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4 years ago

A wave that can travel with or without a medium is called a(n)

Physics

1 answer:

denis23 [38]4 years ago

5 0

Answer:

Explanation:

Answer A: A surface wave is a wave that travels along the surface of a medium.

Answer B, C: Electromagnetic waves are waves that have no medium to travel whereas mechanical waves need a medium for its transmission.

Answer D: The sentence in the answer D does not fit to the blank in the definition ( of the question )

......

Hope this answer can help you.

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4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The

GalinKa [24]

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time $t$ , the horizontal position $x$ and vertical position $y$ of the ball are given respectively by

$x = v_i \cos(\theta_i) t$

$y = v_i \sin(\theta_i) t - \dfrac g2 t^2$

and the horizontal velocity $v_x$ and vertical velocity $v_y$ are

$v_x = v_i \cos(\theta_i)$

$v_y = v_i \sin(\theta_i) - gt$

The ball reaches its maximum height with $v_y=0$ . At this point, the ball has zero vertical velocity. This happens when

$v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g$

which means

$y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g$

At the same time, the ball will have traveled half its horizontal range, so

$x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g$

Solve for $v_i$ and $\theta_i$ :

$\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0$

Since $0^\circ$ , we cannot have $\sin(\theta_i)=0$ , so we're left with (e)

$3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}$

Now,

$\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}$

so it follows that (d)

$R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}$

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

$v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}$

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

$v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}$

Then with $v_y=0$ , the ball's speed $v$ is

$v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}$

Finally, in the work leading up to part (e), we showed the time to maximum height is

$t = \dfrac{v_i \sin(\theta_i)}g$

but this is just half the total time the ball spends in the air. The total airtime is then

$2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}$

and the ball is in the air over the interval (a)

$\boxed{0 < t < 2\sqrt{\frac R{3g}}}$

7 0

3 years ago

A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific he

Gekata [30.6K]

The specific heat capacity of the block is $508J/kg^{\circ}C$

Explanation:

As the block is placed into the water, heat energy is transferred from the water (which is at higher temperature) to the block (which is at lower temperature), until the block and the water are in thermal equilibrium (= same temperature).

Therefore, we can write:

$Q_{water}=Q_{block}$

Where

$Q_{water}=m_w C_w (T_w-T_{eq})$ is the heat energy released by the water, where

$m_w = 0.217 kg$ is the mass of the water

$C_w = 4186 J/kg^{\circ}C$ is the water heat specific capacity

$T_w = 25.0^{\circ}$ is the initial temperature of the water

$T_{eq}=16.4^{\circ}$ is the temperature at equilibrium

Substituting,

$Q_{water}=(0.217)(4186)(25.0-16.4)=7812 J$

Now we can write the heat energy absorbed by the block as

$Q_{block}=m_b C_b(T_{eq}-T_b)$

where

$m_b=0.350 kg$ is the mass of the block

$C_b$ is the specific heat capacity of the block

$T_b = -27.5^{\circ}$ is the initial temperature of the block

And solving for $C_b$ ,

$C_b=\frac{Q_{block}}{m_b(T_{eq}-T_b)}=\frac{7812}{(0.350)(16.4-(-27.5))}=508J/kg^{\circ}C$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

6 0

3 years ago

A uniform conducting rod of length 34 cm has a potential difference across its ends equal to 39 mV (millivolts). What is the mag

lozanna [386]

Electric field is the change in voltage per unit distance

E = ΔV / d

E = 39/34

= 1/147 mV/cm

hope this helps

3 0

4 years ago

Read 2 more answers

_______ are considered to be fluids.

skad [1K]

Answer:

liquids and gases

Explanation:

Liquids and gases are considered to be fluids because they yield to shearing forces, whereas solids resist them.

4 0

3 years ago

Black holes form when____ overcomes the outward____produced by nuclear reactions within a star.

OleMash [197]

"a black hole forms when any object reaches a certain critical density, and its gravity causes it to collapse to an almost infinitely small pinpoint."

I know its not straight forward with the answer

7 0

3 years ago

Read 2 more answers