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lilavasa [31]
1 year ago
5

A box of mass m is pulled with a constant acceleration a along a horizontal frictionless floor by a wire that makes an angle of

15° above the horizontal. If T is the tension in this wire, then _____________.
A) T = ma.
B) T > ma.
C) T < ma.
Physics
1 answer:
emmainna [20.7K]1 year ago
6 0

Answer:

B) T > ma.

Explanation:

To solve this problem, we have to analyze the forces acting in the horizontal direction.

In the horizontal direction, we have:

- The horizontal component of the tension in the wire, Tcos \theta, where T is the magnitude of the tension and \theta the angle that the wire makes with the horizontal

Since this is the only force acting on the box in the horizontal direction, this is also the net force, so it is equal to the product of mass and acceleration (Newton's second law of motion):

Tcos \theta = ma

where

m is the mass of the box

a is the acceleration

We can rewrite the equation as

T=\frac{ma}{cos \theta}

The angle in this problem is \theta=15^{\circ}, so

T=\frac{ma}{cos 15^{\circ}}=\frac{ma}{0.966}=1.035 ma

Therefore, the correct option is

B) T > ma.

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How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J

Final velocity = 40 m/s

\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

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1 year ago
A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
choli [55]
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
3 0
2 years ago
Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12
Lesechka [4]

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.

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-

In case all three capacitors are connected in parallel, the 2.0\;\mu\text{F} 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}]-

In case the 2.0\;\mu\text{F} capacitor is connected in parallel with the 1.5\;\mu\text{F} capacitor, and the two capacitors in parallel is connected to the 3.0\;\mu\text{F} 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,

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}.

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.

Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}.

Voltage is the same across two capacitors in parallel.As a result,

\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}.

<h3>Case C: 2.76 V.</h3>

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Similarly,

  • the effective capacitance of the two capacitors in parallel is 5.0 μF;
  • the effective capacitance of the three capacitors, combined: \displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}.

Charge stored:

Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}.

Voltage:

\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}.

<h3 /><h3>Case D: 4.00 V</h3>

-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-.

Connect all three capacitors in series.

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}.

For each of the three capacitors:

Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}.

For the 2.0\;\mu\text{F} capacitor:

\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}.

6 0
1 year ago
Describe a theme-park ride that has constant speed but changing velocity.<br><br> Thank you!
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8 0
2 years ago
Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una
EleoNora [17]

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

5 0
1 year ago
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