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FrozenT [24]
4 years ago
12

Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going

30 mph. In the second option you are driving 30 mph and crash head-on into a stationary brick wall. In neither case does your car bounce off the thing it hits, and the collision time is the same in both cases. Which of these two situations would result in the greatest impact force?

Physics
2 answers:
11Alexandr11 [23.1K]4 years ago
6 0

These two situations would result in the greatest impact force: The force would be the same in both cases.

<h3>Explanation: </h3>

Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going 30 mph. In the second option you are driving 30 mph and crash head-on into a stationary brick wall. In neither case does your car bounce off the thing it hits, and the collision time is the same in both cases. Which of these two situations would result in the greatest impact force?

F \Delta t= m \Delta V\\ where

  • The force would be the same in both cases.
  • Hitting the brick wall
  • Hitting the other car
  • None of the above choices are correct

Newton 3rd law stated that for every action, there is an equal and opposite reaction.

For example the forces on the trucks are equal but opposite in the direction. Newton 3rd law of impulse stated about momentum formula.  Impulse is force multiplied by time, whereas the time of contact is the same for both, therefore the impulse is the same in magnitude for the two trucks.

Let's look on the attached picture. The case of hitting the brick wall have the same impact force as hitting the other car, because they have the same change in momentum and it doesn't matter what the source is. In first case both the cars are identical and have same velocity whereas in the second case, the wall is stationary.  Also, in both the cases the car does not bounces off the things it hit.

Learn more about  impact force brainly.com/question/13828718

#LearnWithBrainly

DerKrebs [107]4 years ago
5 0

Answer:

The impact force will be same for both the cases.

Explanation:

The rate of change of momentum is known as the Impulse and is given by:

I = \frac{\Delta p}{\Delta t}

where

I = Impulse

\Delta p = change in momentum

\Delta t = time interval

Now,

In first case both the cars are identical and have same velocity and in the second case, the wall is stationary.

Also, in both the cases the car does not bounces off the things it hit.

Thus

\Delta p = 0 - m\times v = - mv

Thus

Impact force, F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t}

Therefore, impact force is same for both the cases.

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Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha
Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N.  Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.  

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation  is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

a_B = a_A-0.5 \frac{m}{s^2}

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0
4 years ago
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Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
4 years ago
PLEASE HELP ASAP!!
Stells [14]

Its B: reduce the amount of energy needed to do the work by putting the work onto something else

3 0
3 years ago
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After the body has begun shivering, what explains why it would stop shivering?
Oksana_A [137]
The body shivers to produce energy and it uses the energy to keep it warm. The body would stop shivering when it has produced enough energy to keep it warm and the atmosphere around it has got warmer
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3 years ago
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If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?
Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

y=y_{o}+V_{oy}t-\frac{g}{2}t^{2} (1)

Where:

y=0 is the final height of the ball

y_{o}=h is the initial height of the ball

V_{oy}=V_{o}sin(0\°)=0 is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

t=0.5 s is the time at which the ball lands

g=9.8 m/s^{2} is the acceleration due gravity

So, with these conditions the equation is rewritten as:

h=\frac{g}{2}t^{2} (2)

h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2} (3)

Finally:

h=1.22 m

4 0
3 years ago
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