Conceptual Question 8.09osama wadi


Part A
In a perfectly ELASTIC collision between two perfectly rigid objects
the momentum of each object is conserved.
both the momentum and the kinetic energy of the system are conserved.
the kinetic energy of the system is conserved, but the momentum of the system is not conserved.
the kinetic energy of each object is conserved.
the momentum of the system is conserved but the kinetic energy of the system is not conserved.
...?

The work done by nonconservative forces in stopping the 17,000-kilogram airplane landing at a speed of 82 m/s is 57,062,000 Joules. This is calculated by the change in kinetic energy of the airplane when it lands and comes to a stop.

The question refers to the concept of work-energy theorem in Physics, especially involving non-conservative forces. The airplane is initially moving and finally comes to rest. Its initial kinetic energy (KE) gets transferred to work done by nonconservative forces, which in this scenario includes friction due to the aircraft carrier deck and air resistance.

The initial kinetic energy of the plane is calculated using the formula 1/2 * m * v^2 where 'm' is the mass of the plane and 'v' is its speed. So, the initial kinetic energy of the plane is 1/2 * 17,000 kg * (82 m/s)^2 = 57,062,000 Joules. When the plane comes to rest, its final kinetic energy is 0. As per the work-energy theorem, the work done by nonconservative forces is equal to the change in the kinetic energy. Therefore, the work done by nonconservative forces in stopping the plane = Initial KE - Final KE = 57,062,000 Joules - 0 = 57,062,000 Joules.

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The work done by nonconservative forces in stopping the airplane is  [tex]{57,154,000 \, \text{J}}[/tex].

To find the work done by nonconservative forces (like friction and air resistance) in stopping the airplane, we can use the work-energy principle. The work done by the nonconservative forces is equal to the change in the kinetic energy of the airplane.

Step-by-Step Solution

1. Calculate the initial kinetic energy ([tex]KE_{\text{initial}}[/tex]):

[tex]KE_{\text{initial}} = \frac{1}{2} m v^2[/tex]

where:

- m is the mass of the airplane (17,000 kg),

- v is the initial speed (82 m/s).

[tex]KE_{\text{initial}} = \frac{1}{2} \times 17,000 \, \text{kg} \times (82 \, \text{m/s})^2 \\\\KE_{\text{initial}} = \frac{1}{2} \times 17,000 \times 6,724 \\\\KE_{\text{initial}} = 57,154,000 \, \text{J}[/tex]

2. Calculate the final kinetic energy ([tex]KE_{\text{final}}[/tex]):

Since the airplane comes to a stop, its final speed is 0 m/s.

[tex]KE_{\text{final}} = \frac{1}{2} m (0)^2 = 0 \, \text{J}[/tex]

3. Calculate the change in kinetic energy (ΔKE):

[tex]\Delta KE = KE_{\text{final}} - KE_{\text{initial}} \\\\\Delta KE = 0 \, \text{J} - 57,154,000 \, \text{J} \\\\\Delta KE = -57,154,000 \, \text{J}[/tex]

4. The work done by nonconservative forces (W):

The work done by nonconservative forces is equal to the negative of the change in kinetic energy (since they are doing work to stop the airplane).

[tex]W = -\Delta KE \\\\W = -(-57,154,000 \, \text{J}) \\\\W = 57,154,000 \, \text{J}[/tex]

Therefore, the work done by nonconservative forces in stopping the airplane is [tex]{57,154,000 \, \text{J}}[/tex] .

The time taken by the ball to reach the batter is 0.42 seconds

The baseball pitcher throws a fastball at 42 m/s

The batter is about 18 meters from the pitcher

Therefore the time for the ball to reach the batter can be calculated as follows;

= 18/42

= 0.42 secs

Hence the time taken by the ball to reach the batter is 0.42 seconds

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Answer:

Horizontal component: [tex]V_{x}=+5m/s[/tex]

Vertical component: [tex]V_{y}=+8.66m/s[/tex]

Explanation:

We have the launch speed [tex]V =+ 10m/s[/tex] and the angle of 60°, so the components of the speed  [tex]V_{x}[/tex] and  [tex]V_{y}[/tex] are those shown in the attached image.

[tex]V_{x}[/tex] which is the horizontal component of the velocity can be found by multiplying the cosine of the angle by the initial velocity:

[tex]V_{x}=+10m/s(cos60)\\V_{x}=+10m/s(0.5)\\V_{x}=+5m/s[/tex]

[tex]V_{y}[/tex] which is the verticalcomponent of the velocity is found by multiplying the sine of angle by the initial velocity

[tex]V_{y}=+10m/s(sin60)\\V_{y}=+10m/s(0.866)\\V_{y}=+8.66m/s[/tex]

In summary:

Horizontal component: [tex]V_{x}=+5m/s[/tex]

Vertical component: [tex]V_{y}=+8.66m/s[/tex]

The correct answer to the question is gamma radiation . Therefore shielding that blocks gamma radiation, then it will most likely also block the other two types.

