Answer:
<h2>
<em>True</em></h2>
Forests helps in binding the soil together. Thus, prevent soil from erosion.
Answer:
13.1
Explanation:
thats what i put in for acellus and its right
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.
Answer:
v = 6.45 10⁻³ m / s
Explanation:
Electric force is
F = q E
Where q is the charge and E is the electric field
Let's use Newton's second law to find acceleration
F- W = m a
a = F / m - g
a = q / m E g
Let's calculate
a = -1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8
a = 0.228 10² -9.8
a= 13.0 m / s²
Now we can use kinematics, knowing that the resting parts electrons
v² = v₀² + 2 a y
v =√ (0 + 2 13.0 1.6 10⁻⁶)
v = 6.45 10⁻³ m / s
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
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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.
