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3 years ago

Two mugs are on a table. What would you need to know in order to determine which mug contains the most thermal energy?

1 answer:

6 0

Change in thermal energy = mass × specific heat capacity × change in temperature.
Answer: you would need to know the mass of each mug

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Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

iris [78.8K]

Answer:

Explanation:

1. $V_{x}$ = $V_{0}$ * cos $\alpha$ ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. $V_{y}$ = $V_{0}$ * sin $\alpha$ ⇒ 16* sin32 ≈ 9.4 m/s

3. $y_{max}$ = $\frac{v_{0}^2*sin^2\alpha}{2g}$ = $\frac{16^2*sin^232}{2*9.8}$ (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

$y_{max}$ ≈ 3.6677+1.5 ≈ 5.2m

4. $x_{max}$ = $\frac{v_{0}^2*sin(2\alpha)}{g}$ = $\frac{16^2*sin(2*32)}{9.8}$ ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

3 0

3 years ago

Lymph ducts are more similar to arteries than veins.<br> A. <br> True<br> B. <br> False

Strike441 [17]

Answer:

False

Explanation:

5 0

4 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

NEED ANSWER PLEASE!!!!

olganol [36]

Answer:A, Concave

Explanation:

3 0

4 years ago

The driver of a car traveling at 31.3 m/s applies the brakes and undergoes a constant deceleration of 1.6 m/s2.How many revoluti

lisov135 [29]

Answer:

$R=156.99\operatorname{Re}vs$

Explanation: The equations used are as follows:

$\begin{gathered} x(t)=x_o+v_ot+\frac{1}{2}at^2\Rightarrow(1) \\ v(t)=v_o+at\Rightarrow(2) \end{gathered}$

By using equation (2), the time needed for the car to come to rest is calculated as follows:

$\begin{gathered} v(t)=(31.3ms^{-1})_{}+(-1.6ms^{-2})t=0 \\ t=\frac{31.3ms^{-1}}{1.6ms^{-2}}=19.56s \\ t=19.563s \end{gathered}$

By using equation (1), The total distance traveled in that time would be as:

$\begin{gathered} x(19.563s)=_{}(31.3ms^{-1})\cdot(19.563s)+\frac{1}{2}(-1.6ms^{-2})\cdot(19.563s)^2\Rightarrow(1) \\ x(19.563s)=612.31-306.17=306.14m \\ \therefore\Rightarrow \\ x(19.563s)=306.14m \end{gathered}$

The revolutions taken by the tire before the car comes to rest would be:

$\begin{gathered} C=2\pi\cdot(0.31m)=1.95m \\ R=\frac{x(19.563s)}{C}=\frac{306.14m}{1.95m}=156.99\operatorname{Re}v \\ R=156.99\operatorname{Re}vs \end{gathered}$

3 0

2 years ago