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brilliants [131]
3 years ago
5

According to Newton's third law of motion, if this snowy owl wants to increase its altitude, what must it do?

Physics
2 answers:
erma4kov [3.2K]3 years ago
6 0

Answer: Correct answer is A.

For USATestprep.

Irina-Kira [14]3 years ago
3 0

Answer:

The owl must push down with its wings in order to give its body lift.

Explanation:

Got this exact question on USATestPrep, hope it helps.

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What is the equation for the potential energy stored in a spring when it is stretched or compressed?
gregori [183]

Answer:

if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).

8 0
3 years ago
Acceleration is positive when you are speeding up.
Vikki [24]

Answer:

B Negative

Explanation:

Its negative because when your going 25 mph your moving faster and when your coming up to the stop sign you will start to slow down going 10 mph then 0 mph when you reach the stop sign so it will be negative.

8 0
2 years ago
How are mathematics and science related?
____ [38]
The relationship between math and science is very complicated, yet at the same time very simple. In fact all scientific equations are expressed into some form of mathematical equations. Science is a body of knowledge about the Universe. Mathematics is a language that can describe relationships and change in relationships in a rational way. Science generally uses mathematics as a tool to describe science and vice versa.
6 0
3 years ago
What is tan for the given triangle?
Rama09 [41]

Answer:

tan is 15 for that triangle

5 0
2 years ago
A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph
Marina86 [1]

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
where: ω = 2.87 rad/s 

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
12.64 J = 2.3 kg * 9.8m/s² * h 
h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ 
Θ = arccos((1-0.29)/1) = 44.77 º 

8 0
3 years ago
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