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2 years ago

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]2 years ago

8 0

The answer to that is the rock will sink

KATRIN_1 [288]2 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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A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

4 0

2 years ago

Can anyone please help me on this question (DUE TOMORROW) (13 POINTS GO AWAY)

julia-pushkina [17]

4.) Quantitative: notice the root word 'quantity', which is a certain percentage of something.

a.) Quantitative, since it represents a certain percentage. The weight before the camp could also be written as an improper fraction.

b.) This is also quantitative because it can be written as a ratio $(2.99:500)$ , which can also be written as a fraction $(\frac{2.99}{500}$ or percentage.

c.) This is qualitative. Joan has a higher percentage than Emile.

d.) Qualitative. It judges the quality of something instead of amount. To make it quantitative requires a fraction of percentage. In this case, maybe you could add how he can run y miles every x seconds.

5.)

a.)This depends on whether or not the air is moving. If the classroom's AC is on, or the window is open, then the correct unit to use is CFM (cubic feet per minute). If not, the common unit is milliliters, but must also include temperature and air pressure as well as the volume of the room.

b.) Simply calculate the volume of the can using inches cubed, since volume is the measure of how much space an object takes up.

c.) The unit would be $\frac{kg}{ft^3}$ . If $\frac{mass}{volume} = density$ , then simply multiply your density by your mass.

d.) I personally would use centimeters. Inches would work too, but centimeters would be much more accurate.

e.) Use the same formula for c.), which means kilograms per cubic foot.

Hope this helped.

8 0

2 years ago

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-hi

Vlada [557]

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

$E_A} =E_{B}$

Spring potential energy =potential energy due to elevation

0.5*k*x²= mg $(h_{B}-h_{A} )$ =mgh

0.5*k*2.3²= 430*9.81*6

k=9568.92 N/m

For safety reason

k"=1.13 *k= 1.13*9568.92

k"=10812.88 N/m

agsin from conservation of energy

$E_A} =E_{C}$

spring potential energy=change in kinetic energy

0.5*k"*x²=0.5*m* $V_{max}^{2}$

10812.88 *2.3²=430* $V_{max}^{2}$

$V_{max}$ =11.53 m/s

5 0

2 years ago

A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th

Snowcat [4.5K]

Answer:

$\boxed{\text{\sf \Large 3.0 s}}$

Explanation:

Use distance formula

$\displaystyle d=ut+\frac{1}{2} at^2$

$u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}$

$\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2$

$t=3$

4 0

2 years ago

You are standing on a skateboard, initially at rest. a friend throws a very heavy ball towards you. you can either catch the obj

PIT_PIT [208]

<span>You should deflect the ball in order to maximize your speed on the skateboard.

Since this creates a larger impulse, you want to deflect the ball. Splitting it up into catching and throwing the ball may by something you can think of deflecting the ball. First, you need to catch the ball, which in turn would push you forward with some speed. (The speed we are talking about should obviously be equal to option A, where you catch the ball). Now, throw the ball back to him since these two processes are equal to deflecting the ball. Throwing a mass away from you would cause or enable you to move even fast.</span>

7 0

2 years ago