Answer:
Explanation:
Amount of gold deposited = 0.5 g
Gold's molar mass = 197 g/mol
Time duration, t = 6 hours
= 6 × 3600
= 12600 s
Calculation of moles: mass/molar mass
= 0.5/197
= 0.00254 mole
Assuming
Au --> Au+ + e-
Faraday's constant = 9.65 x 10^4 C mol-1
Charge, Q = 96500 × 0.00254
= 244.924 C
Relation: Q = I × t
Thus, I = 244.924/12600
= 0.011 A
= 11.34 mA.
Which statement can never be true for athletes in team sports? The statement that is always false among the listed options for team sports athletes is choice C) Conflict resolution indicates a lack of sportsmanship. The other statements are valid in the context of team sports.
<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack.
centripetal force = weight of the ball
m v^2 / r = m g
v^2 / r = g
v^2 = g r
v = sqrt { g r }
v = sqrt { (9.80~m/s^2) (0.7 m) }
v = 2.62 m/s
Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>
Answer:
At this position, the magnetic field equals ZERO
Explanation:
The magnetic field produced by a moving charge is described as

Here, we determine the direction of the magnetic field using

Thus, we find

Leading to a magnetic field of ZERO
Consequently, when the charge moves in the same line as the given position vector, the magnetic field will be nonexistent
Answer:
A. Yes, the ball clears the crossbar by 2.83 meters
Explanation:
This situation pertains to projectile motion.
The horizontal velocity component of the ball is calculated as 26 cos 35 = 21.3 m/s
The vertical velocity component is 26 sin 35 = 14.9 m/s
The time taken to travel the horizontal distance to the goalpost, which is 54.9 m, is:
= distance / horizontal speed
= 54.9 / 21.3
= 2.577 seconds.
The vertical distance achieved during this time is:
h = ut - 1/2 gt², where u is the initial vertical velocity, and t = 2.577 seconds.
h = 14.9 x 2.577 - 0.5 x 9.8 x (2.577)²
= 38.39 - 32.54
= 5.85 m
Thus, the ball surpasses the crossbar by 5.85 - 3.05 = 2.8 m