Ask question

Ask question

All categories

2 years ago

12

A gas in a cylinder with a moveable piston has an initial volume of 89.2 ml . if we heat the gas from 35 âc to 153 âc, what is i

ts final volume (in ml)? express the volume to three significant figures and include the appropriate units.

1 answer:

Korvikt [17]2 years ago

4 0

<span>Use Charles' Law:
v1/T1 = v2/T2
88.2/(273+35) = v2/(273+155)
where v2 is the new volume.

I hope this helps!!!</span>

You might be interested in

How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 39 j of work?

sdas [7]

The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
$W=q \Delta V$
where q is the charge and $\Delta V$ is the potential difference.

In our problem, the work done is W=39 J while the potential difference of the battery is $\Delta V = 9.0 V$ , so we can find the charge transferred by the battery:
$q= \frac{W}{\Delta V}= \frac{39 J}{9.0 V}=4.33 C$

3 0

3 years ago

A ball is thrown vertically up from the edge of a cliff with a speed of 8 m/s, how high is

anygoal [31]

Explanation:

s=d/t

d=s×t

d=8×16.4

d=131.2

distance is 131.2m.

7 0

2 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

2 years ago

Group of physics students are asked to demonstrate Ohm’s Law in parallel circuits. Using their knowledge of Ohm’s Law and the r

astra-53 [7]

The student’s suggestion who provides enough evidence to be able to determine the value of each resistor is student D.

<h3>What is current?</h3>

The current is the stream of charges which flow inside the conductors when connected across the end of voltage.

For the given set of parallel resistors, we need to find the resistance of each resistor.

From the Ohm's law, V =IR

R = V/I

Resistance value depends upon the voltage difference across the resistor and the current flowing through that resistance.

Thus, the student D gives enough evidence to find resistance of the circuit is

Learn more about current.

brainly.com/question/10677063

#SPJ1

8 0

1 year ago

Please help me asdrtyuio

marta [7]

Answer:

9. The Sun's Gravity

10. The core is the densest layer

4 0

3 years ago