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pychu [463]
3 years ago
13

How many mL of water would be displaced by 408 g of lead

Physics
1 answer:
Natasha2012 [34]3 years ago
6 0
For this, we must know the density of lead. This is 11.36 grams per cubic centimeter. The mass present is 408 grams. Therefore, the volume occupied will be:
408 / 11.36 = 35.9 cubic centimeters. This is the volume of water that will be displaced.
Now, we know that one cubic centimeter is equivalent to one mL. Therefore,
35.9 ml of water will be displaced.
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Bezzdna [24]

Answer:

There are many topics related to physics such as :

Kinematics

Dynamics

Light

Sound

7 0
2 years ago
Which of these properties of an object best quantifies its inertia: velocity, acceleration, volume, mass, or temperature?
LuckyWell [14K]

Answer:

Mass

Explanation:

Inertia is essentially an object's tendency to stay in motion or at rest unless it is forced to do otherwise (pun intended). It only makes sense to me that mass would best quantify an object's inertia, because an object with more mass would be harder to move and/or stop from moving.

3 0
3 years ago
A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an
Mademuasel [1]

Answer:

The angular acceleration is 0.209\ rad/s^2

Explanation:

Given that,

Angular velocity, \omega_{i} = 30.0\ rpm

Angular velocity, \omega_{f} = 50.0\ rpm

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

\alpha=\dfrac{\Delta \omega}{\Delta t}

\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}

\alpha=\dfrac{50.0-30.0}{10.0}

Now, we change the angular velocity in rad/s.

\omega=20\times\dfrac{2\pi}{60}

\omega=2.09\ rad/s

\alpha=\dfrac{2.09}{10.0}

\alpha=0.209\ rad/s^2

Hence, The angular acceleration is 0.209\ rad/s^2

5 0
3 years ago
Read 2 more answers
A metal pot feels hot to the touch, but the plastic handle does not. Which type of material is the plastic handle? A. A thermal
gtnhenbr [62]
D is the answer..........

3 0
2 years ago
Read 2 more answers
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
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