To solve this problem, we will use the equations of motion.

Step 1: Convert the distance penetrated by the bullet into meters, because the standard unit of distance in physics is meter. So, 6 cm = 0.06 m.

Step 2: We know that the final velocity of the bullet is 0 m/s (because it stops in the sandbox), the initial velocity is 20 m/s, and the distance (s) is 0.06 m.

Step 3: We use the second equation of motion, which is v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, deceleration), and s is the distance.

Step 4: Substituting the known values into the equation, we get 0 = (20)^2 + 2*a*0.06.

Step 5: Simplifying the equation, we get -400 = 0.12a.

Step 6: Solving for a, we get a = -400 / 0.12 = -3333.33 m/s^2.

So, the deceleration of the bullet in the sandbox is -3333.33 m/s^2. The negative sign indicates that it is a deceleration.

Question

To solve this problem, we will use the equations of motion.

Step 1: Convert the distance penetrated by the bullet into meters, because the standard unit of distance in physics is meter. So, 6 cm = 0.06 m.

Step 2: We know that the final velocity of the bullet is 0 m/s (because it stops in the sandbox), the initial velocity is 20 m/s, and the distance (s) is 0.06 m.

Step 3: We use the second equation of motion, which is v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, deceleration), and s is the distance.

Step 4: Substituting the known values into the equation, we get 0 = (20)^2 + 2*a*0.06.

Step 5: Simplifying the equation, we get -400 = 0.12a.

Step 6: Solving for a, we get a = -400 / 0.12 = -3333.33 m/s^2.

So, the deceleration of the bullet in the sandbox is -3333.33 m/s^2. The negative sign indicates that it is a deceleration.

Knowee AI · Accepted Answer

To solve this problem, we will use the equations of motion.

Step 1: Convert the distance penetrated by the bullet into meters, because the standard unit of distance in physics is meter. So, 6 cm = 0.06 m.

Step 2: We know that the final velocity of the bullet is 0 m/s (because it stops in the sandbox), the initial velocity is 20 m/s, and the distance (s) is 0.06 m.

Step 3: We use the second equation of motion, which is v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, deceleration), and s is the distance.

Step 4: Substituting the known values into the equation, we get 0 = (20)^2 + 2*a*0.06.

Step 5: Simplifying the equation, we get -400 = 0.12a.

Step 6: Solving for a, we get a = -400 / 0.12 = -3333.33 m/s^2.

So, the deceleration of the bullet in the sandbox is -3333.33 m/s^2. The negative sign indicates that it is a deceleration.

A bullet hits a Sandbox with a velocity of 20 m/s and penetrates it up to a distance of 6 cm. Find the deceleration of the bullet in the sandbox.A

Question

A bullet hits a Sandbox with a velocity of 20 m/s and penetrates it up to a distance of 6 cm. Find the deceleration of the bullet in the sandbox.

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