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olga_2 [115]
3 years ago
13

When a space shuttle was launched, the astronauts onboard experienced an acceleration of 29.0 m/s2 . If one of the astronauts ha

d a mass of 60.0 kg, what net force in newtons did the astronaut experience?
Physics
1 answer:
yan [13]3 years ago
6 0
F = m*a when F is force
m is mass and a is acceleration
F = 60 kg * 29 m/s2
F = 1740 N
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Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra
sergey [27]

Answer:

\frac{di}{dt}  = 7.31 \  A/s

Explanation:

From the question we are told that  

     The  number of turns is  N =  450 \  turns

      The  radius is  r =  1.17 \ cm =  0.0117 \ m

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  e  =  8.20 *10^{-6} \  V/m

Generally according to Gauss law

        \int\limits { e } \, dl  =  \mu_o *  N  *  \frac{di}{dt }  *  A

=>    e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  A

Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                A =  \pi r ^2

=>      e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  \pi r^2

=>       \frac{di}{dt}  =  \frac{2e * x  }{\mu_o * N  * r^2}ggl;

Here  \mu_o is the permeability of free space with value

          \mu_o  =  4\pi * 10^{-7} \  N/A^2

=>     \frac{di}{dt}  =  \frac{2 *  8.20*10^{-6} *  0.0345  }{ 4\pi * 10^{-7} * 450  * (0.0117)^2}

=>      \frac{di}{dt}  = 7.31 \  A/s

6 0
4 years ago
Chemistry<br> determine the pressure
sashaice [31]
I think it's 1.03412969 or 1.03
8 0
4 years ago
When is the world going to end? Give me your opinions and the reasons why you think that.
Novay_Z [31]

Answer:

probably when the next pandemic happens

5 0
3 years ago
The index of refraction of a clear plastic is listed as 1.89 in the book, but you measured the angle of incidence 63.5° and the
lakkis [162]

Answer:

<h2>index of refraction = 1.69</h2><h2>percentage error = 10.58%</h2>

Explanation:

According to Snell's law, the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. The constant is known as the refractive index.

Mathematically \frac{sin i}{sin r} = n

i = angle of incidence measured = 63.5°

r = angle of refraction measured = 32°

n = refractive index

n = \frac{sin 63.5^{0} }{sin 32^{0} } \\n = \frac{0.8949}{0.5299}\\ n = 1.69

The index fraction calculated is approx. 1.69.

If the index of refraction of a clear plastic as listed in the book is 1.89 and the calculated is 1.69, the percentage error will be calculated as thus;

%error = \frac{1.89-1.69}{1.89} * 100

%error = \frac{0.2}{1.89}*100

%error  = \frac{20}{1.89}

%error  = 10.58%

6 0
4 years ago
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