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Verdich [7]
3 years ago
9

Mars completes one orbit around the Sun in approximately two Earth years. Mars orbits at an average distance to the Sun of about

1.5 AU, and Mars' mass is about 1/10 of the Earth's mass. Therefore Mars' orbital speed is _____ the orbital speed of the Earth.
Physics
1 answer:
satela [25.4K]3 years ago
3 0

Answer:

about 0.8 times

Explanation: MARS is a planet named after a Roman God of War, it is the fourth planet from the sun and the Second smallest planet following Mercury.

The orbital Speed of a planet is the speed with which the planet uses to completely go round the Sun.

The orbital Speed of the Earth is 29.78kilmeters per Second.

The orbital Speed of Mar is 0.8*29.78kilometers per second/1

=23.824kilmeters per second.

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Calculate the work done when a 100-W lightbulb is lit for 30 seconds.
Mila [183]

Answer:

3000 J

Explanation:

8 0
2 years ago
4. A woman releases one egg every month for 37 years. Calculate how many
BigorU [14]

so, 444 eggs would have been released in 37yrs

5 0
2 years ago
The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed
LenaWriter [7]

Answer:

83.67 m/s

Explanation:

Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km     1000 m         83.67m

-----------  X ----------  =  ---------------   the km will cancel out and you are left

 12.0 s          1 km              s              with m/s

6 0
3 years ago
5. A person fishing from a pier observes that four wave crests pass by in 7.0 s and estimates that the distance between two succ
TiliK225 [7]

Answer:

v= 1.71 m/s

Explanation:

Given that

Distance between two successive crests = 4.0 m

 λ = 4 m

T= 7 sec

T is the time between 3 waves.

3 waves = 7 sec

1 wave = 7 /3 sec

So t= 7/3 s

We know that frequency f

f= 1/t= 3/7 Hz

Lets take speed of the wave is v

v= f λ

f=frequency

λ=wavelength

v= 3/7 x 4 = 12 /7

v= 1.71 m/s

3 0
3 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
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