Answer:
The average magnitude of magnetic field B= 0.0433/ d Tesla
(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)
Explanation:
Induced emf
where B= magnetic field
d= breadth of rectangular piece
V= velocity with which the rectangular piece = o.o6m/s
n= no of turns = 10
EMF = 26mV
since d (breadth of the frame) is not given, I will use it as a variable
EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)
From eq 1, we get
B= (EMF)/(n d V)
B= (26 X 0.001) / (10 d 0.06)
B= 0.0433/ d Tesla
Answer:
THE FIRST ONE YOU SHOULD TELL HIM AND THE LAST ONE YOU SHOUDENT DO BECAUSE HE WILL DO IT AGAIN AND EXPECT OTHERS TO CLEAN UP AFTER HIM
Explanation:
Explanation:
yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .
Answer:
e*P_s = 11 W
Explanation:
Given:
- e*P = 1.0 KW
- r_s = 9.5*r_e
- e is the efficiency of the panels
Find:
What power would the solar cell produce if the spacecraft were in orbit around Saturn
Solution:
- We use the relation between the intensity I and distance of light:
I_1 / I_2 = ( r_2 / r_1 ) ^2
- The intensity of sun light at Saturn's orbit can be expressed as:
I_s = I_e * ( r_e / r_s ) ^2
I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2
I_s = 11 W / e*a
- We know that P = I*a, hence we have:
P_s = I_s*a
P_s = 11 W / e
Hence, e*P_s = 11 W
Answer:
times
Explanation:
First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.
Age of the universe:
(1 followed by 17 zeroes)
Lifetime of the top quark:
(we moved the decimal point 24 places to the right)
Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:
