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3 years ago

Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of

1 answer:

5 0

I don't know if you need to complete this question or do it otherwise, however, I managed to find on the Internet on several places this completion of your sentence:
<span>Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of q = 2 x 10</span>⁶ W/m³.
I'm not sure whether that is the answer you were looking for, but that's what I found.

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What is physics? Why is it important in our life

-Dominant- [34]

Answer:

It helps us to know or be aware of some things that happen regular and teaches us about laws that guide us

8 0

4 years ago

Read 2 more answers

When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

gayaneshka [121]

Answer:

The value is $E = 1.35 *10^{14} \ J$

Explanation:

From the question we are told that

The mass of matter converted to energy on first test is $m = 1 \ g = 0.001 \ kg$

The mass of matter converted to energy on second test $m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg$

Generally the amount of energy that was released by the explosion is mathematically represented as

$E = m * c^2$

=> $E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2$

=> $E = 1.35 *10^{14} \ J$

7 0

3 years ago

A 1.25 kg block is attached to a spring with spring constant 17.0 N/m . While the block is sitting at rest, a student hits it wi

Free_Kalibri [48]

Answer:

The amplitude of the subsequent oscillations is 13.3 cm

Explanation:

Given;

mass of the block, m = 1.25 kg

spring constant, k = 17 N/m

speed of the block, v = 49 cm/s = 0.49 m/s

To determine the amplitude of the oscillation.

Apply the principle of conservation of energy;

maximum kinetic energy of the stone when hit = maximum potential energy of spring when displaced

$K.E_{max} = U_{max}\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\where;\\\\A \ is \ the \ maximum \ displacement = amplitude \\\\mv^2 = kA^2\\\\A^2 = \frac{mv^2}{k} \\\\A = \sqrt{\frac{mv^2}{k}} \\\\A = \sqrt{\frac{1.25\ \times \ 0.49^2}{17}} \\\\A = 0.133 \ m\\\\A = 13.3 \ cm$

Therefore, the amplitude of the subsequent oscillations is 13.3 cm

6 0

3 years ago

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0

3 years ago

A 2-kg sample of a metal increases in temperature by 20°C when heated on a stove for 3 minutes. Can you predict what will happen

padilas [110]

Answer:

Explanation:

7 0

3 years ago