The moon completes one orbit of earth every 27.3 earth days that is its

il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

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8 0

1 year ago

The table represent the thickness, top density, and bottom density of the different layers of the Earth. In most of the layers,

Nostrana [21]

Answer:inner core?

Explanation:

7 0

3 years ago

Read 2 more answers

Compare the strengths of UV light and microwaves (in Hz). Which type of light is more powerful and how do you know?

TEA [102]

Answer:

UV light is more powerful as it has greater energy.

Explanation:

The energy propagated by electromagnetic waves ( light ) through vacuum or medium is known as electromagnetic radiation.

The frequency/wavelength range of electromagnetic radiation is known as electromagnetic spectrum. The electromagnetic spectrum ranging from gamma ray to radio waves.

Frequency range of UV light = ( 8 x 10¹⁴ to 3 x 10¹⁶ ) Hz

Frequency range of Microwaves = ( 300 x 10⁶ to 300 x 10⁹ ) Hz

Ratio of UV light to Microwaves = ( $\frac{8\times10^{14} }{300\times10^{6} }$ to $\frac{3\times10^{16} }{300\times10^{9} }$ )

= ( 2.66 x 10⁶ to 1 x 10⁸ )

Energy of electromagnetic radiation is given by the relation:

E = hν

Here h is plank's constant and ν is frequency.

UV light is more powerful than Microwaves as frequency of UV light is greater than frequency of microwaves. Thus, by the above equation, the energy of UV light is more than energy of Microwaves.

5 0

3 years ago

A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away

Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d) Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it. Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball. Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.

Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

x = x(0) + V_i,x*t_c

D = 0 + V_i,x*t_c

V_o,x + v_r = D/t_c

V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

V_o = sqrt ( V_o,x^2 + V_o,y^2)

V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

- The angle with which it should be thrown can be evaluated by trigonometric relation:

tan(Q) = ( V_o,y / V_o,x )

tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

6 0

3 years ago

Periodic waves with a wavelength of 0.50 meter move with a speed of 0.30 meter in medium A. When the waves enter medium B, they

iris [78.8K]

Answer:

0.25 m

Explanation:

Refraction occurs when the velocity or wavelength of a wave changes at the interface between two media.

We know that refractive index=

Wavelength in medium A/wavelength in medium B = velocity in medium A/velocity in medium B

Let the wavelength of medium B be a

0.5/a = 0.3/0.15

0.5 × 0.15 = 0.3 × a

a= 0.5 × 0.15/0.3

a= 0.25 m

7 0

3 years ago