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2 years ago

What is the mass of the object being measured?

Physics

1 answer:

ioda2 years ago

6 0

Answer:

m = 375 [gram]

Explanation:

A triple Beam balance is an instrument very easy to use, since we only have to perform the arithmetic sum of each of the weights that are recorded in each beam

m = 300 + 70 + 5 = 375 [gram]

For a better understanding, the following image is attached, with values on each beam, which should be read.

The largest mass is in the indicator of 100 [gram], the second mass is in the indicator of 20 [gram] and the third is in the indicator of 5.8 [gram]. Thus the arithmetic sum corresponds to:

M= 100 + 20 + 5.8 = 125.8 [gram]

Note: it is important that when the instrument is in balance, the opposite end of the beam should indicate a position of zeros.

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2 years ago

At any angular speed, a certain uniform solid sphere of diameter D has half as much rotational kinetic energy as a certain unifo

Trava [24]

Answer:

d. 6/5 M.

Explanation:

let the angular velocity = $\omega$

the radius of sphere (r) = $\frac{D}{2}$

moment of inertia of solid sphere $I_s$ = $2/5 M(\frac{D}{2})^2$

rotational K.E of solid sphere= $\frac{1}{2}* I_s* \omega ^2$

= $\frac{1}{2} * 2/5 M(\frac{D}{2})^2 * \omega^2$

we represent the mass of hallow sphere= $M_{hs}$

moment of inertia of solid sphere $I_{ss}$ = $2/3 M_{hs}(\frac{D}{2})^2$

rotational K.E of hollow sphere= $\frac{1}{2}* I_{ss} * \omega^2$

= $1/2 *2/3 M_{hs}(\frac{D}{2})^2* \omega^2$

NOW,the kinetic energy of solid sphere= 1/2 kinetic energy of hallow sphere

= $1/2 [2/5 M(D/2)^2 ]\omega^2= 1/2 (1/2 (2/3 M_{hs}(D/2)^2 )\omega^2 }$

2/5M= 1/3 $M_{hs}$

$M_{hs}$ = 3(2/5M)

$M_{hs}$ = 6/5 M

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3 years ago

Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mas

timurjin [86]

Answer:

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

= 0.135 J + 0.11220 J

= 0.2472 J

≅ 0.247 J = 247 mJ

5 0

2 years ago