalent resistance R can be calculated using this equation:

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/9.666 + 1/9.666 + 1/9.666

1/R = 3/9.666

R = 9.666/3 = 3.222 ohms

The resistance between A and B will be 3.222 ohms

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800 joules of work were done with a force of 200 newtons. Over what
Svetach [21]

Answer:

4m

Explanation:

800/200

7 0
3 years ago
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0
2 years ago
What are the steps to extract metal from the earths crust
Pani-rosa [81]
<span>Methods of extraction include: extract by electrolysis, extract by reaction with carbon or carbon monoxide, and extracted by various chemical reactions.</span>
4 0
2 years ago
A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two
Andrew [12]
A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 newtons and points along the due east/west line. Find the magnitude and direction of F2. Note that there are two answers. <span>The given values are
F1 = 200 N</span> F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N
F2 = 400 - 200
F2 = 200 N



4 0
2 years ago
A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w
kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = ?

v_{1i} = initial velocity of the first body before collision = v

v_{2i} = initial velocity of the second body before collision = 0 m/s

v_{1f} = final velocity of the first body after collision =

using conservation of momentum equation

m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}

Using conservation of kinetic energy

m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35

b)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = 1.35 kg

v_{1i} = initial velocity of the first body before collision = 4 ms⁻¹

v_{2i} = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67 ms⁻¹

8 0
2 years ago
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