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kotegsom [21]
3 years ago
15

A Ping-Pong ball has a diameter of 2.55 cm and average density of 0.116 g/cm3 . What force would be required to hold it complete

ly submerged under water?
Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

F = 0.075 N

Explanation:

given,

diameter of ping pong ball = 2.55 cm

average density of the ball = 0.116 g/cm³

Force require to hold the ball completely submerged= ?

Volume of the ball

V = \dfrac{4}{3} \pi r^3

V = \dfrac{4}{3} \pi (\dfrac{2.55}{2})^3

    V = 8.63 cm³

density of water = 1 g/cm³

force required to keep the ball submerged is equal to

F = \rho_w V_w g - \rho_b V g

F = 1\times 8.63\times g - 0.116 \times 8.63 \times g

F = 7.63 x 9.8 x 10⁻³

F = 0.075 N

Force required to keep the ball inside the water is equal to 0.075 N

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6 0
2 years ago
The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0
2 years ago
You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh
Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

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r₁= 10 m  , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

\beta =10\ log\dfrac{I}{10^{-12}}

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20=10\ log\dfrac{I_1}{10^{-12}}

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

\beta _2=10\ log\dfrac{I_2}{10^{-12}}

60=10\ log\dfrac{I_1}{10^{-12}}

10⁶ x  10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity  ,A=Area

\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}

\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}

r₂=0.1 m

4 0
3 years ago
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