Answer:
Juno scientific payload includes:
- A gravity/radio science system (Gravity Science)
- A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
- A vector magnetometer (MAG)
- Plasma and energetic particle detectors (JADE and JEDI)
- A radio/plasma wave experiment (Waves)
- An ultraviolet imager/spectrometer (UVS)
- An infrared imager/spectrometer (JIRAM)
Explanation:
Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.
Answer:
trees
Explanation:
referring to the tree to prove his/her point.
The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, 
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:
where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s
Substituting,
Learn more about projectile motion:
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Answer:
<h2>
d₂ = 3d</h2><h2>
The diameter of the second wire is 3 times that of the initial wire.</h2>
Explanation:
Using the formula for calculating the resistivity of an object to find the diameter.
Resistivity P = RA/L
R is the resistance of the material
A is the cross sectional area
L is the length of the material
Since A = πd²/4
P = R( πd²/4)/L
P = Rπd²/4L ... 1
If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;
P₂ = (R/9)A₂/L₂
P₂ = (R/9)(πd₂²/4)/L₂
P₂ = (Rπd₂²/36)/L₂
P₂ = (Rπd₂²)/36L₂
Since the length and resistivity are the same;
P = P₂ and L =L₂
Equating 1 and 2;
Rπd²/4L = (Rπd₂²)/36L₂
Rπd²/4L = (Rπd₂²)/36L
d² = d₂²/9
d₂² = 9d²
Taking the square root of both sides;
√d₂² = √9d²
d₂ = 3d
Therefore the diameter of the second wire is 3 times that of the initial wire
Answer:
The diagram assigned B
explanation:
Check the direction of the two vectors, their resultant must be in the same direction.
