Answer:
105.8 m
46 m/s
Explanation:
From the time the rocket is launched to the time it reaches its maximum height:
v = 0 m/s
a = -10 m/s²
t = 9.2 s / 2 = 4.6 s
Find: Δy and v₀
Δy = vt − ½ at²
Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²
Δy = 105.8 m
v = at + v₀
0 m/s = (-10 m/s²) (4.6 s) + v₀
v₀ = 46 m/s
The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as
Here,
= Angular displacement
t = time
The angular velocity is given as
PART A) The angular velocity in SI Units will be,
PART B) From our first equation we can rearrange to find the angular displacement then
Replacing,
In order to solve this problem, we must first find out the value of each line on the number line. However, we can make this problem more simple by ignoring every interval except for the ones between 0 and 6. There are three total intervals in between 0 and 6 (including 6 and excluding 0). Therefore, we can do 6/2, and get an interval value of 2. This means that each line adds a value of 2. Since the car is only one line past zero, we only have to add one value of 2. Since 0 + 2 = 2, our final answer is C. 2.
Hope this helps!
Answer:
The current in the circuit increases
Explanation:
The ohm's law states that the potential across a circuit is proportional to the current in the circuit.
V ∝ I
Where 'V' is the potential difference across the circuit and 'I' is the current in the circuit.
The proportionality constant present in the equation is the resistance of the circuit. Hence, the equation becomes
V = IR
According to the equation, when V is directly proportional to 'I' where 'R' remains as constant, then the change in 'V is brings change in 'I' to make the equation valid.
So, when there is an increase in the voltage, the current on the circuit increases.
Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.
