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3 years ago

Help me with number 28.

Physics

1 answer:

Ray Of Light [21]3 years ago

3 0

What you wrote in the space under the question is correct, but
you can go a lot farther than that.

The question gives you the mass of the object, and enough information
to calculate its volume.

So go on from where you left off ... take the formula you wrote (correctly),
plug in the numbers for mass and volume, and find the density. You can do it !

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How do the discovery of gravity and the invention of electronic satellites most likely relate to the processes of scientific inv

Leya [2.2K]

Answer:

Explanation:

because i said so

5 0

3 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.

LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = $\frac{MR^2}{2}$

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia $(I_y_y )$ of the system with respect to the axis yy.

⇒ $I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g$

⇒ $I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2$

⇒ $I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2$

⇒ $I_y_y=289.088\ kg.m^2$

The moment of inertia of the system is 289.088 kg.m^2

7 0

3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago

What is the net force at the equilibrium point? Derive an equation for the location of the equilibrium point based on the accele

Vitek1552 [10]

To find a general equilibrium point for a spring based on the hook law, it is possible to start from the following premise:

Hook's law is given by:

$F = k\Delta X$

Where,

k= Spring Constant

$\Delta X =$ Change in Length

F = Force

When there is a Mass we have two force acting at the System:

W= mg

Where W is the force product of the weigth. Then the force net can be defined as,

$F_{net} = F+W$

But we have a system in equilibrium, so

$0 = K\Delta X -mg$

We find the equilibrium for any location when

$\Delta X = \frac{mg}{k}$

4 0

3 years ago

You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double

dimulka [17.4K]

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

$\tan\theta=\dfrac{y}{D}$

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

$\tan\theta=\dfrac{1.52}{120}$

$\theta=\tan^{-1}\dfrac{1.52}{120}$

$\theta=0.725$

We need to calculate the wavelength

Using formula of wavelength

$d\sin\theta=n\lambda$

Put the value into the formula

$0.025\times\sin0.725=5\times\lambda$

$\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}$

$\lambda=6.326\times10^{-7}\ m$

$\lambda=633\ nm$

Hence, The wavelength of the light is 633 nm.

4 0

2 years ago