PHYSICS COLLEGE

The International Space Station (ISS) is a space station orbiting the earth above the ground. If the radius of the earth is 3,958.8 miles, mass of earth is 5.972 x 10 24 kg, the period of the ISS at the orbit around the earth is 7.84 hours, can you calculate what is the distance from the ISS to the surface of the earth, in unit of miles

Answers

Answer 1

Answer:

Answer:

8488 miles

Explanation:

The orbital period around an earth is given as:

$T=2\pi \sqrt{(r^3)/(Gm) }$

Where G = constant = 6.67 x 10ˉ¹¹ N m² kgˉ², m = mass of object, T = period taken to round the earth, r = distance from the center of the earth to the orbiting object = radius of earth + orbital altitude.

Given that T = 7.84 hours = 28224 seconds, m = 5.972 x 10²⁴ kg, radius of earth = 3,958.8 miles = 6371071 m

$T=2\pi \sqrt{(r^3)/(Gm) }\n\nsquaring:\n\nT^2=4\pi^2 ((r^3)/(Gm) )\n\nr^3=(GmT^2)/(4\pi^2) \n\nr=\sqrt[3]{(GmT^2)/(4\pi^2) } \n\nr=\sqrt[3]{(6.67*10^(-11)*5.972*10^(24)*(28224)^2)/(4\pi^2) } \n\nr=20031232.62\ meters$

r = radius of earth + distance from the ISS to the surface of the earth

distance from the ISS to the surface of the earth = r - radius of earth

distance from the ISS to the surface of the earth = 20031232.62 meters - 6371071 meters = 13660161.62 meters

distance from the ISS to the surface of the earth = 13660161.62 meters = 8488 miles

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

$\huge{\gray{\sf Question:} }$ A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm and (c) 0 cm.

Answers

Answer:

$here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \n t = .2sec \n omega = (2\pi)/(t) \n = (2\pi)/(.2 ) = 10\pi (rad)/(s) \n we \: want \: find \: a \: and \: v \n we \: know \: that \: a = - {omega}^(2) x \n v = omega \sqrt{ {r}^(2) - {x}^(2) } \n(1)x = .05m \n a = - {10\pi}^(2) * .05 = - 5 {\pi}^(2) \frac{m}{ {s}^(2) } \n v = 10\pi \sqrt{ {.05}^(2) - {.05}^(2) } = 0 \n 2)x = 3cm = .03m \n a = {(10\pi)}^(2) * .03 = - 3 {\pi}^(2) \frac{m}{ {s}^(2) } \n v = 10\pi \sqrt{ {.05}^(2) - {.03}^(2) } = 10\pi * .04 = .4\pi (m)/(s) \n 3)x = 0 \n a = - {(10\pi)}^(2) * 0 = 0 \n v = 10\pi * \sqrt{ {.05}^(2) - {0}^(2) } = - 10\pi * .05 = .5\pi (m)/(s) \n thank \: you$

Answer:

ANSWER

A = 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

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An object is constrained by a cord to move in a circular path of radius 1-m on a frictionless, horizontal surface. The cord will break if the tension exceeds 26.9-N. The maximum kinetic energy that this object can have is _____ J. Round your answer to the nearest whole number.

Answers

Answer:

Explanation:

Let m be the mass of the object and v be the maximum velocity . The tension will provide centripetal force for the circular motion .

T = mv² / R where R is radius of circular path . T is tension .

putting the values given in the equation above

26.9 = m v² / 1

m v² = 26.9

kinetic enrgy = 1/2 m v²

= 26.9 / 2

= 13.45 J

13 J .

Maximum kinetic energy = 13 J .

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Answers

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^(-9)m^3$

The potential energy per unit volume is defined as the energy density.

$u = (U)/(V)$

$u= ((13.0 J))/((6.00*10^(-9) m^3))$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = (1)/(2) \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{(2u)/(\epsilon_0)}$

$E = \sqrt{(2(2.167109))/(8.85*10^(-12))}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

Final answer:

To calculate the electric field strength that would store 12.5 Joules of energy in every 6.00 mm^3 of space, we use the energy density formula. We firstly find the energy density and input it into the formula to solve for the electric field strength. The result is approximately 6.87 X 10^6 N/C.

Explanation:

The energy stored in an electric field is given by the formula U = 1/2 ε E^2. Here, U is the energy density (energy per unit volume), E is the electric field strength, and ε is the permittivity of free space.

Given that the energy stored U is 12.5 joules, and the volume is 6.00 mm^3 or 6.00X10^-9 m^3, the energy density (U) can be computed as 12.5 J/6.00X10^-9 m^3 = 2.08X10^12 Joule/meter^3.

We can solve the formula for E (electric field strength): E = sqrt ((2U)/ε). Substituting the value of ε (8.85 × 10^-12 m^-3 kg^-1 s^4 A^2), we can find E to be approximately 6.87 X 10^6 N/C.

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A toy rocket is launched straight up by using a spring. The rocket is initially pressed down on the spring so that the spring is compressed by 9 cm. If the spring constant is 1050 N/m and the mass of the rocket is 50 g, how high will the rocket go? You may neglect the effects of air resistance.

Answers

Answer:

Rocket will go to a height of 8.678 m

Explanation:

Mass of the rocket m = 50 gram = 0.05 kg

Spring constant k = 1050 N /m

Spring is stretched to 9 cm

So x = 0.09 m

Work done in stretching the spring

$E=(1)/(2)kx^2=(1)/(2)* 1050* 0.09^2=4.2525J$

From energy conservation this energy will convert into potential energy

Potential energy is equal to $E=mgh$ , here m is mass, g is acceleration due to gravity and h is height

So $0.05* 9.8* h=4.2525$

$h=8.678m$

So rocket will go to a height of 8.678 m

Answer:

8.68 m

Explanation:

compression in spring, x = 9 cm = 0.09 m

Spring constant, K = 1050 N/m

mass of rocket, m = 50 g = 0.05 kg

Let it go upto height h.

Use conservation of energy

Potential energy stored in spring = potential energy of the rocket

$(1)/(2)kx^(2)=mgh$

0.5 x 1050 x 0.09 x 0.09 = 0.05 x 9.8 x h

h = 8.68 m

Thus, the rocket will go upto height 8.68 m.

A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?

Answers

Final answer:

The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

Explanation:

To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

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Find an expression for the center of mass of a solid hemisphere, given as the distance R from the center of the flat part of the hemisphere. Express your answer in terms of R. Express the coefficients using three significant figures.

Answers

Answer:

z_c = ⅜R

Explanation:

If we assume that the hemisphere has uniform density, we can express the centre of mass as;

z_c = (ρ/M)∫∫∫ z•dV

We know that density(ρ) = mass(M)/volume(V)

Thus, Vρ = M

And volume of hemisphere = 2πr³/3

Thus;

2Vρπr³/3 = M

So;

z_c = (ρ/(2Vρπr³/3))∫∫∫ z•dV

Where r = R in this case.

ρ will cancel out to give;

z_c = (3/(2πr³))∫∫∫_V (z•dV)

In spherical coordinates,

r is radius

Φ = angle between the point and the z − axis

θ = azimuthal angle

Therefore, the integral becomes what it is in the attached image.

I've completed the explanation as well in the attachment.

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