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3 years ago

What is the most current NASA mission on Mars?

2 answers:

8 0

The mars rover bot i.e curiosity has been send latestly by NASA on mars. You can also see some pics on net if are doing some project or something.

ivanzaharov [21]3 years ago

3 0

The most recent NASA mission on Mars is Mars Science Laboratory. Its key component was Curiosity rover.
Rover has celebrated his first landing anniversary yesterday (Martian year, that is 687 Earth days).
Curiosity managed to find traces of water on Mars, so it's possible that there used to be some life on Mars.
Since landing Curiosity moved only about 5 miles form the landing site (it suffered wheel damage).

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Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force

marusya05 [52]

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

$E=W\frac{a}{b}$

From the figure, you can observe that a/b < 1, thus E < W

8 0

3 years ago

The energy transferred from the water in the kettle to the surroundings in 2 hours is

nignag [31]

The Last one ☝️ I think

6 0

3 years ago

Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert

Rainbow [258]

Answer:

a) The flea's speed when it leaves the ground is $v=1.88m/s$

b) The flea move $18cm$ upward while it is pushing off

Explanation:

<u>Knwons</u>

Mass $m=2.1\times 10^{-4} kg$ , Work $W=3.7\times 10^{-4} J$ and Force $F=0.41N$

a) Here we are going to use $W=\frac{1}{2} mv^{2}$ , so $v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s$

a) Here we are going to use $W=mgh$ , so $h=\frac{W}{mg}= \frac{3.7\times 10^{-4} J}{(2.1\times 10^{-4} kg)(9.8m/s^{2})} =0,1797m$ or $18cm$ approx.

7 0

3 years ago

A wave has an amplitude of 0.2 cm.

Anettt [7]

The distance between the resting point and maximum height of the wave is 0.2 cm.

The amplitude is measured from the resting point up to the highest point of the wave.

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3 years ago

Select the correct answer.

Alexeev081 [22]

For the scale od the paraboliczne curviczne Ur

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3 years ago

Read 2 more answers