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iogann1982 [59]
4 years ago
10

What is the most current NASA mission on Mars?

Physics
2 answers:
JulsSmile [24]4 years ago
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]4 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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Yoon Ki investigates electromagnetic induction by moving a bar magnet into a coil of wire. His experimental setup is shown.
Karolina [17]

A change that would most improve his results would be; D) Connecting the galvanometer to the coil

4 0
3 years ago
Read 2 more answers
Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 2 m. What is your acceleration?
likoan [24]

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed) = (15 m/s - 7 m/s) = 8 m/s

time for the change = 2 minutes = 120 seconds

Acceleration = (8 m/s) / (120 seconds)

Acceleration = 0.067 m/s²

7 0
3 years ago
A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor
Mumz [18]

Answer:

The maximum induced emf in the rotating coil  = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

emf_max = NBAω ; if (A = l × w) , we have:

emf_max  = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

emf_max  = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf =   44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e emf_max = N\frac{d∅}{dt}

since emf_max = 29.66 and N = 44; we have:

29.66 =  44\frac{d∅}{dt}

\frac{d∅}{dt} = \frac{29.66}{44}

= 0.674 Wb/s

5 0
3 years ago
A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must i
kiruha [24]

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

7 0
3 years ago
Define the term overload.
ryzh [129]

Answer:

Verb. Load with too great a burden or cargo.

"Both boats were overloaded and low in the water"

Noun. An excessive load or amount.

"an overload of stress"

Explanation:

Similar words are strain, excess, and overburden.

Have a good day and stay safe!

4 0
3 years ago
Read 2 more answers
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