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andrew-mc [135]
3 years ago
15

A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 1

5.0 V across its ends, while an ammeter measures the current through it. We observe that at room temperature (20.0°C) the ammeter reads 18.5 A, while at 92.0°C it reads 17.2 A. We can ignore any thermal expansion of the rod. Find
a. the resistivity at 20.0°C and
b. the temperature coefficient of resistivity at 20.0°C for the material of the rod.

Physics
1 answer:
nydimaria [60]3 years ago
5 0

Answer:

a) 1.06*10^-5

b) 0.00105 °C^-1

Explanation:

Given that

Length of the cylinder, L = 1.5 m

Radius of the cylinder, r = 0.25 cm

Voltage across the rod, V = 15 V

I• at Temperature T• = 20° C is 18.5 A

I at Temperature T = 90° C is 17.2 A

See attachment for calculations

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An electrically neutral model airplane is flying in a horizontal circle on a 2.0-m guideline, which is nearly parallel to the gr
amm1812

Answer:

q=3.5*10^-4

Explanation:

<u>concept:</u>

The force acting on both charges is given by the coulomb law:

F=kq1q2/r^2

the centripetal force is given by:

Fc=mv^2/r

The kinetic energy is given by:

KE=1/2mv^2

<u>The tension force:</u>

<u><em>when the plane is uncharged </em></u>

T=mv^2/r

T=2(K.E)/r

T=2(50 J)/r

T=100/r

<u><em>when the plane is charged </em></u>

T+k*|q|^2/r^2=2(K.E)charged/r

100/r+k*|q|^2/r^2=2(53.5 J)/r

q=√(2r[53.5 J-50 J]/k)                                          √= square root on whole

q=√2(2)(53.5 J-50 J)/8.99*10^9

q=3.5*10^-4

5 0
2 years ago
Cobaltâ’60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this
Crank

Energy of gamma rays is given by equation

E = h\nu

here we know that

h = Planck's constant

\nu = frequency

now energy is given as

E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}

E = 7.52 \times 10^{-13} J

now by above equation

E = h\nu

7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu

\nu = 1.14 \times 10^{21} Hz

now for wavelength we can say

\lambda = \frac{c}{\nu}

\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}

\lambda = 2.63 \times 10^{-13} m

3 0
2 years ago
Consider a single-slit diffraction pattern for λ=589nm , projected on a screen that is 1.00 m from a slit of width 0.25 mm. How
const2013 [10]

Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R×(mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589×10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

y = 589×10^-9/0.00025

y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

y = 4.712 × 10^-3 m = 4.712mm

8 0
3 years ago
a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
kobusy [5.1K]

The rocket travelled a maximum height at 1.0102 km.

Given,

The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) =  0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

(v)= Δx/Δt

Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

v² = u² +2as

Where, v  is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.

Then putting the value ,

v² = 0 + ( 2× 10 × 0.5×10³)m/s

v² = \sqrt{10000} m/s

v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

To know more about acceleration

brainly.com/question/15135960

#SPJ4

4 0
1 year ago
Explain the term "avalanche state"
vagabundo [1.1K]

Answer:

The downwind side of an obstacle such as a ridge. The addition of weight on top of a snowpack, usually from precipitation, wind drifting, or a person. An avalanche that releases from a point and spreads downhill collecting more snow - different from a slab avalanche. Also called a point-release or sluff.

Explanation:

5 0
3 years ago
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