Radiation shielding is crucial to protect against the three types of radiation emitted by radioactive elements: alpha, beta, and gamma rays. Different materials are used for shielding based on the type of radiation and its penetrating power. For example, a sheet of paper can stop alpha particles, while thick lead or concrete is needed to absorb gamma rays.

Complete question:

You want to be shielded from all three types of nuclear radiation. If you find shielding that blocks ______ radiation, then it will most likely also block the other two types.

Final answer:

The horizontal distance traveled by the bullet before hitting the ground is 600 meters.

Explanation:

To determine how far forward the bullet goes before it hits the ground, we can use the fact that both the horizontally fired bullet and the dropped bullet hit the ground after a certain time. The dropped bullet takes 4 seconds to reach the ground, so we can consider its vertical motion using the equation h = 0.5 * g * t^2, where h is the height, g is the acceleration due to gravity, and t is the time. Plugging in the values, we get 0 = 0.5 * 9.8 * 4^2, which gives us h = 78.4 meters.

Since the horizontally fired bullet has the same horizontal velocity as the dropped bullet, it would take the same time to reach the ground. This means that the horizontally fired bullet travels a horizontal distance equal to its horizontal velocity multiplied by the time it takes to reach the ground. Plugging in the values, we get d = 150 * 4 = 600 meters.

Therefore, the first bullet travels 600 meters forward before hitting the ground.

The bullet fired horizontally travels 600 meters before hitting the ground because it takes the same 4 seconds as the dropped bullet to reach the ground, and it travels at a horizontal velocity of 150 m/s.

Step-by-Step Explanation:

Conclusion:

The bullet fired horizontally travels 600 meters before it hits the ground.

Answer:

absorption

Explanation:

Final answer:

Using Newton's law of cooling, we can calculate that the beer will be approximately 60.4ºF if left out for 20 minutes.

Explanation:

To calculate the final temperature of the beer after being left out for 20 minutes, we can use Newton's law of cooling. This law states that the rate of heat loss of an object is proportional to the temperature difference between the object and its surroundings. In this case, the initial temperature of the beer is 40ºF and the room temperature is 70ºF. Let's determine the constant of proportionality, k, first:

k = (T2 - T1) / t = (70 - 40) / 3 = 10

Now we can use the formula to find the final temperature, T3, after 20 minutes:

[tex]T3 - 70 = (40 - 70)e^(-10(20/3))[/tex]

[tex]T3 - 70 = -30e^(-20/3)[/tex]

[tex]T3 = 70 - 30e^(-20/3)[/tex]

Using a calculator, we can find that T3 is approximately 60.4ºF.

a. The value for the centripetal acceleration is g downward.

(correct answer on plato just did the mastery for this one)

At the top of the loop, where a roller coaster car is just maintaining contact with the track, its centripetal acceleration equals the acceleration due to gravity, which is 9.8 m/s².

The value of the centripetal acceleration at the top of the loop, where a roller coaster car just maintains contact with the track, is equal to the acceleration due to gravity, which is 9.8 m/s². This can be deduced from the balancing forces at the top of the loop. In this situation, the force due to gravity is acting downwards while the centripetal force is also acting downwards - directed towards the center of the circular path. Therefore, these forces must match at the top of the loop for the car to just maintain contact with the track. This gives us the equation for centripetal acceleration, ac = g , where ac is the centripetal acceleration and g is the acceleration due to gravity.

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The centripetal acceleration of the ball will be 38.68 meter per sq.second.

The acceleration needed to move a body in a curved way is understood as centripetal acceleration.

The direction of centripetal acceleration is always in the path of the center of the course.

The given data in the problem;

r is the radius= 2.00m

frequency (f) = 0.7 rev/s

The angular velocity is found as;

[tex]\rm \omega = 2 \pi f \\\\ \omega = 2 \times 3.14 \times 0.7 \\\\ \omega = 4.398 \ rad/sec[/tex]

The centripetal acceleration is given by;

[tex]\rm a_c= \omega^2r \\\\\ a_c= (4.398)^2 \times 2.00 \\\\ a_c=38.68 \ m/sec^2[/tex]

Hence, the centripetal acceleration of the ball will be 38.68 meter per sq.second.

To learn more about the centripetal acceleration refer to the link;

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Answer: The final velocity is 33.78 mi/h, so the driver did reduced his speed enough.

Explanation: The kinetic energy of an object can be calculated as:

K = (1/2)m*v^2

We know that the mass of the car is m=1100kg

and the initial velocity is 22m/s

The initial kinetic energy is:

K = (1/2)*1100*(22)^2 = 266,200 joules.

