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3 years ago

The energy your body gets from food is originally provided by which nuclear

Physics

1 answer:

Naily [24]3 years ago

4 0

Answer:

Nuclear Fusion

Explanation:

The process releases energy because the total mass of the resulting single nucleus is less than the mass of the two original nuclei. The leftover mass becomes energy.

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A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

4 years ago

Determine the distance above Earth's surface to a satellite that completes four orbits per day.

serg [7]

This question involves the concepts of the time period, orbital radius, and gravitational constant.

The distance of the satellite above the Earth's Surface is "10400 km ".

The theoretical time period of the satellite around the earth can be found using the following formula:

$\frac{T^2}{R^3}=\frac{4\pi^2}{GM}\\\\$

where,

T = Time Period of Satellite = $\frac{1}{frequency} = \frac{1}{4\ orbits/day}\frac{3600*24\ s}{1\ day} = 21600\ s$

R = Orbital Radius = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

Therefore,

$\frac{(21600\ s)^2}{R^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\R = \sqrt[3]{\frac{4.66\ x\ 10^8\ s^2}{9.91\ x\ 10^{-14}\ s^2/m^3}} \\R = 1.675\ x\ 10^7\ m = 1.68\ x\ 10^4\ km$

Now, this orbital radius is the sum of the radius of the earth (r) and the distance of satellite above earth's (h) surface:

R = r + h

1.68 x 10⁴ km = 0.64 x 10⁴ km + h

h = 1.68 x 10⁴ km - 0.64 x 10⁴ km

Learn more about the orbital time period here:

brainly.com/question/14494804?referrer=searchResults

The attached picture shows the derivation of the formula for orbital speed.

8 0

3 years ago

50POINTS! What is the escape velocity for lunar module? Lunar module mass 15,200 kg radius of moon 1.74x106m, mass of moon 7.34x

Airida [17]

Answer:

2.73 km/s

Explanation:

The escape velocity of an object in the gravitational field of the moon is (on the surface of the planet)

$v=\sqrt{\frac{2GM}{r}}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=7.34\cdot 10^{22} kg$ is the mass of the Moon

$r=1.74\cdot 10^6 m$ is the radius of the Moon

As we can see, the escape velocity does not depend on the mass of the lunar module.

Substituting the numbers into the formula, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(7.34\cdot 10^{22}kg)}{(1.74\cdot 10^6 m)}}=2732 m/s=2.73 km/s$

7 0

3 years ago

Read 2 more answers

1. How long would the car in Sample Problem C take to come to a stop from its initial velocity of 20.0 m/s to the west? How far

Brilliant_brown [7]

The time taken, and the distance travelled by the car before stop, will be 5.3 second and -53.3 m west.

<h3>What is the distance?</h3>

The length of the path traveled by the body is known as the distance covered by the body.

Distance is a 1-dimensional phenomenon. its unit is a meter(m). The distance can be found by the product of velocity and time.

The given data in the problem will be

u is the initial velocity =20m/sec

t is the time =?

d is the distance =?

From the Newtons second law;

$\rm F \triangle t = \triangle P \\\\ \triangle t = \frac{\triangle P }{F} \\\\ \triangle t = \frac{m(v_f-v_i)}{F} \\\\ \ \triangle t = \frac{2240 (0-20))}{8410} \\\\ \triangle t = 5.3 \ sec \\\$

The distance travelled before the car stop is,

$\rm \triangle t = \frac{1}{2} (v_f+v_i)\traingle t \\\\ \traingle x = \frac{1}{2} (-20+0)5.3 \\\\\ \triangle x= -53.3 m \ west$

Hence,the time taken, and the distance travelled by the car before stop, will be 5.3 second and -53.3 m west.

To learn more about the distance, refer to the link;

brainly.com/question/989117

#SPJ1

6 0

2 years ago

Why is it important for the path of a circuit to be circular?

vredina [299]

Answer:

When the platform rotates, the rotating mass will travel in a circular path due to the force exerted on it by the string (by way of the tension in the spring). Since it is not possible to have an instantaneous readout of this tension force while the platform is rotating, an indirect measurement of this force will be made using the weight of the static mass as shown and explained below.

4 0

3 years ago

Read 2 more answers