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Sophie [7]
3 years ago
6

If a calorie is equivalent to 4.184 joules how many joules are contained in 250 kg calories slice of pizza

Physics
1 answer:
Andreyy893 years ago
6 0
In order to answer this, we will set up a simple ratio as such:

1 calorie = 4.184 joules

1 kilocalorie = 1000 calories

1 kilocalorie = 4,184 joules

250 kilocalories = x joules

Cross multiplying the second and third equations, we get:

x joules = 4,184 * 250

250 kilocalories are equivalent to 1,046 kJ
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a sealed cylinder contains a sample of ideal gas at a pressure of 2.0 atm. The rms speed of the molecules is v0. If the rms spee
fredd [130]

Answer:

P_2 = 1.62 atm

Explanation:

We know the formula for the rms speed of the ideal gas is  given by

v_{rsm}=\sqrt{\frac{3PV}{m} }

P= pressure of the surrounding

V= volume of the vessel

m= mass of the gas

Now, From this formula rms speed (v_rms) is directly proportional to square root is pressure.

Then  

\frac{v_{rsm,1}}{v_{rsm,2}} =\sqrt{\frac{P_1}{P_2} }

given that v_rsm,1= v0

and v_rsm,2=0.9v0

putting these values we get

\frac{v0}{0.9v0} =\sqrt{\frac{2}{P_2} }

P_2 = 1.62 atm

8 0
3 years ago
Read 2 more answers
A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s.
svet-max [94.6K]

Answer:

A) The distance between the stations is 1430m

B) The time it takes the train to go between the stations is 80s

Explanation:

First we will calculate the distance covered for the first 20s.

From one the equations of kinematics for linear motion

S = ut + \frac{1}{2}at^{2}  \\

Where S is distance traveled

u is the initial velocity

t is time

and a is acceleration

Since the train starts from rest, u = 0 m/s

Hence,  for the first 20s

a = 1.1 m/s²; t = 20s,  u = 0 m/s

∴ S = ut + \frac{1}{2}at^{2}  \\ gives

S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}

S =  \frac{1}{2}(1.1)(20)^{2}

S = 220m

This is the distance covered in the first 20s.

  • The train then proceeds at constant speed for 1100m.

Now, we will calculate the speed attained here

From

v = u +at

Where v is the final velocity

Hence,

v = 0 + 1.1(20)

v = 1.1(20)

v = 22 m/s

This is the constant speed attained when it proceeds for 1100m

  • The train then slows down at a rate of 2.2 m/s² until it stops

We can calculate the distance covered while slowing down from

v^{2} = u^{2} + 2as

The initial velocity, u here will be the final velocity before it started slowing down

∴u = 22 m/s

The final velocity will be 0, since it came to a stop.

∴ v = 0 m/s

a = -2.2 m/s² ( - indicates deceleration)

Hence,

v^{2} = u^{2} + 2as gives

0^{2} =22^{2} +2(-2.2)s

0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m

This is the distance traveled while slowing down.

A) The distance between the stations is

220m + 1100m + 110m

= 1430m

Hence, the distance between the stations is 1430m

B) The time it takes the train to go between the stations

The time spent while accelerating at 1.1 m/s² is 20s

We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,

From,

Speed =\frac{Distance}{Time}\\

Then,

Time = \frac{Distance}{Speed}

Time = \frac{1100}{22}

Time = 50 s

And then, the time spent while decelerating (that is, while slowing down)

From,

v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s

This is the time spent while slowing down until it stops at the station.

Hence, The time it takes the train to go between the stations is

20s + 50s + 10s = 80s

The time it takes the train to go between the stations is 80s

6 0
3 years ago
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