The mass of Earth is 5.972×1024 kg and its orbital radius is an average of 1.496×1011 m . Calculate its linear momentum.

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

Strike441 [17]

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer $f_{obs}$ is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

$f_{obs} = f_s (\frac{v_w}{v_w-v_s})$

$f_s$ = Frequency of the source

$v_w$ = Speed of sound

$v_s$ = Speed of source

The velocity of the ambulance is

$v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$v_s = 30.55m/s$

Replacing at the expression to frequency of observer we have,

$f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})$

$f_{obs} = 878Hz$

Therefore the frequency receive by observer is 878Hz

8 0

3 years ago

A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i

Effectus [21]

Answer:

s = 1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}

• s = $\sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}$

s = $\sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}$

s = 1.7 m

7 0

3 years ago

An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution

givi [52]

With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
= (15π rad/s)*(1 rev/2π rad)
= 7.5 rev/s

Answer: 7.5 revolutions per second.

7 0

3 years ago

1. A car starting from rest accelerates uniformly at 3

amid [387]

The car's final speed in m/s after the acceleration is 30.

Given the following data:

Initial velocity, U = 0 m/s (since the cars starts from rest)

Time, t = 10 seconds

Acceleration, a = 3 meter per seconds square.

To find the car's final speed in m/s after the acceleration, we would use the first equation of motion;

Mathematically, the first equation of motion is given by the formula;

$V = U + at\\\\V = 0 + 3(10)$

Final speed, V = 30 m/s

Therefore, the car's final speed in m/s after the acceleration is 30.