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2 years ago

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The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes

of the elevator stop the elevator shortly before it reaches the bottom of the shaft. If the elevator fell a distance of 100 m starting from rest, the heat that the brakes must dissipate to bring the elevator safely to rest is (A) 100 J (B) 1,000 J (C) 10,000 J (D) 100,000 J (E) 1,000,000 J

1 answer:

tester [92]2 years ago

4 0

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

or

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

or

Work done by brake = 1000 × 10 × 100

or

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

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The amusement park ride shown above takes riders straight up a tall tower and then releases an apparatus holding seats. This app

Natasha2012 [34]

The gravitational force on the rider is:

- Way up: equal to mg

- Way down: equal to mg

Explanation:

The choices are missing: find them in the attached figure.

The force of gravity acting on a body is a force directed downward (towards the Earth's centre) and whose magnitude is

$F=mg$

where

m is the mass of the body

g is the acceleration due to gravity

The value of g is approximately constant near the Earth's surface and it is

$g=9.8 m/s^2$

During the ride, the mass of the rider, m, remains constant. This means that the magnitude of the gravitational force, mg, exerted on the rider remains constant during the rider.

Therefore, the correct answer is

- Way up: equal to mg

- Way down: equal to mg

Learn more about force of gravity:

brainly.com/question/1724648

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#LearnwithBrainly

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2 years ago

An astronaut is floating happily outside her spaceship, which is orbiting the earth at a distance above the earths surface equal

serg [7]

Answer:

The astronaut's weight will be one-forth of her normal weight on earth.

Explanation:

From Newton's law of gravitation, we can write the acceleration due to gravity (g) on Earth's surface is given by

$g = \dfrac{GM_{e}}{R^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)\\$

where 'G' is gravitational constant, ' $M_{e}$ ' is Earth's mass and 'R' is Earth's radius.

As shown in the figure, if the astronaut is at a height 'h' from earth's surface and if ' $g'$ ' be the value of the acceleration due to gravity at that height, then

$g' = \dfrac{G M_{e}}{(R + h)^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Taking the ratio of both the equations, and as given h = R.

$\dfrac{g'}{g} &=& \dfrac{g' = \dfrac{G M_{e}}{(R + h)^{2}}}{g' = \dfrac{G M_{e}}{(R + h)^{2}}}\\&=& \dfrac{R^{2}}{(R + h)^{2}}\\&=& R^{2}(R + R)^{2}\\&=& \dfrac{1}{4}\\$

So,

$&& g' = \dfrac{g}{4}\\&or,& mg' = \dfrac{mg}{4}$

where 'm' is the mass of the astronaut.

So the weight of the astronaut will be one-forth her normal weight on earth.

7 0

2 years ago

If c1=c2=4.00μf and c4=8.00μf, what must the capacitance c3 be if the network is to store 2.70×10−3 j of electrical energy?

Akimi4 [234]

Missing detail in the text: total voltage of the circuit $\Delta V = 46.0 V$
Missing figure: https://www.physicsforums.com/attachments/prob-24-68-jpg.190851/

Solution:

1) The energy stored in a circuit of capacitors is given by
$U= \frac{1}{2} C_{eq} (\Delta V)^2$
where $C_{eq}$ is the equivalent capacitance of the circuit. We can find the value for $C_{eq}$ by using $\Delta V=46.0 V$ and the energy of the system, $U=2.7\cdot 10^{-3} J$
$C_{eq}= \frac{2U}{(\Delta V)^2}=2.55\cdot 10^{-6} F=2.55\mu F$

2) Then, let's calculate the equivalente capacitance of C1 and C2. The two capacitors are in series, so their equivalente capacitance is given by
$\frac{1}{C_{12}}= \frac{1}{C_1}+ \frac{1}{C_2}= \frac{1}{4 \mu F} + \frac{1}{4 \mu F}$
from which we find $C_{12}=2 \mu F$

3) Then let's find $C_{123}$ , the equivalent capacitance of $C_{12}$ and C3. $C_{123}$ is in series with C4, therefore we can write
$\frac{1}{C_{eq}}= \frac{1}{C_{123}}+ \frac{1}{C_4}$
Since we already know $C_4=8 \mu F$ and $C_{eq}=2.55 \mu F$ , we find
$C_{123}=3.70 \mu F$

4) Finally, we can find $C_{3}$ , because it is in parallel with $C_{12}$ , and the equivalent capacitance of the two must be equal to $C_{123}$ :
$C_{123}=C_{12}+C_3$
So, using $C_{123}=3.70 \mu F$ and $C_{12}=2 \mu F$ , we find
$C_3=1.70 \mu F$

7 0

2 years ago

A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m Ã 3.00 m Ã 5.00 m. 1. what is the

cupoosta [38]

1) The volume of the raft is the product between the lenghts of its three dimensions:
$V = (2.00 m)(3.00m)(5.00m)=30 m^3$

2) The mass of the raft is the product between its density, d, and its volume, V:
$m=dV=(650 kg/m^3)(30 m^3)=19500 kg$

3) The weight of the raft is the product between its mass m and the gravitational acceleration, $g=9.81 m/s^2$ :
$W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N$

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
$W_a = W- \rho_W V_{disp} g$
where $\rho _W = 1000 kg/m^3$ is the water density and $V_{disp}$ is the volume of displaced fluid.
The density of the raft ( $650 kg/m^3$ ) is smaller than the water density ( $1000 kg/m^3$ ), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
$W=B=\rho _W V_{disp} g$
and therefore, the apparent weight will be zero:
$W_a = W-B=W-W=0$

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
$B=\rho_W V_{disp} g$
When the raft is completely immersed in the water, the volume of fluid displaced $V_{disp}$ is equal to the volume of the raft, $V_{disp}=V$ . Therefore the buoyancy in this situation is
$B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N$
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
$B=W=1.91 \cdot 10^5 N$

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
$mg = m_{disp} g$
where $m_{disp}= \rho_W V_{disp}$ is the mass of the displaced water. From the previous equation, we obtain that $m_{disp}=m=19500 kg$ .

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship $m=dV$ we can rewrite $m=m_{disp}$ as:
$d V =d_W V_{disp}$
and so
$V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3$

8) The volume of water displaced is (point 7) $19.5 m^3$ . This volume is now "filled" with part of the volume of the raft, therefore $19.5 m^3$ is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, $30 m^3$ :
$\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%$
Viceversa, the volume of raft above the water level is $30 m^3-19.5 m^3 = 10.5 m^3$ . Therefore, the fraction of volume of the raft above water level is
$\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%$

9) Let's repeat steps 5-8 replacing $\rho _W$ , the water density, with $\rho_E=806 kg/m^3$ , the ethanol density.

9-5) The buoyant force is given by:
$B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N$
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
$m_E = m = 19500 kg$

9-7) Using the relationship $d= \frac{m}{V}$ , we can find the volume of displaced ethanol:
$V_E = \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3$

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: $24.2 m^3$ . Therefore, the fraction of raft's volume below the ethanol level is
$\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%$
Consequently, the raft's volume above the ethanol level is
$30 m^3 - 24.2 m^3 = 5.8 m^3$
and the fraction of volume above the ethanol level is
$\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%$

8 0

2 years ago

Which of the following substances will show the smallest change in temperature when equal amounts of energy are absorbed?

SSSSS [86.1K]

It would be water because if you freeze it than you will still be able to see it and if you boil it than you will be able to see it disappear.

3 0

2 years ago

Read 2 more answers