The length of second, s pendulum at a place where gravitational acceleration ]g] is 9.8 m/s
1 answer:
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Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut + a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s = a
= x 14 x
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.
Explanation:
It is given that,
Area of nickel wire,
Resistance of the wire, R = 2.4 ohms
Initial value of magnetic field,
Final magnetic field,
Time, t = 1.12 s
Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :
Induced current in the loop of wire is given by :
So, the induced current in the loop of wire over this time is . Hence, this is the required solution.
Answer:
#_electrons = 2 10¹⁰ electrons
Explanation:
For this exercise we can use a direct rule of three proportions rule. If an electron has a charge of 1.6 10⁻¹⁹ C how many electrons have a charge of 3.2 10⁻⁹ C
#_electrons = 3.2 10⁻⁹ ( )
#_electrons = 2 10¹⁰ electrons