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Anestetic [448]
3 years ago
12

(96 Points )A bus and a bicycle have a head-on collision. Compare the force of impact between the bus and the bicycle; compare t

he accelerations of the bus and bicycle.
Physics
1 answer:
lidiya [134]3 years ago
6 0

Answer:

Small acceleration for the bus and major for the bicycle

Explanation:

In a head-on collision of this type, i order to find force involve in a collision, one has to study the change in linear momentum of each object. Because Force s defined as the change of momentum with time.

Recall that momentum is the product of the mass of the object times its velocity. The mass of the bus (M) is much larger than that of the bicycle (m), and even if their speeds are similar, the momentum of the bus will be much larger than that of the bike due to this mass difference. We could assume an inelastic collision (where the two objects stay together after the collision),

The initial total momentum of the system (imagining that the bus is pointing to the right (positive convention), while the bike is pointing in the opposite (to the left), can be written as: Pi=Mv-mv (we are assuming that both vehicles go at the same speed for simplicity)

The final total momentum of the system that underwent an inelastic collision will be: P_f=(M+m)v_f

Therefore, since the mass M is much larger than mass m, the final momentum will be pretty similar to the initial momentum of the bus.

The change in momentum for the bus in the fraction of time of the collision will be almost unnoticeable for the bus (minimal change in its momentum, and therefore minimal acceleration accounting for the change).

For the bicycle, the change in momentum involves change in the direction of motion, (going to the left before the collision, and to the right afterwards) which will include a mayor acceleration backwards.

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Decreases. Air resistance will slow a falling object to its terminal velocity, placing a limit on its acceleration. 
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Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th
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<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

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3 years ago
If a box is pulled with a force of 100 N at an angle of 25
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The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

<u>Given the following data:</u>

  • Force = 100 Newton.
  • Angle of inclination = 25°

To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

F_x = Fcos \theta\\\\F_x =100 \times cos25\\\\F_x =100 \times 0.9063

Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

F_y = Fsin \theta\\\\F_y =100 \times sin25\\\\F_y =100 \times 0.4226

Fy = 42.26 Newton.

Read more on horizontal component here: brainly.com/question/4080400

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Answer:

25°C

Explanation:

Using the linear expansivity formula expressed as;

∝ = ΔL/lΔθ

∝ is coefficient of lineat expansion = 1.2 ∙ 10⁻⁵ °C⁻¹

ΔL is the change in length = 6.00036-6

ΔL = 0.00036m

l is the original length = 6m

Δθ is the change in temperature =θ₂-20

Substituting into the formula;

1.2 ∙ 10⁻⁵ °C⁻¹  = 0.00036/6(θ₂-20)

cross multiply

1.2 ∙ 10⁻⁵ * 6  = 0.00036/(θ₂-20)

7.2 ∙ 10⁻⁵= 0.00036/(θ₂-20)

0.00036 = 7.2 ∙ 10⁻⁵(θ₂-20)

0.00036 = 7.2 ∙ 10⁻⁵θ₂-144∙ 10⁻⁵

7.2 ∙ 10⁻⁵θ₂ = 0.00036+0.00144

7.2 ∙ 10⁻⁵θ₂ = 0.0018

θ₂ = 0.0018/0.000072

θ₂ = 25°C

Hence the temperature at which this bar must be acidic for its compression is 6,00036 m is 25°C

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