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Fittoniya [83]
1 year ago
6

Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87×106km fr

om the earth and traveling at 1.20×104km/h relative to the earth.At this time, what were the following. (a) the spacecraft's kinetic energy relative to the earth J (b) the potential energy of the earth-spacecraft system J
Physics
1 answer:
Andreas93 [3]1 year ago
4 0

Answer:

3494444444.44444 J

-87077491.39453 J

Explanation:

M = Mass of Earth = 6.371\times 10^{6}\ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

R = Radius of Earth = 6.371\times 10^{6}\ m

h = Altitude = 2.87\times 10^9\ m

m = Mass of satellite = 629 kg

v = Velocity of spacecraft = 1.2\times 10^4\ km/h

The kinetic energy is given by

K=\frac{1}{2}629\times \left(1.2\times 10^4\times \dfrac{1000}{3600}\right)^2\\\Rightarrow K=3494444444.44444\ J

The spacecraft's kinetic energy relative to the earth is 3494444444.44444 J

Potential energy is given by

U=-\dfrac{GMm}{R_e+h}\\\Rightarrow U=-\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 629}{2.87\times 10^9+6.371\times 10^{6}}\\\Rightarrow U=-87077491.39453\ J

The potential energy of the earth-spacecraft system is -87077491.39453 J

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Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h
ruslelena [56]

Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

Y+F=600....(I)

X=T.....(II)

On taking moment about base

X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0

Put the value into the formula

X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0

3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30

1.598T=450-1.5(600-F)....(III)

We need to calculate the force for ladder

2F=600\trimes  2

F=600\ N

We need to calculate the tension in the rope

From equation (3)

1.598T=450-1.5(600-600)

1.598T=450

T=\dfrac{450}{1.598}

T=281.60\ N

Hence, The tension in the rope is 281.60 N.

7 0
2 years ago
Read 2 more answers
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
1 year ago
Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.
babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

d\sin \theta =m\lambda

For Destructive interference

d\sin \theta =(m+\frac{1}{2})\lambda

where \lambda =wavelength

d=slit\ width

m=order of maxima and minima

for second order maxima i.e. m=2

For smallest separation taking \lambda =400 nm, \theta =90^{\circ}

d\sin 90=2\times 400\times 10^{-9}

d=0.8\times 10^{-6}

d=0.8\mu m

   

6 0
1 year ago
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

\theta_{net} = 0.108*2\pi=0.216\pi

Therefore the angle of the speck at a time 8.1s is 0.216\pi

4 0
2 years ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
1 year ago
Read 2 more answers
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