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1 year ago

A 25.0-meter length of platinum wire with a cross-sectional area of 3.50 × 10^−6 meter^2 has a resistance

1 answer:

Nookie1986 [14]1 year ago

4 0

R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m

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A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul

Rashid [163]

The Young modulus is given by:
$E= \frac{F /A}{\Delta L / L_0}$
where
F is the force applied
$L_0$ is the initial length of the wire
$A$ is the cross-sectional area of the wire
$\Delta L$ is the stretch of the wire

The wire in the problem stretches by $0.5%$ of its length, this means
$\frac{\Delta L}{L_0} = 0.005$

We can also calculate the area of the wire; its radius is in fact half the diameter:
$r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m$
and so the area is
$A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2$

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

$E= \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2$

3 0

2 years ago

Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball

Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e) y = 0 + ½ g t²

t = √ (2y / g)

t = √(2 125 / 9.8)

t = 5.05 s

6 0

1 year ago

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is 3.57m

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

8 0

2 years ago

Read 2 more answers

The average kinetic energy of the molecules of an ideal gas at 10∘C has the value K10. At what temperature T1 (in degrees Celsiu

Westkost [7]

Answer:

A) T1 = 566 k = 293°C

B) T2 = 1132 k = 859°C

Explanation:

The average kinetic energy of the molecules of an ideal gas is givwn by the formula:

K.E = (3/2)KT

where,

K.E = Average Kinetic Energy

K = Boltzman Constant

T = Absolute Temperature

At 10°C:

K.E = K10

T = 10°C + 273 = 283 K

Therefore,

K10 = (3/2)(K)(283)

FOR TWICE VALUE OF K10:

T = T1

Therefore,

2 K10 = (3/2)(K)(T1)

using the value of K10:

2(3/2)(K)(283) = (3/2)(K)(T1)

T1 = 566 k = 293°C

The average kinetic energy of the molecules of an ideal gas is given by the formula:

K.E = (3/2)KT

but K.E is also given by:

K.E = (1/2)(m)(vrms)²

Therefore,

(3/2)KT = (1/2)(m)(vrms)²

vrms = √(3KT/m)

where,

vrms = Root Mean Square Velocity of Molecule

K = Boltzman Constant

T = Absolute Temperature

m = mass

T = 10°C + 273 = 283 K

vrms = √[3K(283)/m]

FOR TWICE VALUE OF vrms:

T = T2

Therefore,

2 vrms = √(3KT2/m)

using the value of vrms:

2√[3K(283)/m] = √(3KT2/m)

2√283 = √T2

Squaring on both sides:

(4)(283) = T2

T2 = 1132 k = 859°C

8 0

2 years ago

The sun radiates most strongly at a wavelength of about 550 nm. a star that radiated most strongly at 400 nm would be

OLEGan [10]

550 nm is in the visible spectrum. 400 nm is in the ultra-violet spectrum. The shorter the wavelength, the bigger the energy of the light photon emitted. Therefore a star that emits 400 nm photons, is hotter than the Sun. A star that is hotter is either bigger than the Sun (and has about the same age), either younger than the Sun (and has about the same dimension).
Therefore a star that radiates at 400 nm would be hotter, bigger and younger than the Sun.

5 0

2 years ago