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Natasha_Volkova [10]
2 years ago
7

What is the unit measurement of the surface? physics​

Physics
2 answers:
svp [43]2 years ago
5 0

Answer:

square metre is the answer

Explanation:

Hope this helps u

Crown me as brainliest:)

Mamont248 [21]2 years ago
3 0
Square meter is the SI unit of surface
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A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe
Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

I_R=\frac{1}{12}ML^2

where

M=3.30\cdot 10^{-2} kg is its mass

L = 0.450 m is its length

Substituting,

I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2

The moment of inertia of the two rings at the beginning is

I_r = 2mr^2

where

m = 0.200 kg is the mass of each ring

r=5.20\cdot 10^{-2} m is their distance from the center of the rod

Substituting,

I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2

So the total moment of inertia at the beginning is

I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2

The initial angular velocity of the system is

\omega_1 = 35.0 rev/min

The angular momentum must be conserved, so we can write:

L=I_1 \omega_1 = I_2 \omega_2 (1)

where I_2 is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

r=\frac{0.450 m}{2}=0.225 m

so the moment of inertia of the rings is

I_r=2(0.200)(0.225)^2=0.0203 kg m^2

and the total moment of inertia is

I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2

Substituting into (1), we find the final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

I_2 = I_R = 5.57\cdot 10^{-4}kg m^2

So using again equation of conservation of the angular momentum:

L=I_1 \omega_1 = I_2 \omega_2

We find the new final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min

7 0
3 years ago
Regardless of their frequency, wavelength, or energy, all electromagnetic waves: A. travel only through the vacuum of space. B.
agasfer [191]
All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 10^5 (300,000) kilometres per second. some electromagnetic waves are part of the visible light spectrum and some do emit harmful radiation, but certainly not all. they travel fine on earth without the vacuum of space too. 
6 0
3 years ago
Does any one know the answer
valina [46]
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6 0
3 years ago
How many neutrons does element x have if it's atomic number is 27 and it's mass number is 74
Makovka662 [10]

Atomic number=Proton count

Atomic mass=Proton count+ neutron count

Neuton Count=Atomic mass-Proton count

Proton count=Atomic number=27

Mass number=74

Neuton count= 74-27=47

3 0
3 years ago
) By observing that the centripetal acceleration of the Moon around the Earth is ac = 2.7 × 10-3 m/s2, what is the gravitatonal
Sedbober [7]

Answer:

G = 6,786 10⁻¹¹ m³ / s² kg

Explanation:

The law of universal gravitation is

         F = G m M/ r²

Where G is the gravitational constant, m and M are the masses of the bodies and r is the distance from their centers

Let's use Newton's second law

         F = m a

The acceleration is centripetal

          a = a_{c}  

We replace

         G m M / r² = m  a_{c}  

         G =  a_{c}   r² / M

Let's replace and calculate

         G = 2.7 10⁻³ (3.88 10⁸)² / 5.99 10²⁴

         G = 6,786 10⁻¹¹ m³ / s² kg

Let's perform a dimensional analysis

[N m²/kg²] = [kg m/s²   m² / kg²] = [m³ / s² kg]

4 0
3 years ago
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