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lorasvet [3.4K]
3 years ago
11

Metre is the unit of dash.​

Physics
2 answers:
luda_lava [24]3 years ago
8 0

Answer:

From the meter, several other units of measure are derived such as the: unit of speed is the meter per second (m/s).

...

Units of Length

10 millimeters (mm) = 1 centimeter (cm)

10 decimeters = 1 meter (m)

10 decimeters = 1000 millimeters

10 meters = 1 dekameter (dam)

djverab [1.8K]3 years ago
7 0

Answer:

Length is the unit of metre

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Determine la densidad de un objeto cuya masa es de 10 kg y tiene un volumen de 120 cm3
NNADVOKAT [17]

Answer:

dttydghcghcghdytdftygfvghxk,ghc

Explanation:

7 0
2 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
A cart on an air track is moving at 0.5m/s when the air issuddenly turned off. THe cart comes to rest in 1m. The experimentis re
Licemer1 [7]

Answer:

D = 4 m

Explanation:

Speed of cart in air track v₁ = 0.5 m/s

Speed of cart moved when air is turned off v₂= 1 m/s

The distance travelled by the cart is d₁ = 1 m

Work done (W) = F x d

Work done is equal to the kinetic energy

F x d = 1/2mv²

velocity is directly proportional to distance

therefore,

v₁²/ v₂²  = d₁ / d₂

d₂ = d₁v₂² / v₁²

= 1 m x (1 m /s)² / (0.5 m/s)²

= 4 m

5 0
3 years ago
Unseen objects in our galaxy have been found by the bending effect they have on more distant stars' light passing near them. The
Dimas [21]

Answer:

a

Explanation:

5 0
3 years ago
A potter's wheel is a uniform disk of mass of 10.0 kg and radius 20.0 cm. A 2.0-kg lump of clay, roughly cylindrical with radius
Lera25 [3.4K]

Answer:

b. 29.2 rev/min

Explanation:

  • Assuming no external torques acting during the process, total angular momentum must be conserved, as follows:

       L_{0} = L_{f}  (1)

  • The initial angular momentum L₀, can be expressed as follows:

        L_{0} = I_{0} * \omega_{0} (2)

        where I₀ = initial moment of inertia = moment of inertia of the disk +

        moment of inertia of the cylinder and ω₀ = initial angular velocity  =

       30.0 rev/min.

  • Replacing by the values, we get:I_{0} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{c} *r_{c} ^{2}  = 0.2 kg*m2 +9e-4 kg*m2 = 0.2009 kg*m2 (3)⇒ L₀ = I₀* ω₀ = 0.2009 kg*m² * 30.0 rev/min = 6.027 kg*m²*rev/min
  • The final angular momentum can be written as follows:

       L_{f} = I_{f} * \omega_{f} (4)

       where If = final moment of inertia = moment of the inertia of the solid

      disk + moment of  inertia of the clay flattened on a disk, and ωf = final

      angular velocity.

  • Replacing by the values, we get:

   I_{f} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{fd} *r_{fd} ^{2}  = 0.2 kg*m2 +6.4e-3 kg*m2 = 0.2064 kg*m2 (5)

       ⇒ Lo =Lf = If*ωf

  • Replacing (2) in (1), and solving for ωf, we get:

        \omega_{f} = \frac{L_{o}}{I_{f} } = \frac{6.027kg*m2*rev/min}{0.2064kg*m2} = 29.2 rev/min (6)

3 0
3 years ago
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