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Musya8 [376]
3 years ago
10

Define :density٬archimedes principle

Physics
1 answer:
olga2289 [7]3 years ago
5 0

Answer:

density is defined as the amount of mass contained in unit volume of a body .its si unit is kg/m*3

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In your own words, explain how solar heating works
mestny [16]

Answer:

active solar heating systems use solar energy to heat a fluid either liquid or air and then transfer the solar heat directly to the interior space or to a storage system for later use. If the solar system cannot provide adequate space heating, an auxiliary or back-up system provides the additional heat.

hope this helps : )

4 0
3 years ago
Read 2 more answers
A car is moving at a rate of 72 km/hr .How far does car move when it stop after 4 seconds? ​
Zanzabum

Answer:

Assuming it starts at 72 kmph and hits a dead stop: Divide 72 by 60 for distance per minute. So, 1.2km per minute. 1.2km is 1200m and 4 seconds is one fifteenth of a minute.

Explanation:

5 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
2 years ago
A 3.6kg mass is accelerated at 2.5m/s. Calculate the resultant force<br> acting on it.
AleksandrR [38]
Force = mass*acceleration so
3.6*2.5 =9 Newtons
5 0
3 years ago
A basketball is thrown up into the air. It is released with an initial velocity of 8.5 m/s. How long does it take to get to the
ehidna [41]
<h2>It takes 0.867 seconds to get to the top of its motion</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 8.5 m/s

     Final velocity, v = 0 m/s    - At maximum height

     Time, t = ?

     Acceleration , a = -9.81 m/s²

     Substituting

                      v = u + at  

                      0 = 8.5 + -9.81 x t

                      t = 0.867 s

  It takes 0.867 seconds to get to the top of its motion

4 0
3 years ago
Read 2 more answers
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