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hichkok12 [17]
2 years ago
11

A carbon rod with a radius of 2.1 mm is used to make a resistor. The resistivity of this material is 3.2 × 10−5 Ω · m. What leng

th of the carbon rod should be used to make a 13.7 Ω resistor?
Physics
1 answer:
Eduardwww [97]2 years ago
5 0

Answer:

5.93 m

Explanation:

The carbon rod has a radius of 2.1 mm; it can be converted to meter by

1 m = 1000 mm

1 m ÷1000 = 1 mm

2.1 mm = 2.1 × 1 mm = 2.1 ×1÷1000 m = 0.0021 m

using the formula for resistance of a wire

R = ρl÷A, where R is the resistance of the carbon rod in ohms, ρ is resistivity in ohms.m, l is the length of the rod and A is the cross sectional area in m².

A = π r²

Make l the subject of the formula by cross multiplying

RA = ρl

divide both side by ρ

RA÷ρ = ρl÷ρ

RA÷ρ = l

substitute the values in

13.7 × 0.0021 × 0.0021 × 3.142 ÷ 3.2 × 10^-5 = 5.93 m

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If the speed of a particle is doubled, what happens to its kinetic energy?
Norma-Jean [14]

Answer:

it increases

Explanation:

maybe for example we have 5pets then you multiple them by two u get 10

4 0
2 years ago
Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s

The velocity of the flea when leaving the ground is 1.327363 m/s

W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m

The flea will travel 0.00090243 m upward

8 0
3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
The escape speed from the moon is much smaller than from earth. True or False
Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

v_{e} = \sqrt{\frac{2GM}{r}}  (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is 5.972x10^{24}kg and its radius is 6371000m

Then, replacing those values in equation 1 it is gotten.

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}  

v_{e} = 11.18m/s

For the case of the Moon:

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}  

v_{e} = 2.38m/s

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

4 0
3 years ago
A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat
postnew [5]

Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let 'F_A' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

F_A × 10.0 m - W × 4.0 m = 0

∴ F_A × 10.0 m = W × 4.0 m = 600 N × 4.0 m

F_A × 10.0 m = 600 N × 4.0 m

∴  F_A = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', F_A = 240 N.

8 0
2 years ago
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