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3 years ago

What is the energy conversion in a hair dryer?

Physics

1 answer:

wel3 years ago

4 0

Answer:

C.Electrical energy to heat and kinetic energy

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Help me PLEASE! 25 points!!!!! :)

NeTakaya

Answer:

Explanation:

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7 0

2 years ago

Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's displacement?

tresset_1 [31]

Answer:

Joes displacement is 54m

4 0

2 years ago

What is the expected wavelength (in cm) of 10.5 ghz microwaves in free space?

Sindrei [870]

Wavelength = (speed) / (frequency)

Wavelength = (300 thousand km per second) / (10.5 billion per second)

Wavelength = (300 / 10.5) (thousand km per second) / (billion per second)

Wavelength = (28.57) (million meters / second) / (thousand million / second)

Wavelength = (28.57) (meters / second) / (thousand / second)

Wavelength = (28.57) (meters / thousand)

Wavelength = (28.57) (millimeters)

5 0

3 years ago

A current of 12 amps is measured in a circuit with a total resistance of 9.0 ohms. What is the size of the voltage source that s

Serjik [45]

Data:
i (current) = 12 A
R (resistance) = 9.0 Ω
V (voltage) = ? (volts)

Formula:
$V = R*i$

Solving:
$V = R*i$
$V = 9.0*12$
$\boxed{\boxed{V = 108\:volts}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At t = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive x-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

2 years ago