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2 years ago

The average induced emf will be given by the time rate of change and the product of the and the .

1 answer:

8 0

<span>The average induced emf will be given by the time rate of change and the product of the number of coils and the rate of change of magnetic field</span>

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago

the Earth exerts a gravitational force of 500 newtons on an object what is the mass of the object in kilograms

MAXImum [283]

Use Newton's second law F = mass * acceleration
In your problem F = 500, and we know gravity is working on it so use a = 9.81
Substitute into the equation
500 = m * 9.81
m = 50.97 kg

7 0

2 years ago

Lemonade is best described as _______. A. an element B. a compound C. a mixture D. a pure substance

Minchanka [31]

C is the answer because it mix and water combining together

7 0

2 years ago

Read 2 more answers

The acceleration of gravity is -9.8 m/s2. A sling shot fires a rock straight up into the air with a speed of +39.2 m/s. 1. what

Vanyuwa [196]

Given that,

The acceleration of gravity is -9.8 m/s²

Initial velocity, u = 39.2 m/s

Time, t = 2 s

To find,

The final velocity of the shot.

Solution,

Let v is the final velocity of sling shot. Using first equation of motion to find it.

v = u +at

Here, a = -g

v = u-gt

v = (39.2)-(9.8)(2)

v = 19.6 m/s

So, its velocity after 2 seconds is 19.6 m/s.

4 0

2 years ago

An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.

rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = $\frac{1}{2}$ $m_{p}$ v²

v = √( 2K / $m_{p}$ )

lets relate the cross-sectional area A of the beam to its diameter D;

A = $\frac{1}{4}$ πD²

now, we substitute for v and A

n = I / $\frac{1}{4}$ πeD² ×√( 2K / $m_{p}$ )

n = 4I/π eD² × √( $m_{p}$ / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵

n = 3.2 × 10¹³ m⁻³

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0

2 years ago