Select the correct answer.

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium

Kaylis [27]

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

w² = $\frac{k}{m}$

to find the constant (k) of the spring, we use Hooke's law with the initial data

F = - kx

where the force is the weight of the body that is hanging

F = W = m g

we substitute

m g = - k x

k = $- \frac{m g}{x}$

we calculate

k = $- \frac{9.8 m}{- 2.6 \ 10^{-2}}$

k = 3.769 10² m

we substitute in the first equation

w² = $\frac{ 3.769 \ 10^2 \ m }{m}$

w = 19.415 rad / s

angular velocity and frequency are related

w = 2πf

f = $\frac{w}{2\pi }$

f = 19.415 / 2pi

f = 3.09 Hz

7 0

2 years ago

An engineer has the task of producing an aluminum alloy with a density of 3.0 grams per cubic centimeter. She comes up with the

pochemuha

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

$Ro =m/V$

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

For m = 7 [g] and V = 2.3 [cm³]

$Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]$

For m = 10 [g] and V = 7 [cm³]

$Ro=10/7\\Ro=1.42[g/cm^{3} ]\\$

For m = 15 [g] and V = 5 [cm³]

$Ro=15/5\\Ro=3[g/cm^{3} ]\\$

For m = 21 [g] and V = 8 [cm³]

$Ro=21/8\\Ro=2.625[g/cm^{3} ]\\$

5 0

2 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a circular orbit and a low orbit near the surface as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago

The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed

LenaWriter [7]

Answer:

83.67 m/s

Explanation:

Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km 1000 m 83.67m

----------- X ---------- = --------------- the km will cancel out and you are left

12.0 s 1 km s with m/s

6 0

3 years ago

PLEASE HELP!!! WILL GIVE 30 POINTS!! HAS TO BE CORRECT!

belka [17]

Answer:

Disruption to electricity power grid

Explanation:

We're looking a a solar flare. This will whip solar particles at high velocity into space and, If they are near earth, will interact with the earth's magnetic field. These magnetic changes will be measurable in the electric grid. Whether they are strong enough to cause "disruption" depends on a huge number of factors such as strength of and angles of the interacting magnetic fields and location of grid infrastructure,

4 0

2 years ago

Read 2 more answers