You're driving down the highway and a bug spatters into your windshield. Which undergoes the greater change in momentum during t
1 answer:
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1) 1.33 s
2) 8.6 m
3) 19.9 m
Explanation:
1)
The motion of the football is a projectile motion, which consists of two independent motions:
- A uniform motion (=constant velocity) along the horizontal direction
- A uniformly accelerated motion (=constant acceleration) along the vertical direction
The hang time of the football is the time it takes for the football to reach its maximum height. It is given by:
where
is the initial vertical velocity of the ball, where
u = 15 m/s is the initial velocity
is the angle of projection of the ball
is the acceleration due to gravity
Substituting, we find the hang time:
2)
The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:
where:
s is the vertical displacement
is the initial vertical velocity
t is the time
is the acceleration due to gravity
In this part, we want to find the maximum height, so the height reached by the ball when the time is
t = 1.33 s
Therefore, by substituting the values in the equation, we can find the maximum height:
3)
Here we want to find the horizontal range covered by the ball during its flight.
The horizontal range for a projectile is given by the equation
where
u is the initial velocity
is the angle of projection
g is the acceleration due to gravity
For the ball in this problem we have:
u = 15 m/s
Substituting into the equation, we find the horizontal distance covered by the ball:
Answer: The time required to deposit such amount of Mg is 886secs.
Explanation: According to Faraday Law of Electrolysis, the mass of a substance deposited is directly proportional to quantity of electricity passed.
He also stated that;
96500C(1Farday) of electricity is required to deposit 1 mole of any metal.
For +2e- ==>Mg
193000C(2Faraday) of electricity is required to deposit 1mole of Magnesium metal (1 mole of Mg=24g)
Which implies;
19300C will liberate 24g
xCwill liberate 0.110g
Where x is the amount of electricity to deposit 0.110g
x = (19300 × 0.110)/24
x = 884.6C
Recall that Q =It
Where Q is the quantity of electricity, I is the current and t is the time taken
884.6= 0.998 × t
t= 884.6/0.998
t= 886.3secs.
Therefore the time require is 886.3s.