Answer:
Energy needed = 1100 kJ
Explanation:
Energy needed = Change in kinetic energy
Initial velocity = 15 m/s
Mass, m = 1600 kg

Final velocity = 40 m/s

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J
Energy needed = 1100 kJ
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:
sin 65 = h / 150
Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
Answer:
Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.
.
Explanation:
Consider four possible cases.
<h3>Case A: 12.0 V.</h3>

In case all three capacitors are connected in parallel, the
capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.
<h3>Case B: 5.54 V.</h3>
![-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-](https://tex.z-dn.net/?f=-3.0%5C%3B%5Cmu%5Ctext%7BF%7D-%5B%5Cbegin%7Barray%7D%7Bc%7D-%7B%5Cbf%202.0%5C%3B%5Cmu%5Ctext%7BF%7D%7D-%5C%5C-1.5%5C%3B%5Cmu%5Ctext%7BF%7D-%5Cend%7Barray%7D%5D-)
In case the
capacitor is connected in parallel with the
capacitor, and the two capacitors in parallel is connected to the
capacitor in series.
The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.
The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,
.
What will be the voltage across the 2.0 μF capacitor?
The charge stored in two capacitors in series is the same as the charge in each capacitor.
.
Voltage is the same across two capacitors in parallel.As a result,
.
<h3>Case C: 2.76 V.</h3>
.
Similarly,
- the effective capacitance of the two capacitors in parallel is 5.0 μF;
- the effective capacitance of the three capacitors, combined:
.
Charge stored:
.
Voltage:
.
<h3 /><h3>Case D: 4.00 V</h3>
.
Connect all three capacitors in series.
.
For each of the three capacitors:
.
For the
capacitor:
.
How about a carousel (merry go round).
For any one horse or rider, Speed is constant but direction keeps changing, so velocity does too.
Answer:
a) f = 615.2 Hz b) f = 307.6 Hz
Explanation:
The speed in a wave on a string is
v = √ T / μ
also the speed a wave must meet the relationship
v = λ f
Let's use these expressions in our problem, for the initial conditions
v = √ T₀ /μ
√ (T₀/ μ) = λ₀ f₀
now it indicates that the tension is doubled
T = 2T₀
√ (T /μ) = λ f
√( 2To /μ) = λ f
√2 √ T₀ /μ = λ f
we substitute
√2 (λ₀ f₀) = λ f
if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change
λ₀ = λ
f = f₀ √2
f = 435 √ 2
f = 615.2 Hz
b) The tension is cut in half
T = T₀ / 2
√ (T₀ / 2muy) = f = λ f
√ (T₀ / μ) 1 /√2 = λ f
fo / √2 = f
f = 435 / √2
f = 307.6 Hz
Traslate
La velocidad en una onda en una cuerda es
v = √ T/μ
ademas la velocidad una onda debe cumplir la relación
v= λ f
Usemos estas expresión en nuestro problema, para las condiciones iniciales
v= √ To/μ
√ ( T₀/μ) = λ₀ f₀
ahora nos indica que la tensión se duplica
T = 2T₀
√ ( T/μ) = λf
√ ) 2T₀/μ = λ f
√ 2 √ T₀/μ = λ f
substituimos
√2 ( λ₀ f₀) = λ f
si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia
λ₀ = λ
f = f₀ √2
f = 435 √2
f = 615,2 Hz
b) La tension se reduce a la mitad
T = T₀/2
RA ( T₀/2μ) = λ f
Ra(T₀/μ) 1/ra 2 = λ f
fo /√ 2 = f
f = 435/√2
f = 307,6 Hz