(6) Module 4 Study Guide

Frame S' passes frame S in the usual way. Two events are simultaneous in S'.

yuradex [85]

Answer:

c)They can also be simultaneous in S if their separation is zero.

Explanation:

By relativity theory, we can say two events when seen from two different reference frames can only be simultaneous when they are at the same space location and occur simultaneously in at least one reference frame, therefore when Frame S′ usually passes Frame S. Two occurrences in S′ are simultaneous, therefore these occurrences can be simultaneous in S when their separation is 0 (that is they are at the same location)

And therefore option c. If their separation is zero, they can also be simultaneous in S.

8 0

3 years ago

An object goes from 10 m/s to 2 m/s in 2 seconds, what's its acceleration?

Mila [183]

Answer:

a=(v-u)÷t

a= (2-10)÷2

a= -8÷2

a= -4 m/s²

6 0

3 years ago

why are we all cheating if we payed attention in what they trying to tell us we wouldn't have to cheat but we are not paying att

sweet-ann [11.9K]

I’ve always been failing since middle school. it’s bcs of quarantine that made me unmotivated. rn my grades are F’s D C and A . I should be paying attention but my phone just keeps me distracted lol.

7 0

3 years ago

Read 2 more answers

A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga

pashok25 [27]

A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
$p_i V_i = nRT_i$
where
$p_i=1.0 atm=1.01 \cdot 10^5 Pa$ is the initial pressure of the gas
$V_i$ is the initial volume of the gas
$n=2.3 mol$ is the number of moles
$R=8.31 J/K mol$ is the gas constant
$T_i=240^{\circ}C=513 K$ is the initial temperature of the gas

By re-arranging this equation, we can find $V_i$ :
$V_i = \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3$

2) Now the gas cools down to a temperature of
$T_f = 14^{\circ}C=287 K$
while the pressure is kept constant: $p_f = p_i = 1.01 \cdot 10^5 Pa$ , so we can use again the ideal gas law to find the new volume of the gas
$V_f = \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3$

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
$W=p \Delta V= p(V_f -V_i)$
by using the data we found at point 1) and 2), we find
$W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J$
where the negative sign means the work is done by the surrounding on the gas.

5 0

3 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$