Britain and France signed an entente and became thea.
Central Forces.
c.
Central Powers.
b.
Allied Forces.
d.
Axis Powers.

Part A: The enmeshed cars were moving at a velocity of approximately 8.66 m/s just after the collision.

Part B: Car A was traveling at a velocity of approximately 8.55 m/s just before the collision.

How to compute the above velocities

To find the speed of car A just before the collision in Part B, you can use the principle of conservation of momentum.

The total momentum of the system before the collision should equal the total momentum after the collision. You already know the total momentum after the collision from Part A, and now you want to find the velocity of car A just before the collision.

Let's denote:

- v_A as the initial velocity of car A before the collision.

- v_B as the initial velocity of car B before the collision.

In Part A, you found that the enmeshed cars were moving at a velocity of 8.66 m/s at an angle of 60 degrees south of east. You can split this velocity into its eastward and southward components. The eastward component of this velocity is:

v_east = 8.66 m/s * cos(60 degrees)

Now, you can use the conservation of momentum to set up an equation:

Total initial momentum = Total final momentum

(mass_A * v_A) + (mass_B * v_B) = (mass_A + mass_B) * 8.66 m/s (the final velocity you found in Part A)

Plug in the known values:

(1900 kg * v_A) + (1500 kg * v_B) = (1900 kg + 1500 kg) * 8.66 m/s

Now, you can solve for v_A:

(1900 kg * v_A) + (1500 kg * v_B) = 3400 kg * 8.66 m/s

1900 kg * v_A = 3400 kg * 8.66 m/s - 1500 kg * v_B

v_A = (3400 kg * 8.66 m/s - 1500 kg * v_B) / 1900 kg

Now, plug in the values from Part A to find v_A:

v_A = (3400 kg * 8.66 m/s - 1500 kg * 8.66 m/s) / 1900 kg

v_A = (29244 kg*m/s - 12990 kg*m/s) / 1900 kg

v_A = 16254 kg*m/s / 1900 kg

v_A ≈ 8.55 m/s

So, car A was going at approximately 8.55 m/s just before the collision in Part B.

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The answer is A. A warm and humid day with a wind moving in the same direction as the sound. hope this helps!!!!

Answer:

1.Cp₁ = 1.2 J/g.⁰C

Explanation:

For new material:

m₁  = 25 g

T₁ = 80⁰C

specific heat of water = Cp₁

For water :

m₂ = 100 g

T₂ = 20⁰C

The final temperature T=24⁰C

We know that specific heat of water Cp₂ = 4.187 kJ/kg.K

The heat lost new material = Heat gain by Water

m₁ Cp₁ ( T₁ - T ) = m₂ Cp₂ (T- T₂)

25 x Cp₁ (80- 24 ) = 100 x 4.817 (24 - 20 )

Cp₁  x 56 = 4 x 4.187 x 4

Cp₁ = 1.19 kJ/kg.K

Cp₁ = 1.2 J/g.⁰C

Answer:

1.1 x 10⁵m/s²

Explanation:

Given parameters:

Velocity  = 452m/s

distance  = 0.93m

Unknown:

Acceleration of the bullet  = ?

Solution:

To solve this problem, we use one of the kinematics equation which is given below:

           V² = U²  + 2aS

V is the final velocity

U is the initial velocity  = 0m/s

a is the unknown acceleration

S is the distance traveled  

 So;

          452²   = 0²  + (2 x a x 0.93)

          204304  = 1.86a

               a  = 1.1 x 10⁵m/s²

Final answer:

The acceleration of the bullet in the gun barrel can be calculated using the kinematic equation for motion. By substituying the given values into the equation, we find the acceleration to be approximately 1.095 x 10^5 m/s^2.

Explanation:

The subject of this question is Physics, specifically a topic under mechanics known as kinematics. The problem given can be solved using kinematic equations which are used to describe the motion of an object without considering the forces that cause it to move. In this case, the final velocity (vf) of the bullet is given as 452 m/s, the initial velocity (vi) is assumed to be 0 as it starts from rest, and the distance (d) is given as 0.93 m. We are asked to determine the value of acceleration (a).

Using the kinematic equation vf2 = vi2 + 2ad and substituting the given values, we get (452 m/s)2 = 0 + 2*a*0.93 m. We can rearrange to solve for acceleration to get: a = (452 m/s)2 / (2*0.93 m) = 109523.66 m/s2.

So, the acceleration of the bullet in the gun barrel is approximately 1.095 x 105 m/s2.

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Answer:

C. The alpha particle will be deflected in a curve path.