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At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)

Answers

Answer 1

Answer:

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

$R_t = R_o[1 + \alpha(T-T_o)]\n\n$

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

$R_t = R_o[1 + \alpha(T-T_o)]\n\n\R_t = 1.59*10^(-8)[1 + 0.0038(T-20)]$

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

$R_t,_(silver) = 2R_o,_(iron)\n\n1.59*10^(-8)[1 + 0.0038(T-20)] =(2 *9.71*10^(-8))\n\n\ \ (divide \ through \ by \ 1.59*10^(-8))\n\n1 + 0.0038(T-20) = 12.214\n\n1 + 0.0038T - 0.076 = 12.214\n\n0.0038T +0.924 = 12.214\n\n0.0038T = 12.214 - 0.924\n\n0.0038T = 11.29\n\nT = (11.29)/(0.0038) \n\nT = 2971.1 \ ^0C$

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

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A typical automobile under hard braking loses speed at a rate of about 7.2 m/s2; the typical reaction time to engage the brakes is 0.55 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 3.6 m. What maximum speed does this imply for a car in the school zone?

Answers

Answer:

4.3 m/s

Explanation:

a = rate at which the automobile loses speed = - 7.2 m/s²

v₀ = initial maximum speed of automobile

t' = reaction time for applying the brakes = 0.55 s

d = distance available for stopping the vehicle = 3.6 m

d' = distance traveled while applying the brakes = v₀ t' = (0.55) v₀

v = final speed after the vehicle comes to stop = 0 m/s

Using the equation

v² = v₀² + 2 a (d - d' )

0² = v₀² + 2 (- 7.2) (3.6 - (0.55) v₀)

v₀ = 4.3 m/s

The measurement of an electron's energy requires a time interval of 1.2×10^−8 s . What is the smallest possible uncertainty in the electron's energy?

Answers

Answer:

$1.05* 10^(-26)J$

Explanation:

The uncertainty in energy is given by $\Delta E=(h)/(2\pi \Delta t)$

here h is plank's constant which value is $6.67* 10^(-34)$ and $\Delta t$ is the time interval which is given as $1.2* 10^(-8)sec$

So using all the parameters the smallest possible uncertainty in electrons energy is $=(6.67* 10^(-34))/(2* \pi * 1.2* 10^(-8))=1.05* 10^(-26)J$

Which is the SI base unit for mass?

Answers

Answer:

kilogram

Explanation:

Answer:

SI base units of mass=KG

Plzzz helppppp!!! I need answers A, B, C & D

Answers

Answer: the answer i for c is yes 0& 10

Explanation:

across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 165 N directed 26 degrees below the horizontal. While you slide the chair a distance of 6.00 m , the chair's speed changes from 1.30 m/s to 2.50 m/s . Find the work done by friction on the chair.

Answers

Answer:

-847.2J

Explanation:

First find the acceleration from v^2= u^2 + 2as

v= 2.5 m/s

u= 1.3 m/s

a???

s=6.00

a= v^2-u^2/2s

a= (2.5)^2-(1.3)^2/2× 6

a= 0.38ms^-2

From Newtons second law:

(Force applied cos Θ) - (Frictional force) = ma

Frictional force = ma- (Force applied cos Θ)

Frictional force= (18.8×0.38) - (165 cos 26°)

Frictional force= 7.144- 148.3= -141.2N

Therefore,

Work done by friction = Frictional force × distance covered

= -141.2N × 6= -847.2J

Answer:

W = –847J

Explanation:

Given m = 18.8kg, F = 165N, θ = -26° (below the horizontal, s = 6.0m, u = 1.30m/s and v = 2.50m/s

In this problem, two forces act on the chair; the forward force F and the frictional force f. We would apply newton's second law to find the frictional force f after which we can calculate the workdone by the frictional force f×s.

But for us to apply newton's second law, we need to know the acceleration of the chair cause by the net force.

From constant acceleration motion equations

v² = u² + 2as

2.5² = 1.30² + 2a×6

6.25 = 1.69 +12a

12a = 6.25 – 1.69

12a = 4.56a

a = 4.56/12

a = 0.38m/s

By newton's second law the net sum of forces equals m×a

The force F has horizontal and vertical and components. It is the horizontal component of this force that pushes the chair against friction.

Fx and f are oppositely directed.

Fx – f =ma

165cos(-26) – f = 18.8×0.38

148.3 – f = 7.14

f = 148.3 – 7.14

f = 141.2N

Workdone = -fs = –141.2×6.00 = –847J

W = –847J

Work is negative because it is done by a force acting on the chair in a direction opposite (antiparallel) to that of the intended motion.

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)a. Find the wavelength of the initial note.
b. Find the wavelength of the final note.
c. Assume the choir sings the melody with a uniform sound level of 70.0 dB. Find the pressure amplitude of the initial note.
d. Find the pressure amplitude of the final note.
e. Find the displacement amplitude of the initial note.
f. Find the displacement amplitude of the final note.

Answers

Answer:

Detailed step wise solution is attached below

Explanation:

(a) wavelength of the initial note 2.34 meters

(b) wavelength of the final note 0.389 meters

(d) pressure amplitude of the final note 0.09 Pa

(e) displacement amplitude of the initial note 4.78*10^(-7) meters

(f) displacement amplitude of the final note 3.95*10^(-8) meters

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