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allsm [11]
2 years ago
15

___ radars show everything in fhe sky from servere weather to birds and mosquitoes to tornado debris

Physics
2 answers:
Sergio [31]2 years ago
6 0

Answer:Geostationary

Explanation:Geo=Geographic=Climate=Enviroment=Animals

olganol [36]2 years ago
6 0

Answer:

doppler

Explanation:

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This problem has been solved!
AleksAgata [21]

Answer:the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward.

Explanation:

When a ball is tossed vertically upwards, it reaches its maximum point. This maximum point is known as the maximum height of the ball. At this maximum height, its velocity is zero, its acceleration is directed downwards and the force of gravity acting in the ball is directed downwards towards the earth.

6 0
3 years ago
WILL UPVOTE EVERY ANSWER! MULTIPLE CHOICE QUESTION!
Nitella [24]
B hardness

Giddy UP!!!!!
4 0
3 years ago
The diagram below shows a light ray from a pencil hitting a mirror.
max2010maxim [7]

Answer:

1

Explanation:

Its 1 since the shape of it is more downwards which results the reflection going downwards.

4 0
3 years ago
Read 2 more answers
How much work must be done to raise a 1100kg car 2m above the ground?​
Xelga [282]

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

4 0
3 years ago
Read 2 more answers
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
4 years ago
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