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antiseptic1488 [7]
3 years ago
5

Mercury is a metallic liquid element. It has a density of 11.3 g/cm3 . If you placed the metals listed above in mercury, which o

f the choices below describes what would happen? A - All would float. B - Gold would float, the others would sink. C - All would float except gold. D - All would sink
Physics
1 answer:
Ksivusya [100]3 years ago
5 0

Answer:

The answer is below

Explanation:

The question is not complete since the liquid density is not given.

Archimedes principle states that a body at rest in a fluid is acted upon by an upward force known as the buoyant force. The buoyant force is equal to the weight of the fluid displaced.

An object floats when it is placed in a liquid only if the density of the object is less than the density of the liquid. Therefore those metals with density less than that of mercury would float while those with density greater than mercury would sink.

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a block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood
geniusboy [140]

Volume= Length X width X height.

Plug in the values for each and solve for the volume.

V= (L)(W)(H)

V=(4cm)(5cm)(10cm).


8 0
3 years ago
Suppose there is a sample of xenon in a sealed rectangular container. The gas exerts a total force of 6.05 N perpendicular to on
shtirl [24]

Answer:

<h2> 0.147136N/m²</h2>

Explanation:

Pressure is defines as force exerts by a body per its unit area.

Pressure = Force/Area

Given the total force exerted by the gas = 6.05N

Area of the rectangular container = 0.121 m * 0.201 m = 0.024321m²

Pressure of the sample = 6.05/0.02432

Pressure of the sample = 0.147136N/m²

6 0
4 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
Use newton's third law to explain why a blown up balloon but untied balloon will fly around the room when you let it go? really
sergij07 [2.7K]
Newton's third law states that for every action, there is an equal and opposite reaction. When you let go of the ballon, you are letting the force out but the force also pushes the balloon back.
7 0
3 years ago
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang
valina [46]

Answer:

work done is -2.8  × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta

W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta

W = -mgl  \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta

W = -mgl[ -cosθ ]^{0.047}_{0.045 }

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8  × 10⁻⁶ J

Therefore, work done is -2.8  × 10⁻⁶ J

6 0
3 years ago
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