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Ede4ka [16]
3 years ago
10

Consider two copper wires. One has twice the length and twice the cross-sectional area of the other. How do the resistances of t

hese two wires compare? A) Both wires have the same resistance. B) The shorter wire has twice the resistance of the longer wire. The longer wire has twice the resistance of the shorter wire. D) The longer wire has four times the resistance of the shorter wire. E) The shorter wire has four times the resistance of the longer wire.
Physics
1 answer:
Makovka662 [10]3 years ago
7 0

Answer:

option (a)

Explanation:

Wire 1:

length = 2L

Area = 2 A

Wire 2: length = L

Area = A

As we know that the resistance of a wire is directly proportional to the length of the wire and inversely proportional to the area of crossection of the wire.

Let ρ be the resistivity of the cooper wire.

Resistance of wire 1

R1 = ρ x 2L / 2 A = ρ L / A

Resistance of wire 2:

R2 = ρ x L /  A = ρ L / A

As R1 = R2

It means both the wires have same resistance.

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Yes, this statement is true. Majority of the fluids in our body is water. So, when we perspire, it comes out as a salt solution. When the water vaporizes and turns into water vapor, it leaves white residues in our garments after a while. The salt can't be vaporized as fast as the water so it is left as a solid.
3 0
3 years ago
A SDOF undamped system is set into free vibration with an initial displacement and an initial velocity. The mass of the system i
loris [4]

Answer:

Explanation:

stiffness k = 160

m = 10

angular frequency ω = \sqrt{\frac{k}{m} }

= \sqrt{\frac{160}{10} }

= 4

ω  = 4

Let x = 4 - A sinωt

when t = 0

x = 4 in

when t = 2 s , x = - 4

- 4 = 4 - A sinωt

8  = A sin 4 x 2

8 = A sin8

A = 8 / sin 8

= 8 / .989

= 8.09 in .

x = 4 - A sinωt

dx / dt = - Aω cosωt

v =  - Aω cosωt

for t = 0

v = - Aω

= - 8.09 x 4

= - 32.36 in / s

initial velocity v = - 32.36 in /s

displacement x for t = 4s

x = 4 - 8.09 sin 4 x 4

= 4 - 8.09 sin 16

= 4 - 8.09 x - .2879

= 4 + 2.33

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c ) Amplitude of vibration A = 8.09 in .as calculated above .

4 0
3 years ago
A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after
Aleks04 [339]

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g=9.81ms^{-2}

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

v=u+at

Using the formula and substitute the values

0=24.1-9.81t

Where g is negative because motion of ball is against gravity

24.1=9.81t

t=\frac{24.1}{9.81}=2.46s

3.y=y_0+ut+\frac{1}{2}at^2

Using the formula

y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2

y=30.1 m

4.Time of flight for red ball=3.77-2.9=0.87s

y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2

y'=22.0m

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2

0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2

0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2

0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t

-2.86105=-2.851t

t=\frac{2.86105}{2.851}=1.004 s

Hence,  1.004 s  after the blue ball is thrown  are the two balls in the air at the same height.

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In Pensacola in June, high tide was at noon. The water level at high tide was 12 feet and 2 feet at low tide. Assuming the next
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<span>To answer this question, the equation that we will be using is:
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8 0
4 years ago
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