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FrozenT [24]
2 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]2 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]2 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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dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
2 years ago
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

3 0
2 years ago
(1 point) Which of the following statements are true?A.The equation Ax=b is referred to as a vector equation.B.If the augmented
Anvisha [2.4K]

Answer:

A. False

B. False

C. True

D. True

E. True

F. True

Explanation:

A. The equation Ax=b is referred to as a matrix equation and not vector equation.

B. If the augmented matrix [ A b ] has a pivot position in every​ row then equation Ax=b may or may not be consistent. It is inconsistent if [A b] has a pivot in the last column b and it is consistent if the matrix A has a pivot in every row.

C. In the product of Ax also called the dot product the first entry is a sum of products. For example the the product of Ax where A has [a11 a12 a13] in the first entry of each column and the corresponding entries in x are [x1 x2 x3] then the first entry in the product is the sum of products i.e.  a11x1 + a12x2 +a13x3

D. If the columns of mxn matrix A span R^m, this states that every possible vector b in R^m is a linear combination of the columns which makes the equation consistent. So the equation Ax=b has at least one solution for each b in R^m.

E. It is stated that a vector equation x1a1 + x2a2 + x3a3 + ... + xnan = b has the same solution set as that of the linear system with augmented matrix [a1 a2 ... an b]. So the solution set of linear system whose augmented matrix is [a1 a2 a3 b] is the same as solution set of Ax=b if A=[a1 a2 a3]  and b can be produced by linear combination of a1 a2 a3 iff the solution of linear system corresponding to [a1 a2 a3 b] takes place.

F.  It is true because lets say b is a vector in R^m which is not in the span of  the columns. b cannot be obtained for some x which belongs to R^m as b = Ax. So Ax=b is inconsistent for some b in R^m and has no solution.

8 0
2 years ago
Which letter correctly identifies the part of the hydrologic cycle that is most directly affected by impervious building materia
masha68 [24]

Infiltration

Explanation:

The component of the hydrologic cycle affected by impervious building such as concrete and asphalt is infiltration.

  • Water infiltration is a major component of the hydrologic cycle.
  • Concretes and other materials can prevent water from going down into the earth.
  • This affects the ground water system in place.
  • It leads to increase in surface run off and might cause inundation of an area.
  • Infiltration is a very important component of water cycle.
  • It takes water to plant root and recharges groundwater systems.
  • Impervious structures takes this capability away.

learn more:

Biogeochemical cycle brainly.com/question/3509510

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3 0
2 years ago
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
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