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navik
1 month ago
10

Three cars are tested in a 16 m/s frontal crash. The results of the crash tests are shown below, with data indicating how much t

he crumple zone and safety cell of each car deformed. In each car the dummy was secured by the same seat belt and airbag. Which car did the best job of protecting the dummy?

Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
4 0

Response:

  Vehicle C

Rationale:

The deformed crumple zone in Vehicle C is considerably lengthier, resulting in a reduced average deceleration. This lower rate of acceleration exerts less force on the dummy. Vehicle C provided superior protection for the dummy.

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Two loudspeakers 42.0 m apart and facing each other emit identical 115 Hz sinusoidal sound waves in a room where the sound speed
ValentinkaMS [3465]

Answer:

She will be located in a loud area.

Explanation:

To evaluate whether Susan is situated in a quiet or loud area, we must first ascertain the difference in the sound waves reaching her from the two speakers.

If we refer to d₁ as the distance to speaker A, which measures 19.5 m, then d₂ to the other speaker is calculated as follows:

d₂ = 42.0 m - 19.5 m = 22.5 m

This reveals that the path difference between both speakers is:

d = d₂-d₁ = 22.5 m -19.5 m = 3.0 m

Next, we relate this distance to the sound wave's wavelength; for constructive interference, the path difference must equal an even multiple of half the wavelength:

d = (2n)*(λ/2)

To determine λ, we know that for any wave, the relationship between speed, frequency, and wavelength holds:

v = λ*f

<pwhere v="345" m="" and="" f="115" we="" can="" calculate="">

λ = v/f = 345 m/s / 115 (1/s) = 3.0 m.

Since the difference between the distances to Susan from both speakers corresponds exactly to one wavelength, this indicates that both waves arrive at the same phase, resulting in constructive interference, thus labeling it as a loud area.

</pwhere>
5 0
2 months ago
A 620-g object traveling at 2.1 m/s collides head-on with a 320-g object traveling in the opposite direction at 3.8 m/s. If the
Keith_Richards [3271]

Answer:

No kinetic energy is lost as the collision is elastic.

Explanation:

Throughout an elastic collision, both momentum and kinetic energy remain conserved.

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6 0
2 months ago
Read 2 more answers
For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the maximum ra
Ostrovityanka [3204]
Refer to the diagram below.

This discussion operates under a basic analysis that overlooks air resistance and variations in the terrain the missile traverses.

Let V₀ be the launch velocity, at an angle θ to the horizontal.
The horizontal velocity component equals V₀ cosθ.
If the flight duration is t_{f}, then
r=V_{o} \, t_{f}
where r represents the missile's range.

The time t at which the missile is at ground level is expressed by
0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2}
where g signifies acceleration due to gravity.

t = 0 signifies the missile's launch. Thus
t_{f} = \frac{2V_{o}sin\theta}{g}

Consequently,
r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g}

Typically, an angle of θ=45° is optimal for achieving maximum range, resulting in
r= \frac{V_{o}^{2}}{g}

This discussion applies more accurately to a scud missile than to a powered, guided missile.

Response:
t_{f} = \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g}
Usually, θ=45°

6 0
2 months ago
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t
ValentinkaMS [3465]

Kinetic energy is represented as

KE = (0.5) m v²

In each scenario, v = the velocity of the bottle set at  4 m/s

with m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

for m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

if m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

6 0
2 months ago
Read 3 more answers
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