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Lynna [10]
3 years ago
13

PLZ HELP GUYS PLZ!!!!! The answer is not motion, particles.

Physics
1 answer:
harina [27]3 years ago
4 0

Answer:

Energy

Explanation:

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A 50-foot flagpole is at the entrance of a building that is 300 feet tall. If the length of the flagpole's shadow is 30 feet at
natka813 [3]
First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.

To solve this, you will want to create a proportion.

Flagpole height/flagpole shadow=building height/building shadow

Therefore, it should look like this:
50/30= 300/x

Solve for x:
50x=9,000
X= 180 feet

YOUR ANSWER IS C.
7 0
3 years ago
A lunar exploration vehicle was created by a research team. It weighs 3,000 kg on the earth. It needs an acceleration of 10 m/s2
Arada [10]
Answer = 30,000 N

EXPLANATION

Applying Newton’s second of law of motion, which in summary, states that t<span>he acceleration of an object... is directly proportional to the magnitude of the net force... and inversely proportional to the mass of the object.
</span><span>
Therefore, Force = Mass * Acceleration
F = ma

Mass, m = </span><span>3,000 kg
</span>Acceleration, a = <span>10 m/s</span>²<span>
</span>Force, F = 3,000 × 10
= 30,000 N
6 0
3 years ago
I lost 1 point on this and I don’t know what I did wrong.
Viktor [21]

Answer:

i think its because u gave the almost every answer the same exact thing. all the questions have different ways of moving which means different forces for each one i hope this helps :)

Explanation:

4 0
4 years ago
Read 2 more answers
What are some of the objects found in space and their characteristics?
RideAnS [48]

Answer:

meteor

Explanation:

big rock or metal maybe

6 0
3 years ago
Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m
OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

E_m = \frac{1}{2} m\omega_1^2A_1^2

The energy of the system having mass 2m is,

E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2

For the two expressions mentioned above remember that the variables mean

m = mass

\omega =Angular velocity

A = Amplitude

The energies of the two system are same then,

E_m = E_{2m}

\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2

\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}

Remember that

k = m\omega^2 \rightarrow \omega^2 = k/m

Replacing this value we have then

\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}

But the value of the mass was previously given, then

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}

\frac{A_1}{A_2} = 1

Therefore the ratio of the oscillation amplitudes it is the same.

5 0
3 years ago
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