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Umnica [9.8K]
3 years ago
8

Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma

ller average density than box B. When the boxes are released, they accelerate upward to the surface. Which box has the greater acceleration?
Physics
2 answers:
Novay_Z [31]3 years ago
5 0

Answer:

Box A

Explanation:

Let mass of each wooden box =m

Density of box A=\rho_A

Density of box B =\rho_B

\rho_A

Density,\rho=\frac{m}{V}

V_A=\frac{m}{\rho_A}

V_B=\frac{m}{\rho_B}

Density of inversely proportional to volume.

The volume of box with smaller density is larger than the volume of box with large density.

V_A>V_B

When the boxes are submerged under water.

Then, the buoyant force=\rho_l Vg

Where V=Volume of displace fluid.

\rho=Density of fluid

Buoyant force of box A=\rho_lV_Ag

Buoyant force of box  B=\rho_lV_Bg

Force=Buoyant force

ma_A=\rho_lV_Ag

ma_B=\rho_lV_Bg

Acceleration is directly proportional to volume.

Therefore,the box with large volume has greater acceleration.

Hence, the box A has greater acceleration.

Crank3 years ago
3 0

Answer:

Explanation:

Let boxes A and B has equal mass of M and volume be Va and Vb

their density be da and db

given da < db

M / Va < M/Vb

Va> Vb

Buoyant force on box A in water = weight of displaced water

= volume of displaced water x density of water x g

= Va x d x g , d is density of water

Aa  = acceleration on box A

= buoyant force on A  / mass of A  

= Va x d x g / M

Similarly acceleration on box B

Ab = Va x d x g / M

Since Va > Vb

Aa > Ab

So box A will have greater acceleration.

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wariber [46]
<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = \frac{V^{2} }{R}

=> R = \frac{V^{2} }{P}             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = \frac{V^{2} }{P}    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = \frac{12.0^{2} }{4}

R₁ = \frac{144}{4}

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = \frac{V^{2} }{P}    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = \frac{12.0^{2} }{4}

R₂ = \frac{144}{4}

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

\frac{1}{R_{X} } = \frac{1}{R_1} + \frac{1}{R_2}       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

\frac{1}{R_X} = \frac{1}{36} + \frac{1}{36}

\frac{1}{R_X} = \frac{2}{36}

Rₓ = \frac{36}{2}

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = \frac{11.7}{18}

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = \frac{0.3}{0.65}

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0
3 years ago
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The y-component of a projectile’s velocity is 12.1 m/s. When the projectile once again passes by the height from which it was la
Nat2105 [25]
It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it. 
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faltersainse [42]

Answer:

AM has longer wavelength

Explanation:

The relation between the wavelength and teh frequency is given by

v = f x λ

Where, f is the frequency and λ be the wavelength.

It shows that the wavelength is inversely proportional to the frequency.

So, higher the frequency, smaller be the wavelength.

So, FM has high frequency than AM, thus, FM has lower wavelength as compared to AM.

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3 years ago
A plane travels 2.5 KM at an angle of 35 degrees to the ground, then changes direction and travels 5.2 km at an angle of 22 degr
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Answer:

7.7 km 26°

Explanation:

The total x component is:

x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87

The total y component is:

y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38

The magnitude is:

d = √(x² + y²)

d = 7.7 km

The direction is:

θ = atan(y/x)

θ = 26°

5 0
3 years ago
A drag racer starts from the rest and accelerates at 7.4m/s^2.how far will he travel in 2.0 seconds?
Vesna [10]

Answer:

14.8 m

Explanation:

S= ut + \frac{1}{2}at^{2}

where u = initial velocity

S= (0 \frac{m}{s})(2s) + \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=  \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=14.8 m

8 0
3 years ago
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