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Katen [24]
3 years ago
13

PLZZ HELP ME ASAP!! If you get it correct I will give brainliest and 50 POINTS!!

Physics
2 answers:
pav-90 [236]3 years ago
6 0

Answer:

cccccccccccccccccccccc

julsineya [31]3 years ago
4 0
Cccccccccccccccccccccc
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Which of the following is not a use for a weather radar?
Elodia [21]
Weather radar is used to locate precipitation<span>, calculate its motion, and estimate its type rain, snow, </span>hail etc

so, d.measuring temperature <span>is not a use for a weather radar

</span>effect of global warming-<span>heavier rainfall and flooding</span>
5 0
3 years ago
Read 2 more answers
A bag of rocks has a mass of 16.4 kg what is it weight here on the earth
Alex73 [517]
Answer
160,72N
Explanation
W=mg
=(16.4)(9.8)
6 0
1 year ago
The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0
3 years ago
A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the
Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

E_i =E_f

so:

Mgh = \frac{1}{2}IW^2 +\frac{1}{2}MV^2

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = \frac{1}{2}MR^2

I = \frac{1}{2}(5kg)(2m)^2

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = \frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2

Replacing the data:

(5kg)(9.8)(0.3m) = \frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2

solving for V:

(5kg)(9.8)(0.3m) = V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))

V = 2 m/s

8 0
3 years ago
The force on a dropped apple hitting the ground depends upon
yulyashka [42]

Answer:

F=ma

Explanation:

Force = mass * acceleration

5 0
2 years ago
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