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Inessa05 [86]
3 years ago
11

How big is the international space station?

Physics
1 answer:
oee [108]3 years ago
3 0
The dimensions of the completed ISS research facility will be approximately 356 feet (109 meters) by 240 feet (73 meters), or slightly larger than a football field. the ISS weighes around 450 tons (408,000 kg), or 450 times the weight of an average car.
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What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?
sveta [45]
Momentum = (mv). 
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>
8 0
3 years ago
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

104.4685 = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

f = 15.76×10³⁴ Hz

Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

Frequency (f) = 15.76×10³⁴ Hz

Wavelength (λ) =?

v = λf

43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

8 0
3 years ago
An attack helicopter is equipped with a 20- mm cannon that fires 88.7 g shells in the forward direction with a muzzle speed of 9
Svetradugi [14.3K]
The impulse imparted to the shells equals the change in the momentum:
Fav*(Delta t)= Delta m*v.
The mass change is
Delta m= n*m= (89.9shells)*(88.7g)=7.97Kg
So the average force is
F=((v)*(Delta m))/t= ((929)*(7.97))/4.84=1529.78 N
Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.



4 0
3 years ago
Calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the mean radius of the earth at
valentinak56 [21]

Answer: M = 5.98\times 10^{24} kg

Explanation:

We know that force acting on an object due to Earth's gravity on the surface is given by:

mg = G\frac{Mm}{r^2}\\ \Rightarrow g = \frac{GM}{r^2}

where g is the acceleration due to gravity, r would be radius of Earth, M is the mass of Earth and G is the gravitational constant.

It is given that at pole, g = 9.830 m/s² and r = 6371 km = 6371 × 10³ m

\Rightarrow M = \frac{g\times r^2}{G}

M = \frac{9.830 m/s^2 \times (6371 \times 10^3 m )^2}{6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}}

\Rightarrow M = 5.98\times 10^{24} kg

Hence, Earth's mass is 5.98\times 10^{24} kg

5 0
3 years ago
A<br> Н20= C=1 h=4 o=4<br> 2.<br> H3PO4 +<br> КОН<br> KзРО І
maks197457 [2]

Answer:

Н20= C=1 h=4 o=4

Explanation:

7 0
3 years ago
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