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2 years ago

Why is mass a better unit for measuring matter then weight

Physics

2 answers:

ivanzaharov [21]2 years ago

3 0

Answer:

your weight changes slightly when you fly in an airplane

Explanation:

Technically, because gravity is weaker the farther you are from a gravitational source, your weight changes slightly when you fly in an airplane. It would change even more if you're in space or on a planet with a different mass than Earth.

VashaNatasha [74]2 years ago

3 0

Explanation:

because gravity is weaker the farther you are from a gravitational source

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A fin whale is swimming at the speed of 35 km/h how many hours will it take to swim 7 km

Aleks04 [339]

Answer:

0.2 hours

Explanation:

$\frac{7}{35} = 0.2$

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2 years ago

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3 years ago

The unit for energy is the joule (j). If the energy content of an object is 1,000 J, but then later is found to be 900 J, what i

Dennis_Churaev [7]

Answer: the object transferred some of its energy to its surroundings.

Explanation:

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2 years ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

2 years ago

A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on

makkiz [27]

Answer:

a) 19.4 m/s

b) 19 m/s

Explanation:

a) In the given question,

the potential energy at the initial point = Ui = 0

the potential energy at the final point = Uf = mgh

the kinetic energy at the initial point = Ki = 1/2 mv₀².

the kinetic energy at the final point = Kf = 0

work done by air= Ea= fh = 0.262 N

Now, using the law of conservation of energy

initial energy= final energy

Ki +Ui = Kf + Uf +Ea

1/2 mv₀² + 0 = 0 + mgh + fh

1/2 mv₀² = mgh + fh

h = v₀²/ 2g (1 +f/w)

calculate m

m= w/g = 5.29 /9.8

= 0.54 kg

h = 20 ²/ (2 x9.80) x (1 0.265/5.29)

h = 19.4 m.

b) 1/2 mv² + 2fh = 1/2 mv₀²

Vg = 19 m/s

6 0

2 years ago