Please Help with this ASAP: A wave vibrates 45 times in 30 seconds, calculate its frequency

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

Setler [38]

Answer:

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R = $R_{m}$ + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s

3 0

3 years ago

Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is

vivado [14]

Answer:

0.045 J

Explanation:

From the question,

The elastic potential energy stored in a spring is given as,

E = 1/2ke²...................... Equation 1

Where E = elastic potential energy, k = spring constant, e = compression.

Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m

Substitute these values into equation 1

E = 1/2(100)(0.03²)

E = 50(9×10⁻⁴)

E = 0.045 J

Hence the right option is 0.045 J

4 0

3 years ago

Which pulls harder the earth or the student

Nana76 [90]

Answer:

wha

Explanation:

5 0

3 years ago

If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave

kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an $n^{th}$ bright fringe from the central maxima is given as:

$y_n=\frac{n \lambda D}{d}$

so for the distance of the second-order fringe when wavelength $\lambda_1$ = 745-nm can be calculated as:

$y_2 = \frac{n \lambda_1 D}{d}$

where;

n = 2

$\lambda_1$ = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

$y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}$

$y_2$ = 0.00276 m

$y_2$ = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength $\lambda_2$ = 660-nm is as follows:

$y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}$

$y^'}_2$ = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

$\delta y = y_2-y^{'}_2$

$\delta y$ = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

$\delta y$ = 10 ⁻³ (2.76 - 1.94)

$\delta y$ = 10 ⁻³ (0.82)

$\delta y$ = 0.82 × 10 ⁻³ m

$\delta y$ = 0.82 × 10 ⁻³ m $(\frac{1.0mm}{10^{-3}m} )$

$\delta y$ = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0

3 years ago

You are sitting on a merry-go-round at a distance of 2m from its center. It spins 15 times in 3 min. What distance do you move a

soldier1979 [14.2K]

Answer:

A) 12.57 m

B) 5 RPM

C) 3.142 m/s

Explanation:

A) Distance covered in 1 Revolution:

The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:

s = rθ

where,

s = distance covered = ?

r = radius of circle = 2 m

θ = Angle = 2π radians (For 1 complete Revolution)

Therefore,

s = (2 m)(2π radians)

s = 12.57 m

B) Angular Speed:

The formula for angular speed is given as:

ω = θ/t

where,

ω = angular speed = ?

θ = angular distance covered = 15 revolutions

t = time taken = 3 min

Therefore,

ω = 15 rev/3 min

ω = 5 RPM

C) Linear Speed:

The formula that gives the the linear speed of an object moving in a circular path is given as:

v = rω

where,

v = linear speed = ?

r = radius = 2 m

ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s

Therefore,

v = (2 m)(1.571 rad/s)

v = 3.142 m/s

8 0

3 years ago