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pashok25 [27]
2 years ago
7

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

t end crumples so much that the truck is 0.62 m shorter than before.
(a) What is the average speed of the truck during the collision (that is, during the interval between the first contact with the wall and coming to a stop)?
(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)
(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?
(d) It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force mg on the truck. This large ratio shows why a collision is so damaging.
(e) What approximations were necessary in making this analysis? (Select all that apply.)

a. Neglect the horizontal component of the force of the road on the truck tires.
b. Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.
c. The deceleration of the truck is approximately equal to g.
Physics
1 answer:
slega [8]2 years ago
5 0

Answer:

a)   v_average = 11 m / s, b)  t = 0.0627 s

, c)    F = 7.37 10⁵ N

, d)   F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

The average speed is

           v_average = (v + v₀) / 2

           v_average = (22 + 0) / 2

           v_average = 11 m / s

b) The average time

          v = v₀ - a t

          t = v₀ / a

          t = 22 / 350.72

          t = 0.0627 s

c) The force can be found with Newton's second law

             F = m a

             F = 2100 350.72

             F = 7.37 10⁵ N

.d) the ratio of this force to weight

             F / W = 7.37 10⁵ / (2100 9.8)

             F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

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A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz,
masha68 [24]

Answer:

a) The car was moving at a speed of 29.167\ m.s^{-1}

b) The negative sign of v_o denotes that the observer is coming towards the police car which is the source of the sound.

c) f_o=1283.33\ Hz

Explanation:

Given:

  • original frequency of the source, f=1200\ Hz
  • speed of the source, v_s=0\ m.s^{-1}
  • velocity of the obstacle car be, v_o
  • speed of sound, s=350\ m.s^{-1}
  • observed frequency, f_o=1100\ Hz

<u>Using the equation from the Doppler's effect:</u>

\frac{f_o}{f} =\frac{(s+v_o)}{(s-v_s)}

\frac{1100}{1200} =\frac{(350+v_o)}{350-0}

v_o=-29.167\ m.s^{-1}

a)

The car was moving at a speed of 29.167\ m.s^{-1}

b)

The negative sign of v_o denotes that the observer is coming towards the police car which is the source of the sound.

c)

Now when, v_s=50\ m.s^{-1}

Then, f_o=?

Using the Doppler's eq.:

\frac{f_o}{1200} =\frac{(350+(-29.167))}{(350-50)}

f_o=1283.33\ Hz

3 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

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2 years ago
As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th
Gre4nikov [31]
Transvere wave because the direction which the particles are being displaced
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3 years ago
Determine the half-life of uranium-238 based on the graph.
Nikitich [7]
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3 years ago
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Which of the statements concerning light are true? The speed of light is the same no matter what material it is traveling throug
wel

Answer:

The statements that are true concerning light are the last three statements:

  • Its propagation direction is perpendicular to both the electric field and the magnetic field.
  • It moves at a constant speed through a vacuum.
  • The speed of light in matter is less than the speed of light in a vacuum.

Explanation:

<em>Light</em> is <em>electromagnetic waves.  </em>

The properties of the electromagnetic waves were established by James Clerk Maxwell.

They included that they are the result of the oscillation of a <em>magnetic field </em>in phase with an <em>electric field</em> which are always is always <em>perpendicular</em> to each other.

Also, the electromagnetic waves propagate at right-angles to the direction of both the magnetic and the electric field,  meaning that they are a type of transverse wave.

Thus, the second statement (<em>"Its propagation direction is parallel to both the electric field and the magnetic field"</em>) is false, and the fourth statement ("Its propagation direction is perpendicular to both the electric field and the magnetic field") is true.

On the other hand, it is a postulate of the special theory of relativity that the speed of light is a constant (absolute value) in vacuum: nothing can travel faster than what light travels in vacuum. Thus, the fifth statement, <em>"It moves at a constant speed through a vacuum"</em> is true.

About the speed of light in matter, it is always less than the speed of light in vacuum. Thus, the first statement, "<em>the speed of light is the same no matter what material it is traveling through</em>", and the third statement "<em>the speed of light in matter is greater than the speed of light in a vacuum"</em> are false; while the last statement, "<em>the speed of light in matter is less than the speed of light in a vacuum</em>" is true.

The explanation on why the speed of light is less in a medium than in vacuum is related with the fact that at nanoscopic level the waves suffer polarization which means deviations from the straighi path, which makes that the net straight propagation is slower.

8 0
3 years ago
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