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Jlenok [28]
3 years ago
11

Instruments on board the trmm (tropical rainfall measuring mission) satellite show 3d images of very tall rain columns called __

___.
Physics
1 answer:
Ket [755]3 years ago
7 0
These columns are called hot towers
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Help!! <br> A. 0.5 s <br> B. 1.0 s<br> C. 1.5 s<br> D. 2.0 s
algol13

Answer:

c.    1.5 s

Explanation:

i had this and i got it right

3 0
3 years ago
A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one
Fittoniya [83]
I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)
8 0
3 years ago
Read 2 more answers
Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k
Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

\frac{GM_{ns}}{R^2}= \omega^2 R

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0
3 years ago
You make the following measurements of an object 42kg and 22m3 what would the objects density be
miv72 [106K]
The density of the object is approximately 1.91 kg per m³.

42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship $$density = mass / volume$$ to solve for the object's density.

42 kg \div 22 m³ \approx 1.91 kg per m³.
4 0
3 years ago
Read 2 more answers
The route followed by a hiker consists of three displacement vectors A with arrow, B with arrow, and C with arrow. Vector A with
kotykmax [81]

Answer:

D_{B}=1173.98m\\D_{C}=675.29m

Explanation:

If we express all of the cordinates in their rectangular form we get:

A = (1404.77 , 655.06) m

B = A + ( -D_{B} *sin(41) , -D_{B} * cos(41) )

C = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )

Since we need C to be (0,0) we stablish that:

C = (0,0) = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )

That way we make an equation system from both X and Y coordinates:

A_{x} + B_{x} + C_{x} = 0

A_{y} + B_{y} + C_{y} = 0

Replacing values:

1404.77 - D_{B}*sin(41) - D_{C}*cos(20) = 0

655.06 - D_{B}*cos(41) + D_{C}*sin(20) = 0

With this system we can solve for both Db and Dc and get the answers to the question:

D_{B}=1173.98m

D_{C}=675.29m

7 0
3 years ago
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