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dezoksy [38]
3 years ago
9

A test car is driving toward a solid crash-test barrier with a speed of 46 mi/h. Two seconds prior to impact, the car begins to

brake, but it is still moving when it hits the wall. After the collision with the wall, the car crumples somewhat and comes to a complete stop. In order to estimate the average force exerted by the wall on the car, what information would you need to collect?
Physics
1 answer:
Mekhanik [1.2K]3 years ago
4 0

Answer:

we have to measure distances and time, possibly with an automated system since the values ​​are very small

Explanation:

For this exercise we can use the relationship between momentum and momentum

        I = ∫ F dt = Δp

        F t = m v_{f} - mv₀

In the exercise they indicate that the final speed is zero

         F t = - m v₀

         F = -m v₀ / t

With this equation we can find what measurements should be carried out.

To find the speed with which the car collides with the wall, less measure the displacement and its time during the braking process before reaching the wall and from here find the speed with which it reaches the wall.

During the impact, we must find the time that the vehicle is in contact with the wall in the first approach is equal to the time that the car takes to reach the final speed of zero.

In summary we have to measure distances and time, possibly with an automated system since the values ​​are very small

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Ede4ka [16]
Work = force x distance. In units, Joules = Newtons x meters.

So: Work = 50 Newtons x 3 meters

Work  = 150 joules. Answer D is correct

FYI - to ace physics, you should learn to identify these values using their fundamental units:
Force = Newtons = Kg·m/s²
Work = joules = kg·m²/s²
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In high school physics, If you learn to arrange equations so the units work out properly for the answer, you'll get most problems correct.
7 0
3 years ago
A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.
g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle=34^{\circ}

Range=240 m

We know that Range =\frac{u^2sin2\theta }{g}

240=\frac{u^2sin(68)}{9.81}

u=50.39 m/s

(b)maximum height of projectile is given by

H=\frac{u^2sin^2\theta }{2g}

H=\frac{50.39^2(sin34)^2}{2\times 9.81}

H=40.46 m

3 0
3 years ago
Read 2 more answers
Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien
otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

  • The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
  • The height at which the block stops rising is approximately 1.1415 m
  • The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m}  = \mathbf{ 1000\, N/m}

Coefficient of kinetic friction, \mu_k = 0.39

Therefore, we have;

Friction force = \mathbf{\mu_k}·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

  • The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

At \ the \ maximum \ height, \ h, \ we \ have ; \  \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x

Which gives;

Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{  7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2  } \approx 1.536

The distance up the inclined, the block rises, at maximum height is therefore;

x_{max} ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

  • The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

  • Length of the incline, x_{max} = 1.536 m

Learn more about conservation of energy here:

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5 0
2 years ago
An ecg is a recording of the electrical activity of the heart
lukranit [14]

Answer:

True

Explanation:

This is a true statement my friend :)

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Describe what happens to the particles of water when it boils.
Nezavi [6.7K]

Answer:

The liquid turns to a gas.

Explanation:

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8 0
3 years ago
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