<span>22.5 newtons.
First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as:
E = 0.5mv^2
where
E = Energy
m = mass
v = velocity
Substituting known values and solving gives:
E = 0.5 3.06 kg (7 m/s)^2
E = 1.53 kg 49 m^2/s^2
E = 74.97 kg*m^2/s^2
Now ignoring air resistance, how much energy should the rock have had?
We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So
3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2
So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is
299.88 J - 74.97 J = 224.91 J
So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division.
224.91 J / 10 m
= 224.91 kg*m^2/s^2 / 10 m
= 22.491 kg*m/s^2
= 22.491 N
Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
Answer:
Explanation:
W = Fd
Engine 500(20) = 10000 J
friction 200(-20) = -4000 J
Energy increase is 10000 - 4000 = 6000 J
Answer:
The dependent variable is the variable that is studied while the independent variable is the variable that is being manipulated
The total energy stored in the capacitors is determined as 2.41 x 10⁻⁴ J.
<h3>What is the potential difference of the circuit?</h3>
The potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
where;
- C is capacitance of the capacitor
- V is the potential difference
For a parallel circuit the voltage in the circuit is always the same.
The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
2U = CV²
V = √2U/C
V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)
V = 12 V
The equivalent capacitance of C1 and C2 is calculated as follows;
1/C = 1/C₁ + 1/C₂
1/C = (1)/(0.9 x 10⁻⁶) + (1)/(16 x 10⁻⁶)
1/C = 1,173,611.11
C = 1/1,173,611.11
C = 8.52 x 10⁻⁷ C
The total capacitance of the circuit is calculated as follows;
Ct = 8.52 x 10⁻⁷ C + 2.5 x 10⁻⁶ C
Ct = 3.35 x 10⁻⁶ C
The total energy of the circuit is calculated as follows;
U = ¹/₂CtV²
U = ¹/₂(3.35 x 10⁻⁶ )(12)²
U = 2.41 x 10⁻⁴ J
Learn more about energy stored in a capacitor here: brainly.com/question/14811408
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Answer:
8.78?m/s
Explanation:
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