The net force on the cart is 100 N in the direction of its motion, so by Newton's second law we can find the acceleration <em>a</em> applied to it:
100 N = (50 kg) <em>a</em>
<em>a</em> = (100 N) / (50 kg)
<em>a</em> = 2 m/s²
The cart starts at rest and travels a distance of 10 m, so that its final velocity <em>v</em> satisfies
<em>v</em> ² - 0² = 2 (2 m/s²) (10 m)
<em>v</em> ²= 40 m²/s²
and so the cart ends up with kinetic energy
KE = 1/2 <em>m</em> <em>v</em> ² = 1/2 (50 kg) (40 m²/s²) = 1000 J
Wnet = net work (done by the net force)
KE = kinetic energy = (1/2)mv2
m = 45 kg
vo = 0 m/s (starts at rest)
vf = 12 m/s (final speed)
Answer:D. The metal in the door expands when it becomes warm due to an increase in kinetic energy, and contracts when it's cooler.
Explanation:
because the metal in the door expands due to warm to an increase in kinetic energy and contracts.
Answer:
56 J
Explanation:
The following data were obtained from the question:
Energy 1 (E₁) = 7 J
Extention 1 (e₁) = 1.8 cm
Extention 2 (e₂) = 1.8 + 3.6 = 5.4 cm
Energy 2 (E₂) =?
Energy stored in a spring is given by the following equation:
E = ke²
Where E is the energy.
K is the spring constant.
e is the extension.
E = ke²
Divide both side by e²
K = E/e²
Thus,
E₁/e₁² = E₂/e₂²
7/ 1.8² = E₂/ 5.4²
7 / 3.24 = E₂/ 29.16
Cross multiply
3.24 × E₂ = 7 × 29.16
3.24 × E₂ = 204.12
Divide both side by 3.24
E₂ = 204.12 / 3.24
E₂ = 63 J
Thus, the additional energy required can be obtained as follow:
Energy 1 (E₁) = 7 J
Energy 2 (E₂) = 63 J
Additional energy = 63 – 7
Additional energy = 56 J
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
Put the value into the formula
The potential on the second plate
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
Put the value into the formula
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
