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Firlakuza [10]
3 years ago
7

Noah drops a rock with a density of 1.73 g/cm3 into a pond. Will the rock float or sink? Explain your answer.

Physics
2 answers:
Rainbow [258]3 years ago
8 0
The answer to that is the rock will sink 
KATRIN_1 [288]3 years ago
7 0

Answer:

The rock will sink in the water because it has high density than water.

Explanation:

As we know that the density of water is 1 gm/cm^{3}

And in the given question the density of the rock is  1.73 gm/cm^{3}.

And also it is known that any object which has higher density than the water it will sink or if it is lower density than water it will float.

Now in the given situation the density of the rock is greater than the density of the water.

Therefore, the rock will sink in the water because of high density than water.

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Impulse = Force * time
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Which source of energy would you prefer for cooking and why?
timurjin [86]

Answer:

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4 years ago
Two 1-kg objects, C and D, increase in temperature by the same amount, but the
Ede4ka [16]

The object D is made up of material Lead. The correct option is D.

<h3>What is specific heat?</h3>

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

Two 1-kg objects, C and D, increase in temperature by the same amount, but the thermal energy transfer of object C is greater than the thermal energy transfer of object D. The object C has a specific heat of 235 J/kg-K.

Q = m C ΔT

Qc > Qd

The energy transfer is proportional to specific heat.

Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.

Thus, the correct option is D.

Learn more about specific heat,

brainly.com/question/11297584

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4 0
2 years ago
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
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