Now, if the kinetic energy decreases by 1.4x10^5 J, the new kinetic energy is:

K = 266,200j - 140,000j = 126,200j

So we now can find the new velocity in m/s.

126,200 = (1/2)*1100*v^2

126,200*2/1100 = v^2

229.45 = v^2

v = (229.45)^(1/2) = 15.1 m/s

We know that the limit is 35 mi/h, so we need to transform our result into miles per hour.

We know that in one hour, there are 3600 seconds, so the velocity per hour is:

15.1*3600 m/h = 54,360 m/h

and we know that one mile is 1609.34 meters, so we need to divide by 1609.34.

v = (54,360/1609.34) mi/h = 33.78 mi/h

this is less than the speed limit, so the driver reduced his speed enough.

After losing kinetic energy, the car's final velocity was approximately 18.99 m/s, exceeding the exit's speed limit of 15.64 m/s, hence the driver did not reduce their speed adequately.

To determine whether the driver reduced their speed enough when exiting the highway, we must calculate the car's speed after its kinetic energy decreases by 1.4×105J. The initial kinetic energy (KE) of the 1100-kg car traveling at 22 m/s can be calculated using the equation KE = ½ mv². Plugging in the values, we can find the initial kinetic energy:

KEinitial = ½ (1100 kg)(22 m/s)² = 5.28×105J

After losing 1.4×105J of energy, the remaining kinetic energy will be:

KEfinal = KEinitial - 1.4×105J = (5.28 - 1.4)×105J = 3.88×105J

We can solve for the final velocity (vfinal) using the remaining kinetic energy:

½ (1100 kg)vfinal² = 3.88×105J

vfinal = √((2×3.88×105J) / 1100 kg)

vfinal ≈ 18.99 m/s

To compare to the speed limit, we convert 35 mi/h to meters per second:

35 mi/h × 0.44704 (conversion factor) = 15.64 m/s

Since the final velocity of the car is 18.99 m/s, which is greater than the exit's speed limit of 15.64 m/s, the driver did not reduce their speed enough.

The inequality x² + 2x - 168 ≤ 0 can be used to determine the possible lengths of the base of a triangle when the height is 4 inches greater than twice the base and the area is no more than 168 square inches.

To find the possible lengths of the base of a triangle where the height is 4 inches greater than twice the base and the area is no more than 168 square inches, we start with the formula for the area of a triangle:

Area = (1/2) × base × height.

Let the base be x. Thus, the height can be expressed as 2x + 4. Therefore, the area equation becomes:

(1/2) × x × (2x + 4) ≤ 168

To simplify this, multiply both sides by 2:

x(2x + 4) ≤ 336

Expanding and rearranging the terms gives:

x²  + 4x ≤ 336

Subtract 336 from both sides to form a quadratic inequality:

x²  + 4x - 336 ≤ 0

Divide the entire inequality by 2 to simplify:

x²  + 2x - 168 ≤ 0

This is the inequality that can be used to find the possible lengths of the base of the triangle.

Answer:

a giant molecular cloud.

Explanation:

Both balls will hit the ground simultaneously because the time it takes to fall is based on the height, not the initial velocity.

The time it takes for an object to fall to the ground depends only on its height and is independent of its initial velocity. In this case, both balls are dropped, so they have an initial velocity of 0 m/s. The ball dropped from 100 ft and the ball dropped from 75 ft will both hit the ground at the same time. This is because the only factor affecting the time it takes to fall is the height, not the initial velocity. Therefore, both balls will hit the ground simultaneously.

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The acceleration of an object in free fall on a planet that has twice the mass of the Earth and twice the Earth's radius is 4.9 m/s², half of the acceleration on Earth.

The acceleration of an object in free fall on a planet depends on the mass and radius of the planet. In this case, the planet has twice the mass of the Earth and twice the Earth's radius. To calculate the acceleration, we can use the formula for gravitational acceleration:

g = (G * M) / R^2

Where g is the acceleration, G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.

Since the planet has twice the mass and twice the radius of the Earth, we can substitute those values into the formula:

g = (G * 2M) / (2R)^2 = (2 * G * M) / (4 * R^2) = G * M / (2 * R^2)

So the acceleration of an object in free fall on this planet is half of the acceleration on Earth, which is 4.9 m/s².

Answer:

Initial speed of football=14.11m/s

Explanation:

Range is the horizontal distance travelled. Therefore using formular for Range,R to determine the speed.

R=V^2Sin2theta/g

Given: R=20m

Theta=50°

20= V^2 Sin(2×50)°/ 9.8

Cross multiply

20×9.8 = V^2 Sin100°

196= V^2×0.9848

V=Sqrt(196/0.9848)

V= 14.11m/s

Radioactive dating, is your answer.