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1 year ago

5

A rod (length = 80 cm) with a rectangular cross section (1.5 mm × 2.0 mm) has a resistance of 0.20 Ω. What is the resistivity of

the material used to make the rod?

1 answer:

Andru [333]1 year ago

7 0

Answer:

The resistivity of the material used to make the rod is ρ= 7.5 * 10⁻⁷ Ω.m

Explanation:

R= 0.2 Ω

L= 0.8 m

S= 1.5mm*2mm= 3 mm² = 3 * 10⁻⁶ m²

ρ = (R*S)/L

ρ= 7.5 * 10⁻⁷ Ω.m

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A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.

kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

$F_f = \mu N = \mu mg$

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

$a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2$

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

$t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s$

3 0

1 year ago

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li

babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

$y=\frac{m\lambda D}{d}$

At m = 0

$y_0=0$

$y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm$

At m = 1

$y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m$

The slit separation is 0.00001266 m

3 0

2 years ago

Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp

MatroZZZ [7]

Answer:

E) True. Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

Em₀ = K

Em₀ = ½ m v²

Final

Emf = U = mg h

Em₀ = emf

½ m v² = mgh

h = v² / 2g

For ball A

h_A = v² / 2g

For ball B

h_B = (2v)² / 2g

h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False has 4 times the gravitational potential energy than ball A

D) False. It goes 4 times higher

E) True.

6 0

1 year ago

Nerve impulses are carried along axons, the elongated fibers that transmit neural signals. We can model an axon as a tube with a

jeka94

Answer:

The resistance of the axon is $1.27\times 10^7\ \Omega$ .

Explanation:

Given that,

Inner diameter of the model of an axon, $d=10\ \mu m$

Radius of the model, $r=5\ \mu m=5\times 10^{-6}\ m$

Resistivity of the fluid inside the tube wall, $\rho=0.5\ \Omega -m$

Length of the axon, l = 2 mm = 0.002 m

We know that the resistance in terms of resistivity of an object is given by :

$R=\rho\dfrac{l}{A}\\\\R=0.5\times \dfrac{0.002}{\pi (5\times 10^{-6})^2}\\\\R=1.27\times 10^7\ \Omega$

So, the resistance of the axon is $1.27\times 10^7\ \Omega$ . Hence, this is the required solution.

8 0

1 year ago

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

6 0

1 year ago