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2 years ago

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65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight

of the body in air.

1 answer:

Anna35 [415]2 years ago

5 0

Answer:

Given, Apparent weight(W₂)=4.2N

Weight of liquid displaced (u)=2.5N

Let weight of body in air = W₁

Solution,

U=W₁-W₂

W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

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3 years ago

Your high-fidelity amplifier has one output for a speaker of resistance 8 Ω. How can you arrange two 8-Ω speakers, one 4-Ω speak

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Answer:

(a) 8Ω (b) Ratio = Parra/P8 ohm = 1

Explanation:

Solution

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(a) How can two 8-Ω speakers be arranged, when one = 4-Ω speaker, and one =12-Ω speaker

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The Lower arm is : 12 ohm, 4 ohm

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(b) compare your arrangement power output of with the power output of a single 8-Ω speaker

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3 years ago

Mary and her younger brother Alex decide to ride the 26 ft diameter carousel at the State Fair. Mary sits on one of the horses i

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Answer:

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Explanation:

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$r_A$ = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

$\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A$

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3 years ago

A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

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The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>

Mass of the box, m = 3.1 kg
Distance moved by the box, d = 2.5 m
Coefficient of friction, = 0.35
Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>

Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

$W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)$

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

$W = (0) d \times cos(30)\\\\W = 0 \ J$

The work done by the force of gravity is calculated as follows;

$W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J$

The work done on the box by the normal force is calculated as follows;

$W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J$

Learn more about work done here: brainly.com/question/8119756

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2 years ago