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2 years ago

Explain why the motto "Do whatever it takes to win!" may not be an ethical guideline to follow.

Physics

2 answers:

Murrr4er [49]2 years ago

4 0

If you did this then it could lead to cheating or someone else getting hurt.

Karolina [17]2 years ago

3 0

Be selfish is pretty much what its telling you

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The headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer who

densk [106]

Answer:

The most correct option is;

B. 10 km

Explanation:

$L = \frac{y \times d}{1.22 \times \lambda} = \frac{1.2 \times 0.004}{1.22 \times 400 \times 10^{-9}} = 9836.066 \ km$

Where:

y = Distance between the two headlights

d = Aperture of observers eye

λ = Wavelength of light

L = Distance between the observer and the headlight

Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.

Also we have

sinθ = y/L = 1.22 (λ/d)

$= 1.22 \times \frac{400 \times 10^{-9}}{0.004}$

sinθ = 1.22×10⁻⁴ rad

6 0

2 years ago

Read 2 more answers

In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

Mandarinka [93]

Answer:

0.22 m

Explanation:

We are told that the driver can survive an acceleration of 50g only if the collision lasts no longer than 30 ms. So,

$t = 30 ms = 0.030 s$

The acceleration is

$a=-50g = -50(9.8)=-490 m/s^2$

where the negative sign is due to the fact that this is a deceleration, since the driver comes to a stop in the collision.

First of all, we can find what the initial velocity of the car should be in this conditions by using the equation:

$v=u+at$

And since the final velocity is zero, v=0, and solving for u,

$u=-at=-490(0.030)=14.7 m/s$

And now we can find the corresponding distance travelled using the equation:

$d=ut+\frac{1}{2}at^2 = (14.7)(0.030)+\frac{1}{2}(-490)(0.030)^2=0.22 m$

8 0

2 years ago

If low CVP precipitates a suction alarm, rapid infusion of volume can remedy the situation after dropping the P-level.

motikmotik

Answer:

Explanation:

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2 years ago

You throw a baseball straight up into the air with a speed of 24.5 m/s. How long does it take the baseball to reach its highest

galben [10]

Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
Time=24.5/9.8=2.5 s

3 0

2 years ago

Read 2 more answers

A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b

trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2$

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$ -------(1)

substitute $P_a_t_m for P_1, (P_a_t_m +pgh) for P_2$

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$

$\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}$

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

$\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4$

substitute

$P_a_t_m for P_3, P_a_t_m for P_4$

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

$\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}$

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

$\sqrt{64.4(20-h)}$ for V₂

$\sqrt{64.4h}$ for V₄

A₂V₂ = A₃V₃

$A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft$

Therefore ,height of the water rise in tank is 10ft

3 0

2 years ago