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umka21 [38]
3 years ago
14

If the area of an iron rod is 10 cm by 0.5 cm and length is 35 cm. Find the value of resistance, if 11x10^-8 ohm.m be the resist

ivity of iron.
Physics
1 answer:
malfutka [58]3 years ago
6 0

Answer:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, \rho=11\times 10^{-8}\ \Omega-m

We need to find the resistance of the iron rod. It is given by :

R=\rho\dfrac{L}{A}

R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}

R=0.000077 \Omega

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.

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Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete
Sonbull [250]

Answer:

<em>work done in pumping the entire fuel is 466587 J</em>

<em></em>

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical,<em> we assume that the fuel within also takes the cylindrical shape.</em>

<em>Also, we assume that the fuel completely fills the tank.</em>

volume of a cylinder = \pi r^{2}l

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x 1.5^{2} x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = <em>466587 J</em>

8 0
3 years ago
Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock
pashok25 [27]

Answer:

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

c) H=294300\ m

d) t_T=544.95\ s

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, a=2g=2\times 9.81=19.62\ m.s^{-2}

time for which the rocket accelerates, t_a=100\ s

<u>For the course of upward acceleration:</u>

using eq. of motion,

s_a=ut+\frac{1}{2}at_a^2

where:

u= initial velocity of the rocket at the launch =0

s_a= height the rocket travels just before its fuel finishes off

so,

s_a=0+\frac{1}{2}\times 19.62\times 100^2

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

v_a=u+at_a

v_a=0+19.62\times 100

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

v^2=v_a^2-2gh

where:

g= acceleration due to gravity

v= final velocity of the rocket at the top height

0^2=1962^2-2\times 9.81\times h

h=196200\ m

c) So the total height at which the rocket gets:

H=h+s

H=196200+98100

H=294300\ m

d)

Time taken by the rocket to reach the top height after the fuel is over:

v=v_a+g.t

0=1962-9.81t

t=200\ s

Now the time taken to fall from the total height:

H=v.t'+\frac{1}{2}\times gt'^2

294300=0+0.5\times 9.81\times t'^2

t'=244.95\ s

Hence the total time taken by the rocket to strike back on the earth:

t_T=t_a+t+t'

t_T=100+200+244.95

t_T=544.95\ s

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0
3 years ago
Select all the correct answers.
sdas [7]
I believe there are two correct answers, and those answers are A and D
5 0
3 years ago
A skier accelerates down the hill at 3m/s2 how fast is he going in 4 seconds​
almond37 [142]

Answer:

Explanation: simple kinematics

we suppose that initially vo= 0 so if the skier moves 4s :

vf = vo +at = 0 + 3*4 = 12 m/s

best wishes from colombia

7 0
3 years ago
A small plastic bead has been charged to -15nC.
marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.  

8 0
3 years ago
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