All categories

3 years ago

What energy does a galloping horse and a speeding bullet represent

Physics

2 answers:

Alex73 [517]3 years ago

8 0

Since both of those things are moving,
they both have Kinetic Energy.

Darya [45]3 years ago

6 0

These are both forms of kinetic energy.

(: I hope this helps! :)

You might be interested in

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

Help!?!? Please and thank you

natita [175]

C is probably the correct one

3 0

3 years ago

Read 2 more answers

234 oz to tons using the table method (US)

Rom4ik [11]

Answer:

234 oz = 0.0066339 t.

Explanation:

Boom Logic...

3 0

3 years ago

During a hurricane in 2008, the Westin Hotel in downtownNew Orleans suffered damage. Suppose a piece of glass dropped near the t

tangare [24]

Answer:

77.8 m/s, downward

Explanation:

For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf

Average speed = (Vf + Vi)/2

Also, by definition, the average speed is the distance divided by the time:

Average speed = distance / time

Then:

(Vf + Vi)/2 = 300m/6.62s

Other kinematic equation for uniform acceleration is:

Vf = Vi + a×t

Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)

Replacing the known values we can set a system of two equations:

From (Vf + Vi)/2 = 300m/6.62s

(Vf + Vi) = 2 × 300m/6.62s

Vf + Vi = 90.634 equation 1

From Vf = Vi + a×t

Vf - Vi = 9.8 (6.62)

Vf - Vi = 64.876 equation 2

Adding the two equations:

2Vf = 155.510

Vf = 77.8 m/s downward (velocities must be reported with their directions)

8 0

2 years ago

Snorkelers breathe through tubes that extend above the surface of the water. In prin- ciple, a snorkeler could go deeper with a

Yanka [14]

Answer:

h = 1.02 m

Explanation:

This is a fluid mechanics exercise, where the pressure is given by

P = $P_{atm}$ + ρ g h

The gauge pressure is

P - $P_{atm}$ = ρ g h

In this case the upper part of the tube we have the atmospheric pressure. and the diver can exert a pressure 10 KPa below the outside pressure, this must be the gauge pressure

$P_{m}$ = P - $P_{atm}$

$P_{m}$ = ρ g h

h = $P_{m}$ / ρ g

calculate

h = 10 103 / (1000 9.8)

h = 1.02 m

This is the depth at which man can breathe

8 0

3 years ago