Yeah, it can be anywhere along the back wallHello,
My team recently went to an invitational and found out that the TP (target point) was off 5 cm from the center line from that particular invitational. The team reviewed the rules once again and it says that event supervisors select a TP that is the same for all teams.
Does this mean that event supervisors can select a TP anywhere along the Target Wall?
Thanks.
Yes, the proctor can make the target anywhere along the target wall. So far, every invitational I have been to this year has had the target off-center. The farthest I've seen so far is 9cm off-centerHello,
My team recently went to an invitational and found out that the TP (target point) was off 5 cm from the center line from that particular invitational. The team reviewed the rules once again and it says that event supervisors select a TP that is the same for all teams.
Does this mean that event supervisors can select a TP anywhere along the Target Wall?
Thanks.
Read rule 3.f. very carefully. It says one of the marks is the TP, It does NOT say a particular mark is the TP.Hello,
My team recently went to an invitational and found out that the TP (target point) was off 5 cm from the center line from that particular invitational. The team reviewed the rules once again and it says that event supervisors select a TP that is the same for all teams.
Does this mean that event supervisors can select a TP anywhere along the Target Wall?
Thanks.
I asked my cousin (an Astronomy Harvard professor) said to use the Red Doppler Shift formula, and then I got the same answer as you. The test writer probably did something wrong lol. Good luck!Can someone check me back on this problem its a Doppler Shift one?
"A spaceship is moving away from an asteroid at a relative velocity of 2.8481 x 10^8 m/s. The spaceship sends a signal with a frequency of 5 x 10^6 Hz to a base located on the asteroid. What is the frequency of the signal measured by the base?"
I keep getting 2.56 x 10^6 Hz but the test key says the answer is 3.1225 x 10^7 Hz.
I asked my cousin (an Astronomy Harvard professor) he said to use the Red Doppler Shift formula, and then I got the same answer as you. The test writer probably did something wrong lol. Good luck!Can someone check me back on this problem its a Doppler Shift one?
"A spaceship is moving away from an asteroid at a relative velocity of 2.8481 x 10^8 m/s. The spaceship sends a signal with a frequency of 5 x 10^6 Hz to a base located on the asteroid. What is the frequency of the signal measured by the base?"
I keep getting 2.56 x 10^6 Hz but the test key says the answer is 3.1225 x 10^7 Hz.
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I think the test writer has the correct answer. If you're using the typical Doppler Shift equation f = f_source * v / (v-v_source), you haven't taken into consideration that the spaceship has a very high relative velocity (close to the speed of light). Using the Doppler Shift equation here will get us an answer of 2.56 x 10^6 Hz.I asked my cousin (an Astronomy Harvard professor) he said to use the Red Doppler Shift formula, and then I got the same answer as you. The test writer probably did something wrong lol. Good luck!Can someone check me back on this problem its a Doppler Shift one?
"A spaceship is moving away from an asteroid at a relative velocity of 2.8481 x 10^8 m/s. The spaceship sends a signal with a frequency of 5 x 10^6 Hz to a base located on the asteroid. What is the frequency of the signal measured by the base?"
I keep getting 2.56 x 10^6 Hz but the test key says the answer is 3.1225 x 10^7 Hz.
Using hyperphysics, you have to specify a negative relative velocity when plugging it into their equation since the spaceship is moving away. They specify this, saying "for red shift calculations, use negative velocities." We know it must be red shift since the spaceship is moving away which would elongate the wavelengths and decrease the frequency.I think the test writer has the correct answer. If you're using the typical Doppler Shift equation f = f_source * v / (v-v_source), you haven't taken into consideration that the spaceship has a very high relative velocity (close to the speed of light). Using the Doppler Shift equation here will get us an answer of 2.56 x 10^6 Hz.I asked my cousin (an Astronomy Harvard professor) he said to use the Red Doppler Shift formula, and then I got the same answer as you. The test writer probably did something wrong lol. Good luck!Can someone check me back on this problem its a Doppler Shift one?
"A spaceship is moving away from an asteroid at a relative velocity of 2.8481 x 10^8 m/s. The spaceship sends a signal with a frequency of 5 x 10^6 Hz to a base located on the asteroid. What is the frequency of the signal measured by the base?"
I keep getting 2.56 x 10^6 Hz but the test key says the answer is 3.1225 x 10^7 Hz.
To get to correct answer, we need to take relativistic effects into consideration by using the Relativistic Doppler Shift equation:
f = f_source * Sqrt(1 - (v_source/c)^2) / (1- (v_source/c))
Using this equation, we get 3.1225 x 10^7 Hz (you can test it out yourself at the bottom of hyperphysics. When the v_source is small compared to the speed of light, this equation can be approximated as the typical Doppler Shift equation that you were using.
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