A sound wave has a frequency of 425Hz. What is the period of this wave? a) 0.00235 b) 0.807 c) 425 d) 850

A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while

marta [7]

Answer:

(c) As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

Energy stored in a parallel plate capacitor is given by:

$U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)\\\\$

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

3 0

3 years ago

Read 2 more answers

El Sol está en promedio a 93 millones de millas de la Tierra. ¿A cuántos metros equivale esto? Expréselo usando potencias de die

mars1129 [50]

Hola!

93 millones de millas equivalen a 1,50 x 10¹¹ metros.

Para calcular este valor, necesitamos saber la equivalencia entre millas y metros (1 milla=1609,34m), y aplicar el siguiente factor de conversión. Para expresar el resultado en potencias de 10, se debe correr la coma a la izquierda hasta obtener un valor de una sola unidad y multiplicar este valor por 10 elevado a la cantidad de espacios que la coma se tuvo que correr a la izquierda:

$93000000millas*\frac{1609,34 metros}{1milla}=150000000000,0 metros=1,50* 10^{11} metros$

Saludos!

5 0

3 years ago

Two constant forces act on an object of mass m = 5.60 kg object moving in the xy plane as shown in the figure below. Force F1 is

Gnoma [55]

Answer:

Part a)

$F_1 = 22.5\hat i + 15.7 \hat j$

$F_2 = -35.5 \hat i + 20.5 \hat j$

Part b)

$F = -10 \hat i + 36.2 \hat j$

Part c)

$a = -1.78 \hat i + 6.46 \hat j$

Part d)

$v_f = -0.64 \hat i + 21.93\hat j$

Part e)

$r_f = 6.09\hat i + 36.72 \hat j$

Part f)

$KE = 1347.7 J$

Part g)

$KE = 1348 J$

Explanation:

Part a)

first force is given as

$F_1 = 27.5 N$ at 35 degree

$F_1 = 27.5 cos35 \hat i + 27.5 sin35 \hat j$

$F_1 = 22.5\hat i + 15.7 \hat j$

Second force is given as

$F_2 = 41 N$ at 150 degree

$F_2 = 41 cos150 \hat i + 41 sin150\hat j$

$F_2 = -35.5 \hat i + 20.5 \hat j$

Part b)

Total force is given as

$F = F_1 + F_2$

$F = (22.5 - 35.5)\hat i + (15.7 + 20.5)\hat j$

$F = -10 \hat i + 36.2 \hat j$

Part c)

as we know

F = ma so object acceleration is given as

$a = \frac{F}{m}$

$a = \frac{-10}{5.60} \hat i + \frac{36.2}{5.60} \hat j$

$a = -1.78 \hat i + 6.46 \hat j$

Part d)

By kinematics we know that

$v_f = v_i + at$

$v_f = (4.70 \hat i + 2.55 \hat j) + (-1.78\hat i + 6.46\hat j)(3)$

$v_f = -0.64 \hat i + 21.93\hat j$

Part e)

As we know that final position is given as

$r_f = v_i t + \frac{1}{2}at^2$

$r_f = (4.70 \hat i + 2.55 \hat j)(3) + \frac{1}{2}(-1.78 \hat i + 6.46 \hat j)(3^2)$

$r_f = 6.09\hat i + 36.72 \hat j$

Part f)

final kinetic energy is given as

$KE = \frac{1]{2}mv_f^2$

$KE = \frac{1}{2}(5.60)(0.64^2 + 21.93^2)$

$KE = 1347.7 J$

Part g)

final kinetic energy is given as

$KE = KE_i + F.r$

$KE = \frac{1}{2}(5.60)(4.70^2 + 2.55^2) + (-10 \hat i + 36.2 \hat j).(6.09\hat i + 36.72\hat j)$

$KE =80 + 1268.4$

$KE = 1348 J$

4 0

3 years ago

What is the effect on acceleration if the velocity is increased by 4 times?

Liula [17]

Answer:

If velocity is higher, then traversed distance during any time increment dt is higher. This tells you that the amount of energy delivered to the object is proportional to the velocity, which is proportional to elapsed time t.

Explanation:

If velocity is higher, then traversed distance during any time increment dt is higher. This tells you that the amount of energy delivered to the object is proportional to the velocity, which is proportional to elapsed time t.

4 0

3 years ago

The resistance of 100 W bulb is less than resistance of 40 W bulb. Explain the reason.

vazorg [7]

Ok so we know that electric power is:

$P=UI=I^2R$

If we express resistance.

$R=\dfrac{P}{I^2}$

Now if you have 100W of power you will probably get a bigger resistance than with 40W of power.

However here it says that the resistance of 40W bulb is bigger than 100W bulb. Which means the statement is incorrect.

Hope this helps.

r3t40

4 0

3 years ago