Answer:
B.It is a satellite that collects data about rain and snow
C.Its orbit covers 90 percent of Earth’s surface
F.The sensors measure microwaves
Answer:
False statement = There must be a non-zero net force acting on the object.
Explanation:
An object is moving at a constant speed along a straight line. If the speed is constant then its velocity must be constant. We know that the rate of change of velocity is called acceleration of the object i.e.

a = 0
⇒ The acceleration of the object is zero.
The product of force and acceleration gives the magnitude of force acting on the object i.e.
F = m a = 0
⇒ The net force acting on the object must be zero.
So, the option (a) is not true. This is because the force acting on the object is zero. First option contradicts the fact.
Answer:
The coefficient of kinetic friction 
Explanation:
From the question we are told that
The length of the lane is 
The speed of the truck is 
Generally from the work-energy theorem we have that

Here N is the normal force acting on the truck which is mathematically represented as
is the change in kinetic energy which is mathematically represented as
=>
=>

=> 
=> 
Answer:
a)
= 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo = 
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s
Answer:
mass = 0.18 [kg]
Explanation:
This is a classic problem where we can apply the definition of density which is equal to mass over volume.
![density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]](https://tex.z-dn.net/?f=density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5C%5C%5Cwhere%3A%5C%5Cvolume%20%3D%201%20%5Bm%5E3%5D%5C%5Cdensity%20%3D%200.18%5Bkg%2Fm%5E3%5D)
mass = 0.18*1
mass = 0.18 [kg]