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Nuetrik [128]
3 years ago
8

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×103m/s, given the collision lasts 6.00×10–8s.
Physics
1 answer:
Eva8 [605]3 years ago
5 0

Answer:

6666.67 Newtons

Explanation:

The formula F=ma (force is equal to mass multiplied by acceleration) can be used to calculate the answer to this question.

In this case:

  • mass= 0.1mg= 1*10^-7 kg
  • velocity= 4.00*10^3 m/s
  • time= 6.00*10^-8 s

Using velocity and time, acceleration can be calculated as:

  • a= 6.667*10^10 m/s²

Substituting these values into the formula F=ma, the answer is:

  • F= (1*10^-7)kg * (6.667*10^10) m/s²
  • F= 6666.67 Newtons of force
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What is the speed of a wave that has a frequency of 45 Hz and a wavelength of 0.1 meters?
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Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir
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Answer:

7560 Joules

Explanation:

m_1 = Mass of first car = 1.5\times 10^5\ kg

m_2 = Mass of second car = 2\times 10^5\ kg

u_1 = Initial Velocity of first car = 0.3 m/s

u_2 = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s

Energy lost is

\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J

The Energy lost in the collision is 7560 Joules

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2 years ago
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3 years ago
A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin
blsea [12.9K]

Answer:

The stress is calculated as 1.003\times 10^{9}\ Pa

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = 0.560\times 10^{- 3}\ m

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, \Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m

Now,

The stress in the wire is given by:

Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = 25.2\times 9.8 = 246.96\ N

Cross  sectional Area, A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}

Using these values in eqn (1):

\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa  

8 0
2 years ago
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