Answer:
Correct answer: t = 2.86 seconds
Explanation:
We first use this formula
V² - V₀² = 2 a d
where V is the final velocity (speed), V₀ the initial velocity (speed),
a the acceleration and d the distance.
We will calculate the acceleration from this formula
a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10
a = 0.525 m/s²
then we use this formula
V = V₀ + a t => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds
t = 2.86 seconds
God is with you!!!
We want to know what does the fact that Mercury has no atmosphere tell us. Since Mercury has no atmosphere it cant reflect a lot of sunlight that is hitting its surface. Its constantly being hit by solar wind. So Mercury reflects a small percentage of the sunlight that strikes it.
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
brainly.com/question/14883923
Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
Answer:
The angle through which the wheel turned is 947.7 rad.
Explanation:
initial angular velocity, 
= 33.3 rad/s
angular acceleration, α = 2.15 rad/s²
final angular velocity, 
= 72 rad/s
angle the wheel turned, θ = ?
The angle through which the wheel turned can be calculated by applying the following kinematic equation;
Therefore, the angle through which the wheel turned is 947.7 rad.
