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2 years ago

8

A sphere moves in simple harmonic motion with a frequency of 4.40 Hz and an amplitude of 3.50 cm. (a) Through what total distanc

e (in cm) does the sphere move during one cycle of its motion

1 answer:

Anna [14]2 years ago

6 0

Answer:

It will move a distance of 4 * 3.5 = 4 cm in one cycle.

zero to max A, back to zero, zero to min A, back to zero (4 * A)

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What is the velocity of an object with a kinetic energy of 800 J and a mass of 12 kg?

Elis [28]

K.E = 1/2 mv²

800 = 1/2 ×12 ×v²

800 = 6 v²

800 / 6 = v²

= 133.4 =v²

√133.4 = √v²

11.5 = v²

I hope this answer is correct.

3 0

3 years ago

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

5 0

3 years ago

find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is

Readme [11.4K]

Explanation:

Let magnitude of the two forces be x and y.

Resultant at right angle R1= √15N) and at

60 degrees be R2= √18N.

Now, R1 = √(x² + y²) = √15,

R2= √(x² + y² +2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x)2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck

7 0

3 years ago

How much must a woman weigh ( force) if the pressure she exerts while standing on one foot has an area of 0.6m squared exerts a

lions [1.4K]

Answer:

W = 9.6 N

Explanation:

Given that,

Area on 1 foot, A = 0.6 m²

Pressure, P = 16 Pa

The pressure is given by force acting per unit area. So,

$P=\dfrac{F}{A}\\\\P=\dfrac{W}{A}\\\\W=16\times 0.6\\\\W=9.6\ N$

So, the required weight is 9.6 N.

4 0

2 years ago

A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu

kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

$\omega f = \omega o-\alpha*t$

Solving for α:

$\alpha=\frac{\omega o-\omega f}{t}$

where ωo = 600*2*π/60; ωf = 0; t=10s

$\alpha=6.283rad/s^2$

The sum of torque is:

$\tau=I*\alpha$

$\tau=M*R^2/2*\alpha$

$\tau=0.060 N.m$

8 0

3 years ago